Department of Mathematics at UTSA - User contributions [en]

Abstract Algebra: Groups

2021-11-19T20:28:14Z

Lila: /* Cancellation Law */

Recall that an operation <math>\cdot</math> on <math>S</math> is said to be associative if for all <math>a, b, c \in S</math> we have that <math>a \cdot (b \cdot c) = (a \cdot b) \cdot c</math> and <math>\cdot</math> is said to be commutative if for all <math>a, b \in S</math> we have that <math>a \cdot b = b \cdot a</math>.
An element <math>e \in S</math> is the identity element of <math>S</math> under <math>\cdot</math> if for all <math>a \in S</math> we have that <math>a \cdot e = a</math> and <math>e \cdot a = a</math>.
We can now begin to describe our first type of algebraic structures known as groups, which are a set <math>G</math> equipped with a binary operation <math>\cdot</math> that is associative, contains an identity element, and contains inverse elements under <math>\cdot</math> for each element in <math>G</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: A Group is a pair <math>(G, \cdot)</math> where <math>G</math> is a set and <math>\cdot</math> is a binary operation on <math>G</math> with the following properties: 
1. For all <math>a, b, c \in G</math>, <math>a \cdot (b \cdot c) = (a \cdot b) \cdot c</math> (Associativity of <math>\cdot</math>). 
2. There exists an <math>e \in G</math> such that for all <math>a \in G</math>, <math>a \cdot e = a</math> and <math>e \cdot a = a</math> (The existence of an Identity Element). 
3. For all <math>a \in G</math> there exists an <math>a^{-1} \in G</math> such that <math>a \cdot a^{-1} = e</math> and <math>a^{-1} \cdot a = e</math> (The existence of inverses). 
Furthermore, if <math>G</math> is a finite set then the group <math>(G, \cdot)</math> is said to be a Finite Group and if <math>G</math> is an infinite set then the group <math>(G, \cdot)</math> is said to be an Infinite Group. More generally, the Order of <math>(G, \cdot)</math> (or **Size of <math>(G, \cdot)</math>) is the size of <math>G</math> and is denoted <math>| G |</math>.</td>
</blockquote>
When we use the multiplication symbol <math>\cdot</math> to denote the operation on <math>G</math>, we often call <math>G</math> a “multiplicative group”. When the operation of the group is instead denoted by <math>+</math> (instead of <math>\cdot</math>) then we often call <math>G</math> an “additive group”, and we write the inverse of each <math>a \in G</math> as <math>-a</math> (instead of <math>a^{-1}</math>).
Some of the sets and binary operations we have already seen can be considered groups. For example, <math>(\mathbb{R}, +)</math> is a group under standard addition <math>+</math> since the sum of any two real numbers is a real number, <math>+</math>, is associative, an additive identity <math>0 \in \mathbb{R}</math> exists and inverse elements exist for every <math>a \in \mathbb{R}</math> (namely <math>-a \in \mathbb{R}</math>).
Furthermore, <math>(\mathbb{Z}, +)</math> is also a group under the operation of standard addition since the sum of any two integers is an integer, addition is associative, the additivity identity is <math>0 \in \mathbb{Z}</math>, and for all <math>a \in \mathbb{Z}</math> we have <math>-a \in \mathbb{Z}</math> as additive inverses.
We will examine many other (more interesting) groups later on, but for now, let's look at an example of a set and a binary operation that does NOT form a group.
===Example 1===
Consider the set of integers <math>\mathbb{Z}</math> and define <math>*</math> for all <math>a, b \in \mathbb{Z}</math> by:
<div style="text-align: center;"><math>\begin{align} \quad a * b = a + 2b \end{align}</math></div>
(Where the <math>+</math> on the righthand side is usual addition of numbers). We will show that <math>(\mathbb{Z}, *)</math> is NOT a group by showing that <math>*</math> is not associative. Let <math>a, b, c \in \mathbb{Z}</math>. Then <math>*</math> is not associative since:
<div style="text-align: center;"><math>\begin{align} \quad a * (b * c) = a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c \end{align}</math></div>
<div style="text-align: center;"><math>\begin{align} \quad (a * b) * c = (a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c \end{align}</math></div>
Clearly <math>a * (b * c) \neq (a * b) * c</math> so <math>\mathbb{Z}</math> does not form a group under the operation <math>*</math>.

==Basic Theorems Regarding Groups==
Recall that a group <math>(G, \cdot)</math> is a set <math>G</math> with a binary operation <math>\cdot</math> such that:
<ul>
<li>1) <math>\cdot</math> is associative, i.e., for all <math>a, b, c \in G</math>, <math>a \cdot (b \cdot c) = (a \cdot b) \cdot c)</math>.</li>
</ul>
<ul>
<li>2) There exists an identity element <math>e \in G</math> such that <math>a \cdot e = a = e \cdot a</math> for all <math>a \in G</math>.</li>
</ul>
<ul>
<li>3) For each <math>a \in G</math> there exists an <math>a^{-1} \in G</math> such that <math>a \cdot a^{-1} = a^{-1} \cdot a = e</math>.</li>
</ul>
We will now look at some rather basic results regarding groups which we can derive from the group axioms above.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Proposition 1: Let <math>(G, \cdot)</math> be a group and let <math>e</math> be the identity for this group. Then: 
a) The identity element <math>e \in G</math> is unique. 
b) For each <math>a \in G</math>, the corresponding inverse <math>a^{-1} \in G</math> is unique. 
c) For each <math>a \in G</math>, <math>(a^{-1})^{-1} = a</math>. 
d) For all <math>a, b \in G</math>, <math>(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}</math>. 
e) For all <math>a, b \in G</math>, if <math>a \cdot b = e</math> then <math>a = b^{-1}</math> and <math>b = a^{-1}</math>. 
f) If <math>a^2 = a</math> then <math>a = e</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>e</math> and <math>e'</math> are both identities for <math>\cdot</math>. Then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad e = e \cdot e = e \cdot e' = e' \cdot e' = e' \end{align}</math></div>
<ul>
<li>Therefore <math>e = e'</math> so the identity for <math>\cdot</math> is unique. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of b) Suppose that <math>a^{-1} \in G</math> and <math>a^{-1'} \in G</math> are both inverses for <math>a \in G</math> under <math>\cdot</math>. Then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a^{-1} = a^{-1} \cdot e = a^{-1} \cdot (a \cdot a^{-1'}) = (a^{-1} \cdot a)*a^{-1} = e \cdot a^{-1'} = a^{-1'} \end{align}</math></div>
<ul>
<li>Therefore <math>a^{-1} = a^{-1'}</math> so the inverse for <math>a</math> is unique. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Let <math>a \in G</math>. Then <math>(a^{-1})^{-1}</math> is the inverse to <math>a^{-1}</math>. However, the inverse to <math>a^{-1}</math> is <math>a</math> and by (b) we have shown that the inverse of each element in <math>G</math> is unique. Therefore <math>a = (a^{-1})^{-1}</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d) If we apply the operation <math>\cdot</math> between <math>b^{-1} \cdot a^{-1}</math> and <math>(a \cdot b)</math> we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad (a \cdot b) \cdot [b^{-1} \cdot a^{-1}] & = a \cdot [(b \cdot b^{-1}) \cdot a^{-1}] \\ \quad &= a \cdot [e \cdot a^{-1}] \\ \quad &= a \cdot a^{-1} \\ \quad &= e \end{align}</math></div>
<ul>
<li>Therefore the inverse of <math>(a \cdot b)</math> is <math>b^{-1} \cdot a^{-1}</math>. We also have that the invere of <math>(a \cdot b)</math> is <math>(a \cdot b)^{-1}</math>. By (b), the inverse of <math>(a \cdot b)</math> is unique and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad (a \cdot b)^{-1} = b^{-1} \cdot a^{-1} \quad \blacksquare \end{align}</math></div>
<ul>
<li>Proof of e) Suppose that <math>a \cdot b = e</math>. Then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a \cdot b &= e \\ \quad (a \cdot b) \cdot b^{-1} &= e \cdot b^{-1} \\ \quad a \cdot (b \cdot b^{-1}) &= b^{-1} \\ \quad a \cdot e &= b^{-1} \\ \quad a &= b^{-1} \end{align}</math></div>
<ul>
<li>Similarly:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a \cdot b &= e \\ \quad a^{-1} \cdot (a \cdot b) &= a^{-1} \cdot e \\ \quad (a^{-1} \cdot a) \cdot b &= a^{-1} \\ \quad e \cdot b &= a^{-1} \\ \quad b &= a^{-1} \quad \blacksquare \end{align}</math></div>
<ul>
<li>Proof of f) Suppose that <math>a^2 = a \cdot a = a</math>. Then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a^2 &= a \\ \quad a \cdot a &= a \\ \quad a^{-1} \cdot (a \cdot a) &= a^{-1} \cdot a \\ \quad (a^{-1} \cdot a) \cdot a &= e \\ \quad e \cdot a &= e \\ \quad a &= e \end{align}</math></div>
<ul>
<li>Hence <math>a = e</math>. Alternatively we see that if <math>a \cdot a = a</math> then the inverse of <math>a</math> with respect to <math>\cdot</math> is <math>e</math>, that is <math>a^{-1} = e</math>. Multiplying both sides of this equation by <math>a</math> gives us that <math>a = e</math>. <math>\blacksquare</math></li>
</ul>

==Cancellation Law==
We will now look at another important property of groups called the cancellation law.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Cancellation Law for Groups): Let <math>(G, \cdot)</math> be a group and let <math>a, b, c \in G</math>. If <math>a \cdot b = a \cdot c</math> or <math>b \cdot a = c \cdot a</math> then <math>b = c</math>.</td>
</blockquote>
<ul>
<li>Proof: Let <math>a^{-1} \in G</math> denote the inverse of <math>a</math> under <math>\cdot</math>. Suppose that <math>a \cdot b = a \cdot c</math>. Then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a \cdot b &= a \cdot c \\ \quad (a^{-1} \cdot a) \cdot b &= (a^{-1} \cdot a) \cdot c \\ \quad e \cdot b &= e \cdot c \\ \quad b &= c \end{align}</math></div>
<ul>
<li>Similarly, suppose now that <math>b \cdot a = c \cdot a</math>. Then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad b \cdot a &= c \cdot a \\ \quad (b \cdot a) \cdot a^{-1} &= (c \cdot a) \cdot a^{-1} \\ \quad b \cdot (a \cdot a^{-1}) &= c \cdot (a \cdot a^{-1}) \\ \quad b \cdot e &= c \cdot e \\ \quad b &= c \quad \blacksquare \end{align}</math></div>
It is very important to note that the cancellation law holds with regards to the operation <math>\cdot</math> for any group <math>(G, \cdot)</math>. We will see that the cancellation law does not necessarily hold with respect to an operation on a set when we look at algebraic structures with two defined operations.
It is also important to note that if <math>a \cdot b = c \cdot a</math> or <math>b \cdot a = a \cdot c</math> then we cannot necessarily deduce that <math>b = c</math> because we would then require the additional property that <math>\cdot</math> is commutative which is not one of the group axioms (but instead one of the Abelian group axioms).

== Licensing ==
Content obtained and/or adapted from:
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Groups

2021-11-19T20:05:01Z

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Abstract Algebra: Groups

2021-11-19T19:57:47Z

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Abstract Algebra: Groups

2021-11-19T19:40:50Z

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Abstract Algebra: Preliminaries

2021-11-19T18:38:48Z

Lila: /* Licensing */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
Let <math>n \in \N</math>. Since the relation of congruence modulo n is an equivalence relation on <math>\Z</math>, we can discuss its equivalence classes. Recall that in this situation, we refer to the equivalence classes as congruence classes.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Definition of the integers modulo <math>n</math>: Let <math>n \in \N</math>. The set of congruence classes for the relation of congruence modulo <math>n</math> on <math>\Z</math> is the set of integers modulo <math>n</math>, or the set of integers mod <math>n</math>. We will denote this set of congruence classes by <math>\Z_n</math>.
</blockquote>

<math>\Z = [0] \cup [1] \cup [2] \cup \cdot\cdot\cdot \cup [n - 1]</math>.

In addition, we know that each integer is congruent to precisely one of the integers 0, 1, 2, ..., <math>n - 1</math>. This tells us that one way to represent <math>\Z_n</math> is

<math>\Z_n = \{[0], [1], [2], ... [n - 1]\}</math>.

Consequently, even though each integer has a congruence class, the set <math>\Z_n</math> has only <math>n</math> distinct congruence classes.

The set of integers <math>\Z</math> is more than a set. We can add and multiply integers. That is, there are the arithmetic operations of addition and multiplication on the set <math>\Z</math>, and we know that <math>\Z</math> is closed with respect to these two operations.

One of the basic problems dealt with in modern algebra is to determine if the arithmetic operations on one set “transfer” to a related set. In this case, the related set is <math>\Z_n</math>. For example, in the integers modulo 5, <math>\Z_5</math>, is it possible to add the congruence classes [4] and [2] as follows?

<math>\begin{array} {rcl} {[4] \oplus [2]} & = & {[4 + 2]} \\ {} & = & {[6]} \\ {} & = & {[1].} \end{array}</math>

We have used the symbol ̊ to denote addition in <math>\Z_5</math> so that we do not confuse it with addition in <math>\Z</math>. This looks simple enough, but there is a problem. The congruence classes [4] and [2] are not numbers, they are infinite sets. We have to make sure that we get the same answer no matter what element of [4] we use and no matter what element of [2] we use. For example,

<math>9 \equiv 4</math> (mod 5) and so [9] = [4]. Also, 
<math>7 \equiv 2</math> (mod 5) and so [7] = [2].

Do we get the same result if we add [9] and [7] in the way we did when we added [4] and [2]? The following computation confirms that we do:

<math>\begin{array} {rcl} {[9] \oplus [7]} & = & {[9 + 7]} \\ {} & = & {[16]} \\ {} & = & {[1].} \end{array}</math>

The left side shows the properties in terms of the congruence relation and the right side shows the properties in terms of the congruence classes.

<table class="mt-responsive-table">
<tr>
<td class="mt-noheading lt-math-7078">
If <math>a \equiv 3</math> (mod 6) and <math>b \equiv 4</math> (mod 6), then

<ul>
<li class="lt-math-7078"><math>(a + b) \equiv (3 + 4)</math> (mod 6);</li>
<li class="lt-math-7078"><math>(a \cdot b) \equiv (3 \cdot 4)</math> (mod 6).</li>
</ul>
</td>
<td class="mt-noheading lt-math-7078">
If <math>[a] = [3]</math> and <math>[b] = [4]</math> in <math>\Z_6</math>, then

<ul>
<li class="lt-math-7078"><math>[a + b] = [3 + 4]</math>;</li>
<li class="lt-math-7078"><math>[a \cdot b] = [3 \cdot 4]</math>.</li>
</ul>
</td>
</tr>
</table>

These are illustrations of general properties that we have already proved in Theorem 3.28. We repeat the statement of the theorem here because it is so important for defining the operations of addition and multiplication in <math>\Z_n</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then

<ol>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>(a + c) \equiv (b + d)</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>ac \equiv bd</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>m \in \N</math>, then <math>a^m \equiv b^m</math> (mod <math>n</math>).</li>
</ol>

</blockquote>

Since <math>x \equiv y</math> (mod <math>n</math>) if and only if <math>[x] = [y]</math>, we can restate the result of this Theorem 3.28 in terms of congruence classes in <math>\Z_n</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then, in <math>\Z_n</math>.

<ol>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a + c] = [b + d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a \cdot c] = [b \cdot d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>m \in \N</math>, then <math>[a]^m = [b]^m</math>.</li>
</ol>
</blockquote>

Now, we know that the following formal definition of addition and multiplication of congruence classes in <math>\Z_n</math> is independent of the choice of the elements we choose from each class. We say that these definitions of addition and multiplication are well defined. 

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license
* [https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7%3A_Equivalence_Relations/7.4%3A_Modular_Arithmetic Modular Arithmetic, LibreTexts: Mathematics] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T18:03:41Z

Lila: /* Algebra on \Z_m */

Abstract Algebra: Preliminaries

2021-11-19T18:01:16Z

Lila: /* Algebra on \Z_m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
Let <math>n \in \N</math>. Since the relation of congruence modulo n is an equivalence relation on <math>\Z</math>, we can discuss its equivalence classes. Recall that in this situation, we refer to the equivalence classes as congruence classes.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Definition of the integers modulo <math>n</math>: Let <math>n \in \N</math>. The set of congruence classes for the relation of congruence modulo <math>n</math> on <math>\Z</math> is the set of integers modulo <math>n</math>, or the set of integers mod <math>n</math>. We will denote this set of congruence classes by <math>\Z_n</math>.
</blockquote>

<math>\Z = [0] \cup [1] \cup [2] \cup \cdot\cdot\cdot \cup [n - 1]</math>.

In addition, we know that each integer is congruent to precisely one of the integers 0, 1, 2, ..., <math>n - 1</math>. This tells us that one way to represent <math>\Z_n</math> is

<math>\Z_n = \{[0], [1], [2], ... [n - 1]\}</math>.

Consequently, even though each integer has a congruence class, the set <math>\Z_n</math> has only <math>n</math> distinct congruence classes.

The set of integers <math>\Z</math> is more than a set. We can add and multiply integers. That is, there are the arithmetic operations of addition and multiplication on the set <math>\Z</math>, and we know that <math>\Z</math> is closed with respect to these two operations.

One of the basic problems dealt with in modern algebra is to determine if the arithmetic operations on one set “transfer” to a related set. In this case, the related set is <math>\Z_n</math>. For example, in the integers modulo 5, <math>\Z_5</math>, is it possible to add the congruence classes [4] and [2] as follows?

<math>\begin{array} {rcl} {[4] \oplus [2]} & = & {[4 + 2]} \\ {} & = & {[6]} \\ {} & = & {[1].} \end{array}</math>

We have used the symbol ̊ to denote addition in <math>\Z_5</math> so that we do not confuse it with addition in <math>\Z</math>. This looks simple enough, but there is a problem. The congruence classes [4] and [2] are not numbers, they are infinite sets. We have to make sure that we get the same answer no matter what element of [4] we use and no matter what element of [2] we use. For example,

<math>9 \equiv 4</math> (mod 5) and so [9] = [4]. Also, 
<math>7 \equiv 2</math> (mod 5) and so [7] = [2].

Do we get the same result if we add [9] and [7] in the way we did when we added [4] and [2]? The following computation confirms that we do:

<math>\begin{array} {rcl} {[9] \oplus [7]} & = & {[9 + 7]} \\ {} & = & {[16]} \\ {} & = & {[1].} \end{array}</math>

The left side shows the properties in terms of the congruence relation and the right side shows the properties in terms of the congruence classes.

<table class="mt-responsive-table">
<tr>
<td class="mt-noheading lt-math-7078">
If <math>a \equiv 3</math> (mod 6) and <math>b \equiv 4</math> (mod 6), then

<ul>
<li class="lt-math-7078"><math>(a + b) \equiv (3 + 4)</math> (mod 6);</li>
<li class="lt-math-7078"><math>(a \cdot b) \equiv (3 \cdot 4)</math> (mod 6).</li>
</ul>
</td>
<td class="mt-noheading lt-math-7078">
If <math>[a] = [3]</math> and <math>[b] = [4]</math> in <math>\Z_6</math>, then

<ul>
<li class="lt-math-7078"><math>[a + b] = [3 + 4]</math>;</li>
<li class="lt-math-7078"><math>[a \cdot b] = [3 \cdot 4]</math>.</li>
</ul>
</td>
</tr>
</table>

These are illustrations of general properties that we have already proved in Theorem 3.28. We repeat the statement of the theorem here because it is so important for defining the operations of addition and multiplication in <math>\Z_n</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then

<ol>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>(a + c) \equiv (b + d)</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>ac \equiv bd</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>m \in \N</math>, then <math>a^m \equiv b^m</math> (mod <math>n</math>).</li>
</ol>

</blockquote>

Since <math>x \equiv y</math> (mod <math>n</math>) if and only if <math>[x] = [y]</math>, we can restate the result of this Theorem 3.28 in terms of congruence classes in <math>\Z_n</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then, in <math>\Z_n</math>.

<ol>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a + c] = [b + d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a \cdot c] = [b \cdot d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>m \in \N</math>, then <math>[a]^m = [b]^m</math>.</li>
</ol>
</blockquote>

Because of Corollary 7.19, we know that the following formal definition of addition and multiplication of congruence classes in <math>\Z_n</math> is independent of the choice of the elements we choose from each class. We say that these definitions of addition and multiplication are well defined. 

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:59:25Z

Lila: /* Algebra on \Z_m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
Let <math>n \in \N</math>. Since the relation of congruence modulo n is an equivalence relation on <math>\Z</math>, we can discuss its equivalence classes. Recall that in this situation, we refer to the equivalence classes as congruence classes.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Definition of the integers modulo <math>n</math>: Let <math>n \in \N</math>. The set of congruence classes for the relation of congruence modulo <math>n</math> on <math>\Z</math> is the set of integers modulo <math>n</math>, or the set of integers mod <math>n</math>. We will denote this set of congruence classes by <math>\Z_n</math>.
</blockquote>

<math>\Z = [0] \cup [1] \cup [2] \cup \cdot\cdot\cdot \cup [n - 1]</math>.

In addition, we know that each integer is congruent to precisely one of the integers 0, 1, 2, ..., <math>n - 1</math>. This tells us that one way to represent <math>\Z_n</math> is

<math>\Z_n = \{[0], [1], [2], ... [n - 1]\}</math>.

Consequently, even though each integer has a congruence class, the set <math>\Z_n</math> has only <math>n</math> distinct congruence classes.

The set of integers <math>\Z</math> is more than a set. We can add and multiply integers. That is, there are the arithmetic operations of addition and multiplication on the set <math>\Z</math>, and we know that <math>\Z</math> is closed with respect to these two operations.

One of the basic problems dealt with in modern algebra is to determine if the arithmetic operations on one set “transfer” to a related set. In this case, the related set is <math>\Z_n</math>. For example, in the integers modulo 5, <math>\Z_5</math>, is it possible to add the congruence classes [4] and [2] as follows?

<math>\begin{array} {rcl} {[4] \oplus [2]} & = & {[4 + 2]} \\ {} & = & {[6]} \\ {} & = & {[1].} \end{array}</math>

We have used the symbol ̊ to denote addition in <math>\Z_5</math> so that we do not confuse it with addition in <math>\Z</math>. This looks simple enough, but there is a problem. The congruence classes [4] and [2] are not numbers, they are infinite sets. We have to make sure that we get the same answer no matter what element of [4] we use and no matter what element of [2] we use. For example,

<math>9 \equiv 4</math> (mod 5) and so [9] = [4]. Also, 
<math>7 \equiv 2</math> (mod 5) and so [7] = [2].

Do we get the same result if we add [9] and [7] in the way we did when we added [4] and [2]? The following computation confirms that we do:

<math>\begin{array} {rcl} {[9] \oplus [7]} & = & {[9 + 7]} \\ {} & = & {[16]} \\ {} & = & {[1].} \end{array}</math>

The left side shows the properties in terms of the congruence relation and the right side shows the properties in terms of the congruence classes.

<table class="mt-responsive-table">
<tbody>
<tr>
<td class="mt-noheading lt-math-7078">
If <math>a \equiv 3</math> (mod 6) and <math>b \equiv 4</math> (mod 6), then

<ul>
<li class="lt-math-7078"><math>(a + b) \equiv (3 + 4)</math> (mod 6);</li>
<li class="lt-math-7078"><math>(a \cdot b) \equiv (3 \cdot 4)</math> (mod 6).</li>
</ul>
</td>
<td class="mt-noheading lt-math-7078">
If <math>[a] = [3]</math> and <math>[b] = [4]</math> in <math>\Z_6</math>, then

<ul>
<li class="lt-math-7078"><math>[a + b] = [3 + 4]</math>;</li>
<li class="lt-math-7078"><math>[a \cdot b] = [3 \cdot 4]</math>.</li>
</ul>
</td>
</tr>
</tbody>
</table>

These are illustrations of general properties that we have already proved in Theorem 3.28. We repeat the statement of the theorem here because it is so important for defining the operations of addition and multiplication in <math>\Z_n</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then

<ol>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>(a + c) \equiv (b + d)</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>ac \equiv bd</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>m \in \N</math>, then <math>a^m \equiv b^m</math> (mod <math>n</math>).</li>
</ol>

</blockquote>

Since <math>x \equiv y</math> (mod <math>n</math>) if and only if <math>[x] = [y]</math>, we can restate the result of this Theorem 3.28 in terms of congruence classes in <math>\Z_n</math>.

<blockquote style="background: white; border: 1px solid black; padding: 1em;">

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then, in <math>\Z_n</math>.

<ol>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a + c] = [b + d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a \cdot c] = [b \cdot d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>m \in \N</math>, then <math>[a]^m = [b]^m</math>.</li>
</ol>
</blockquote>

Because of Corollary 7.19, we know that the following formal definition of addition and multiplication of congruence classes in <math>\Z_n</math> is independent of the choice of the elements we choose from each class. We say that these definitions of addition and multiplication are well defined. 

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:57:05Z

Lila: /* Algebra on \Z_m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
Let <math>n \in \N</math>. Since the relation of congruence modulo n is an equivalence relation on <math>\Z</math>, we can discuss its equivalence classes. Recall that in this situation, we refer to the equivalence classes as congruence classes.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Definition of the integers modulo <math>n</math>: Let <math>n \in \N</math>. The set of congruence classes for the relation of congruence modulo <math>n</math> on <math>\Z</math> is the set of integers modulo <math>n</math>, or the set of integers mod <math>n</math>. We will denote this set of congruence classes by <math>\Z_n</math>.
</blockquote>

<math>\Z = [0] \cup [1] \cup [2] \cup \cdot\cdot\cdot \cup [n - 1]</math>.

In addition, we know that each integer is congruent to precisely one of the integers 0, 1, 2, ..., <math>n - 1</math>. This tells us that one way to represent <math>\Z_n</math> is

<math>\Z_n = \{[0], [1], [2], ... [n - 1]\}</math>.

Consequently, even though each integer has a congruence class, the set <math>\Z_n</math> has only <math>n</math> distinct congruence classes.

The set of integers <math>\Z</math> is more than a set. We can add and multiply integers. That is, there are the arithmetic operations of addition and multiplication on the set <math>\Z</math>, and we know that <math>\Z</math> is closed with respect to these two operations.

One of the basic problems dealt with in modern algebra is to determine if the arithmetic operations on one set “transfer” to a related set. In this case, the related set is <math>\Z_n</math>. For example, in the integers modulo 5, <math>\Z_5</math>, is it possible to add the congruence classes [4] and [2] as follows?

<math>\begin{array} {rcl} {[4] \oplus [2]} & = & {[4 + 2]} \\ {} & = & {[6]} \\ {} & = & {[1].} \end{array}</math>

We have used the symbol ̊ to denote addition in <math>\Z_5</math> so that we do not confuse it with addition in <math>\Z</math>. This looks simple enough, but there is a problem. The congruence classes [4] and [2] are not numbers, they are infinite sets. We have to make sure that we get the same answer no matter what element of [4] we use and no matter what element of [2] we use. For example,

<math>9 \equiv 4</math> (mod 5) and so [9] = [4]. Also, 
<math>7 \equiv 2</math> (mod 5) and so [7] = [2].

Do we get the same result if we add [9] and [7] in the way we did when we added [4] and [2]? The following computation confirms that we do:

<math>\begin{array} {rcl} {[9] \oplus [7]} & = & {[9 + 7]} \\ {} & = & {[16]} \\ {} & = & {[1].} \end{array}</math>

The left side shows the properties in terms of the congruence relation and the right side shows the properties in terms of the congruence classes.

<table class="mt-responsive-table">
<tbody>
<tr>
<td class="mt-noheading lt-math-7078">
If <math>a \equiv 3</math> (mod 6) and <math>b \equiv 4</math> (mod 6), then

<ul>
<li class="lt-math-7078"><math>(a + b) \equiv (3 + 4)</math> (mod 6);</li>
<li class="lt-math-7078"><math>(a \cdot b) \equiv (3 \cdot 4)</math> (mod 6).</li>
</ul>
</td>
<td class="mt-noheading lt-math-7078">
If <math>[a] = [3]</math> and <math>[b] = [4]</math> in <math>\Z_6</math>, then

<ul>
<li class="lt-math-7078"><math>[a + b] = [3 + 4]</math>;</li>
<li class="lt-math-7078"><math>[a \cdot b] = [3 \cdot 4]</math>.</li>
</ul>
</td>
</tr>
</tbody>
</table>

These are illustrations of general properties that we have already proved in Theorem 3.28. We repeat the statement of the theorem here because it is so important for defining the operations of addition and multiplication in <math>\Z_n</math>.

<div class="box-definition"><div mt-section-origin="Bookshelves/Mathematical_Logic_and_Proof/Book:_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7:_Equivalence_Relations/7.4:_Modular_Arithmetic" class="mt-section"><h5 class="box-legend lt-math-7078 editable">Theorem 3.28</h5>

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then

<ol>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>(a + c) \equiv (b + d)</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>ac \equiv bd</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>m \in \N</math>, then <math>a^m \equiv b^m</math> (mod <math>n</math>).</li>
</ol>

<dl>
<dt>Proof</dt>
<dd>
Add proof here and it will automatically be hidden 
</dd>
</dl>
</div></div>

Since <math>x \equiv y</math> (mod <math>n</math>) if and only if <math>[x] = [y]</math>, we can restate the result of this Theorem 3.28 in terms of congruence classes in <math>\Z_n</math>.

<div class="box-definition"><div mt-section-origin="Bookshelves/Mathematical_Logic_and_Proof/Book:_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7:_Equivalence_Relations/7.4:_Modular_Arithmetic" class="mt-section"><h5 class="box-legend lt-math-7078 editable">Corollary 7.19.</h5>

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then, in <math>\Z_n</math>.

<ol>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a + c] = [b + d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a \cdot c] = [b \cdot d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>m \in \N</math>, then <math>[a]^m = [b]^m</math>.</li>
</ol>
</div></div>

Because of Corollary 7.19, we know that the following formal definition of addition and multiplication of congruence classes in <math>\Z_n</math> is independent of the choice of the elements we choose from each class. We say that these definitions of addition and multiplication are well defined. 

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:55:05Z

Lila: /* Algebra on \Z_m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
Let <math>n \in \N</math>. Since the relation of congruence modulo n is an equivalence relation on <math>\Z</math>, we can discuss its equivalence classes. Recall that in this situation, we refer to the equivalence classes as congruence classes.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Definition of the integers modulo <math>n</math>: Let <math>n \in \N</math>. The set of congruence classes for the relation of congruence modulo <math>n</math> on <math>\Z</math> is the set of integers modulo <math>n</math>, or the set of integers mod <math>n</math>. We will denote this set of congruence classes by <math>\Z_n</math>.
</blockquote>

<math>\Z = [0] \cup [1] \cup [2] \cup \cdot\cdot\cdot \cup [n - 1]</math>.

In addition, we know that each integer is congruent to precisely one of the integers 0, 1, 2, ..., <math>n - 1</math>. This tells us that one way to represent <math>\Z_n</math> is

<math>\Z_n = \{[0], [1], [2], ... [n - 1]\}</math>.

Consequently, even though each integer has a congruence class, the set <math>\Z_n</math> has only <math>n</math> distinct congruence classes.

The set of integers <math>\Z</math> is more than a set. We can add and multiply integers. That is, there are the arithmetic operations of addition and multiplication on the set <math>\Z</math>, and we know that <math>\Z</math> is closed with respect to these two operations.

One of the basic problems dealt with in modern algebra is to determine if the arithmetic operations on one set “transfer” to a related set. In this case, the related set is <math>\Z_n</math>. For example, in the integers modulo 5, <math>\Z_5</math>, is it possible to add the congruence classes [4] and [2] as follows?

<math>\begin{array} {rcl} {[4] \oplus [2]} &= & {[4 + 2]} \\ {} &= & {[6]} \\ {} &= & {[1].} \end{array}</math>

We have used the symbol ̊ to denote addition in <math>\Z_5</math> so that we do not confuse it with addition in <math>\Z</math>. This looks simple enough, but there is a problem. The congruence classes [4] and [2] are not numbers, they are infinite sets. We have to make sure that we get the same answer no matter what element of [4] we use and no matter what element of [2] we use. For example,

<math>9 \equiv 4</math> (mod 5) and so [9] = [4]. Also, 
<math>7 \equiv 2</math> (mod 5) and so [7] = [2].

Do we get the same result if we add [9] and [7] in the way we did when we added [4] and [2]? The following computation confirms that we do:

<math>\begin{array} {rcl} {[9] \oplus [7]} &= & {[9 + 7]} \\ {} &= & {[16]} \\ {} &= & {[1].} \end{array}</math>

The left side shows the properties in terms of the congruence relation and the right side shows the properties in terms of the congruence classes.

<table class="mt-responsive-table">
<tbody>
<tr>
<td class="mt-noheading lt-math-7078">
If <math>a \equiv 3</math> (mod 6) and <math>b \equiv 4</math> (mod 6), then

<ul>
<li class="lt-math-7078"><math>(a + b) \equiv (3 + 4)</math> (mod 6);</li>
<li class="lt-math-7078"><math>(a \cdot b) \equiv (3 \cdot 4)</math> (mod 6).</li>
</ul>
</td>
<td class="mt-noheading lt-math-7078">
If <math>[a] = [3]</math> and <math>[b] = [4]</math> in <math>\Z_6</math>, then

<ul>
<li class="lt-math-7078"><math>[a + b] = [3 + 4]</math>;</li>
<li class="lt-math-7078"><math>[a \cdot b] = [3 \cdot 4]</math>.</li>
</ul>
</td>
</tr>
</tbody>
</table>

These are illustrations of general properties that we have already proved in Theorem 3.28. We repeat the statement of the theorem here because it is so important for defining the operations of addition and multiplication in <math>\Z_n</math>.

<div class="box-definition"><div mt-section-origin="Bookshelves/Mathematical_Logic_and_Proof/Book:_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7:_Equivalence_Relations/7.4:_Modular_Arithmetic" class="mt-section"><h5 class="box-legend lt-math-7078 editable">Theorem 3.28</h5>

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then

<ol>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>(a + c) \equiv (b + d)</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>ac \equiv bd</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>m \in \N</math>, then <math>a^m \equiv b^m</math> (mod <math>n</math>).</li>
</ol>

<dl>
<dt>Proof</dt>
<dd>
Add proof here and it will automatically be hidden 
</dd>
</dl>
</div></div>

Since <math>x \equiv y</math> (mod <math>n</math>) if and only if <math>[x] = [y]</math>, we can restate the result of this Theorem 3.28 in terms of congruence classes in <math>\Z_n</math>.

<div class="box-definition"><div mt-section-origin="Bookshelves/Mathematical_Logic_and_Proof/Book:_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7:_Equivalence_Relations/7.4:_Modular_Arithmetic" class="mt-section"><h5 class="box-legend lt-math-7078 editable">Corollary 7.19.</h5>

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then, in <math>\Z_n</math>.

<ol>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a + c] = [b + d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a \cdot c] = [b \cdot d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>m \in \N</math>, then <math>[a]^m = [b]^m</math>.</li>
</ol>
</div></div>

Because of Corollary 7.19, we know that the following formal definition of addition and multiplication of congruence classes in <math>\Z_n</math> is independent of the choice of the elements we choose from each class. We say that these definitions of addition and multiplication are well defined. 

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:52:17Z

Lila: /* Algebra on \Z_m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
Let <math>n \in \N</math>. Since the relation of congruence modulo n is an equivalence relation on <math>\Z</math>, we can discuss its equivalence classes. Recall that in this situation, we refer to the equivalence classes as congruence classes.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Definition of the integers modulo <math>n</math>: Let <math>n \in \N</math>. The set of congruence classes for the relation of congruence modulo <math>n</math> on <math>\Z</math> is the set of integers modulo <math>n</math>, or the set of integers mod <math>n</math>. We will denote this set of congruence classes by <math>\Z_n</math>.
</blockquote>

<math>\Z = [0] \cup [1] \cup [2] \cup \cdot\cdot\cdot \cup [n - 1]</math>.

In addition, we know that each integer is congruent to precisely one of the integers 0, 1, 2, ..., <math>n - 1</math>. This tells us that one way to represent <math>\Z_n</math> is

<math>\Z_n = \{[0], [1], [2], ... [n - 1]\}</math>.

Consequently, even though each integer has a congruence class, the set <math>\Z_n</math> has only <math>n</math> distinct congruence classes.

The set of integers <math>\Z</math> is more than a set. We can add and multiply integers. That is, there are the arithmetic operations of addition and multiplication on the set <math>\Z</math>, and we know that <math>\Z</math> is closed with respect to these two operations.

One of the basic problems dealt with in modern algebra is to determine if the arithmetic operations on one set “transfer” to a related set. In this case, the related set is <math>\Z_n</math>. For example, in the integers modulo 5, <math>\Z_5</math>, is it possible to add the congruence classes [4] and [2] as follows?

\[\begin{array} {rcl} {[4] \oplus [2]} &= & {[4 + 2]} \\ {} &= & {[6]} \\ {} &= & {[1].} \end{array}\]

We have used the symbol ̊ to denote addition in <math>\Z_5</math> so that we do not confuse it with addition in <math>\Z</math>. This looks simple enough, but there is a problem. The congruence classes [4] and [2] are not numbers, they are infinite sets. We have to make sure that we get the same answer no matter what element of [4] we use and no matter what element of [2] we use. For example,

<math>9 \equiv 4</math> (mod 5) and so [9] = [4]. Also, 
<math>7 \equiv 2</math> (mod 5) and so [7] = [2].

Do we get the same result if we add [9] and [7] in the way we did when we added [4] and [2]? The following computation confirms that we do:

\[\begin{array} {rcl} {[9] \oplus [7]} &= & {[9 + 7]} \\ {} &= & {[16]} \\ {} &= & {[1].} \end{array}\]

This is one of the ideas that was explored in Preview Activity <math>\PageIndex{1}</math>. The main difference is that in this preview activity, we used the relation of congruence, and here we are using congruence classes. All of the examples in Preview Activity <math>\PageIndex{1}</math> should have illustrated the properties of congruence modulo 6 in the following table. The left side shows the properties in terms of the congruence relation and the right side shows the properties in terms of the congruence classes.

<table class="mt-responsive-table">
<tbody>
<tr>
<td class="mt-noheading lt-math-7078">
If <math>a \equiv 3</math> (mod 6) and <math>b \equiv 4</math> (mod 6), then

<ul>
<li class="lt-math-7078"><math>(a + b) \equiv (3 + 4)</math> (mod 6);</li>
<li class="lt-math-7078"><math>(a \cdot b) \equiv (3 \cdot 4)</math> (mod 6).</li>
</ul>
</td>
<td class="mt-noheading lt-math-7078">
If <math>[a] = [3]</math> and <math>[b] = [4]</math> in <math>\Z_6</math>, then

<ul>
<li class="lt-math-7078"><math>[a + b] = [3 + 4]</math>;</li>
<li class="lt-math-7078"><math>[a \cdot b] = [3 \cdot 4]</math>.</li>
</ul>
</td>
</tr>
</tbody>
</table>

These are illustrations of general properties that we have already proved in Theorem 3.28. We repeat the statement of the theorem here because it is so important for defining the operations of addition and multiplication in <math>\Z_n</math>.

<div class="box-definition"><div mt-section-origin="Bookshelves/Mathematical_Logic_and_Proof/Book:_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7:_Equivalence_Relations/7.4:_Modular_Arithmetic" class="mt-section"><h5 class="box-legend lt-math-7078 editable">Theorem 3.28</h5>

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then

<ol>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>(a + c) \equiv (b + d)</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>c \equiv d</math> (mod <math>n</math>), then <math>ac \equiv bd</math> (mod <math>n</math>).</li>
<li class="lt-math-7078">If <math>a \equiv b</math> (mod <math>n</math>) and <math>m \in \N</math>, then <math>a^m \equiv b^m</math> (mod <math>n</math>).</li>
</ol>

<dl>
<dt>Proof</dt>
<dd>
Add proof here and it will automatically be hidden 
</dd>
</dl>
</div></div>

Since <math>x \equiv y</math> (mod <math>n</math>) if and only if <math>[x] = [y]</math>, we can restate the result of this Theorem 3.28 in terms of congruence classes in <math>\Z_n</math>.

<div class="box-definition"><div mt-section-origin="Bookshelves/Mathematical_Logic_and_Proof/Book:_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/7:_Equivalence_Relations/7.4:_Modular_Arithmetic" class="mt-section"><h5 class="box-legend lt-math-7078 editable">Corollary 7.19.</h5>

Let <math>n</math> be a natural number and let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be integers. Then, in <math>\Z_n</math>.

<ol>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a + c] = [b + d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>[c] = [d]</math>, then <math>[a \cdot c] = [b \cdot d]</math>.</li>
<li class="lt-math-7078">If <math>[a] = [b]</math> and <math>m \in \N</math>, then <math>[a]^m = [b]^m</math>.</li>
</ol>
</div></div>

Because of Corollary 7.19, we know that the following formal definition of addition and multiplication of congruence classes in <math>\Z_n</math> is independent of the choice of the elements we choose from each class. We say that these definitions of addition and multiplication are well defined. 

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:40:36Z

Lila: /* Algebra on \Z_m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==
We will now look at the additive group of integers modulo <math>n</math> denoted <math>\mathbb{Z}/n\mathbb{Z} = \{ 0, 1, 2, ..., n - 1 \}</math> for integers <math>n > 1</math>. For each of these groups, define the operation of <math>+</math> for <math>x, y \in \mathbb{Z}/n\mathbb{Z}</math> by <math>x + y</math> to equal <math>x + y \pmod n</math> (where the righthand <math>+</math> denotes usual addition).
The operation <math>+</math> above has a very nice interpretation. For any <math>n > 1</math> we can interpret the operation of <math>+</math> on <math>\mathbb{Z}/n\mathbb{Z}</math> for <math>x + y</math> by considering a circle numbered <math>0</math> through <math>n - 1</math> clockwise, starting at <math>x</math>, and then traveling clockwise around the circle a distance of <math>y</math> to get <math>x + y</math> as illustrated in the following diagram:
<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/the-additive-group-of-integers-modulo-n-z-nz/Screen%20Shot%202015-09-05%20at%2011.03.05%20AM.png" alt="Screen%20Shot%202015-09-05%20at%2011.03.05%20AM.png" class="image" /></div>
We will now show that <math>(\mathbb{Z}/n\mathbb{Z}, +)</math> is a group. For <math>x, y \in \mathbb{Z}/n\mathbb{Z}</math> we have the <math>x + y</math> is the remainder of the sum <math>x + y</math> on division by <math>n</math>. The only possible remainders are <math>0, 1, ..., n - 1</math> and so <math>(x + y) \in \mathbb{Z}/n\mathbb{Z}</math>, so <math>\mathbb{Z}/n\mathbb{Z}</math> is closed under <math>+</math>.
It's not hard to intuitively see that the remainder of <math>x + (y + z)</math> upon division by <math>n</math> is equal to the remainder of <math>(x + y) + z</math> for all <math>x, y, z \in \mathbb{Z}/n\mathbb{Z}</math>, so indeed, <math>x + (y + z) = (x + y) + z</math>, so <math>+</math> is associative.
For each <math>x \in \mathbb{Z}/n\mathbb{Z}</math> we note that the remainder of <math>x + 0</math> is equal to the remainder of <math>x</math> upon division by <math>n</math> and the remainder of <math>0 + x</math> is equal to the remainder of <math>x</math> upon division by <math>n</math>. In other words, <math>x + 0 = x</math> and <math>0 + x = x</math>. Furthermore, <math>0 \in \mathbb{Z}/n\mathbb{Z}</math>, so <math>0</math> is the identity element.
Lastly, if <math>x \in \mathbb{Z}/n\mathbb{Z}</math> then <math>-x \in \mathbb{Z}/n\mathbb{Z}</math> is such that <math>x + (-x) = 0</math> and <math>(-x) + x = 0</math>.
Thus we conclude that <math>(\mathbb{Z}/n\mathbb{Z}, +)</math> is a group.

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:26:52Z

Lila: /* Fundamental Theorem of Arithmetic */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product (<math> n = p_1p_2...p_k </math> for unique primes <math> p_1, p_2, ..., p_k </math>).</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:24:14Z

Lila: /* Fundamental Theorem of Arithmetic */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product; that is, <math> n = p_1p_2...p_k </math> for unique primes <math> p_i </math>.</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:18:06Z

Lila: /* Fundamental Theorem of Arithmetic */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product.</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:16:44Z

Lila: /* Fundamental Theorem of Arithmetic */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

The Fundamental Theorem of Arithmetic, also known as the Unique Prime Factorization theorem, is as follows:

<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Unique Prime Factorization Theorem): If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes uniquely apart from the ordering of the primes in the product.</td>
</blockquote>
If <math>n \in \mathbb{Z}</math> and <math>n < -1</math> then <math>n</math> can also be written as a product of primes multiplied by <math>-1</math>.
<ul>
<li>Proof: We first show that if <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then <math>n</math> can be written as a product of primes. We saw on the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that if every <math>n \in \mathbb{Z}</math> and <math>n > 1</math> is divisible by a prime number. Let <math>p_1</math> be this prime number. Then for some <math>n_2 \in \{1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1n_1 \end{align}</math></div>
<ul>
<li>If <math>n_1 = 1</math> then <math>n = p_1</math> and we're done. If <math>n_1</math> is prime then let <math>p_2 = n_1</math> and so <math>n = p_1p_2</math> and we're done. If <math>n_1</math> is composite then <math>n_1 > 1</math> and so once again, <math>n_1</math> is divisible by a prime, say <math>p_2</math>, where <math>n_1 = p_2n_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2n_2 \end{align}</math></div>
<ul>
<li>If <math>n_2 = 1</math> then <math>n = p_1p_2</math> and we're done. If <math>n_2</math> is prime then let <math>p_3 = n_2</math> and so <math>n = p_1p_2p_3</math> and we're done. If <math>n_2</math> is composite then <math>n_2 > 1</math> and again, <math>n_2</math> is divisible by a prime, say <math>p_3</math>, where <math>n_2 = p_3n_3</math>.</li>
</ul>
<ul>
<li>Note that <math>n_1 > n_2 > ... \geq 1</math>. This sequence of integers is bounded below by <math>1</math> and hence must terminate at some point. Therefore, for some <math>k \in \{ 1, 2, ... \}</math> we have that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad n = p_1p_2...p_k \end{align}</math></div>
<ul>
<li>Therefore <math>n</math> can be written as a product of primes. We will now show that this factorization is unique. Suppose that <math>n = p_1p_2...p_k</math> and <math>n = q_1q_2...q_r</math> so that then:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_1p_2...p_k = q_1q_2...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_1 \mid n</math> we must have <math>p_1 \mid q_1q_2...q_r</math>. Therefore <math>p_1 = q_i</math> for some <math>i \in \{1, 2, ... r \}</math>. Without loss of generality, assume that <math>p_1 = q_1</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_2p_3...p_k = q_2q_3...q_r \end{align}</math></div>
<ul>
<li>Since <math>p_2 \mid n</math> we must also have <math>p_2 \mid q_2q_3...q_r</math>. Therefore <math>p_2 = q_i</math> for some <math>i \in \{2, 3, ... r \}</math>. Without loss of generality, assume that <math>p_2 = q_2</math>. Therefore:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_3p_4...p_k = q_3q_4...q_r \end{align}</math></div>
<ul>
<li>We continue in this process. We note that we won't run out of primes <math>p</math> before primes <math>q</math>, otherwise this implies a prime <math>q</math> is equal to <math>1</math>. Furthermore, we won't run out of primes <math>q</math> before primes <math>p</math>, otherwise this implies a prime <math>p</math> is equal to <math>1</math>. Therefore <math>k =r</math> and we eventually get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad p_k = q_r \end{align}</math></div>
<ul>
<li>Therefore the prime factorization <math>n = p_1p_2...p_k</math> is unique. <math>\blacksquare</math></li>
</ul>

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:11:34Z

Lila: /* Licensing */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Euclid%27s_lemma Euclid's lemma, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Prime_number Prime number, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:09:03Z

Lila: /* Euclid's Lemma */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

A common proof of Euclid's lemma involves the Bézout's identity, which was unknown at Euclid's time. Bézout's identity states that if {{math|''x''}} and {{math|''y''}} are relatively prime integers (i.e. they share no common divisors other than 1 and -1) there exist integers {{math|''r''}} and {{math|''s''}} such that
:<math>
rx+sy = 1.
</math>
Let {{math|''a''}} and {{math|''n''}} be relatively prime, and assume that {{math|''n'' <math>=</math>''ab''}}. By Bézout's identity, there are {{math|''r''}} and {{math|''s''}} making
:<math>
rn+sa = 1.
</math>
Multiply both sides by {{math|''b''}}:
:<math>
rnb+sab = b.
</math>
The first term on the left is divisible by {{math|''n''}}, and the second term is divisible by {{math|''ab''}}, which by hypothesis is divisible by {{math|''n''}}. Therefore their sum, {{math|''b''}}, is also divisible by {{math|''n''}}. This is the generalization of Euclid's lemma mentioned above.

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:07:11Z

Lila: /* Euclid's Lemma */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Euclid's lemma''': If a prime {{math|''p''}} divides the product {{math|''ab''}} of two integers {{math|''a''}} and {{math|''b''}}, then {{math|''p''}} must divide at least one of those integers {{math|''a''}} and {{math|''b''}}.</blockquote>

For example, if {{math|''p'' <math>=</math> 19}}, {{math|''a'' <math>=</math> 133}}, {{math|''b'' <math>=</math> 143}}, then {{math|''ab'' <math>=</math> 133 × 143 <math>=</math> 19019}}, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, {{math| 133 <math>=</math> 19 × 7}}.

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T17:03:40Z

Lila: /* Congruence modulo m */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math> m </math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T03:19:17Z

Lila: /* Primes */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
<div style="text-align: center;"><math>
\begin{align} \quad p_1, p_2, ..., p_k \end{align}</math></div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T02:29:48Z

Lila: /* Primes */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
(1)
<div class="math-equation" id="equation-1">\begin{align} \quad p_1, p_2, ..., p_k \end{align}</div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T02:26:02Z

Lila: /* Primes */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
Recall from the <a href="/prime-and-composite-numbers">Prime and Composite Numbers</a> page that an integer <math>p > 1</math> is a prime number if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>, and an integer <math>p > 1</math> that is not a prime number is called a composite number.
We will now look at a very famous, simple, and important theorem which states that there exists infinitely many primes.
<table class="wiki-content-table">
<tr>
<td>Theorem 1 (Euclid's Theorem of the Existence of Infinitely Many Primes): There are infinitely many prime numbers.</td>
</tr>
</table>
<ul>
<li>Proof: We will carry this proof out by contradiction. Suppose that instead there are a finite number of primes. Then we can list all of these primes:</li>
</ul>
(1)
<div class="math-equation" id="equation-1">\begin{align} \quad p_1, p_2, ..., p_k \end{align}</div>
<ul>
<li>Consider the number <math>n = p_1p_2...p_k + 1</math>. We have already seen that every integer <math>n > 1</math> has a prime divisor. Let this divisor be <math>p_i</math> for some <math>i \in \{ 1, 2, ..., k \}</math>. Then <math>p_i \mid n</math> and <math>p_i \mid p_1p_2...p_i...p_k</math>. But this implies that <math>p_i \mid 1</math> which is a contradiction since all <math>p_i \geq 2</math> and the only divisors of <math>1</math> are <math>\pm 1</math>.</li>
</ul>
<ul>
<li>Therefore the assumption that there were finitely many primes was false and so there are infinitely many primes. <math>\blacksquare</math></li>
</ul>

===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T02:19:50Z

Lila: /* Primes */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T02:15:04Z

Lila: /* Primes */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>p > 1</math> is said to be Prime if the only divisors of <math>p</math> are <math>\pm 1</math> and <math>\pm p</math>.</td>
</blockquote>
For a long time it was debated whether the number <math>1</math> should be classified as a prime number or not. The importance of <math>1</math> being classified as a prime number was uninteresting and so the first prime number is officially <math>2</math>.
If <math>\mathbb{P}</math> is the set of all prime numbers, then the first few primes in <math>\mathbb{P}</math> are:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{P} = \{2, 3, 5, 7, 11, 13, ... \} \end{align}</math></div>
We will now look at a very simple theorem that says the greatest common divisor between an integer <math>a</math> and any prime <math>p</math> is either <math>1</math> or <math>p</math> depending on whether <math>p</math> divides <math>a</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, p \in \mathbb{Z}</math> and <math>p</math> is a prime number then <math>\gcd (a, p) = 1</math> if <math>p \not \mid a</math> or <math>\gcd (a, p) = p</math> if <math>p \mid a</math>.</td>
</blockquote>
<ul>
<li>Proof: Since <math>p</math> is prime, the only positive divisors of <math>p</math> are <math>1</math> and <math>p</math> so <math>\gcd (a, p) = 1</math> if <math>p \not \mid a</math> or <math>\gcd (a, p) = p</math> if <math>p \mid a</math>. <math>\blacksquare</math></li>
</ul>
One important aspect of prime numbers is that every integer <math>n > 1</math> is divisible by a prime number <math>p</math> which we prove in the following lemma.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Lemma 1: If <math>n \in \mathbb{Z}</math> and <math>n > 1</math> then there exists a prime <math>p \in \mathbb{P}</math> such that <math>p \mid n</math>.</td>
</blockquote>
Lemma 1 also holds if <math>n \in \mathbb{Z}</math> and <math>n < 1</math> and can be proven in an analogous manner. Furthermore, if <math>n = 0</math> then every number divides <math>0</math> and by extension, every prime <math>p \in \mathbb{P}</math> divides <math>0</math>.
<ul>
<li>Proof: Let <math>S</math> be the set of positive divisors of <math>n</math>, that is:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad S = \{ d \in \mathbb{Z} : d \mid n \} \end{align}</math></div>
<ul>
<li>The set <math>S</math> is nonempty since <math>1, n \in S</math>. If <math>1</math> and <math>n</math> are the only elements in <math>S</math> then this implies that <math>n</math> is prime and so <math>p = n</math> divides <math>n</math>. If there exists more elements in <math>S</math>, then since <math>S \setminus \{1 \}</math> is a nonempty subset of integers we have that by <a href="/the-well-ordering-principle-of-the-natural-numbers">The Well-Ordering Principle of the Natural Numbers</a> that a least element <math>d</math> exists in <math>S \setminus \{ 1 \}</math>.</li>
</ul>
<ul>
<li>There are two subcases to consider. If <math>d_1</math> is prime then <math>p =d_1</math> and <math>d_1 \mid n</math>. If <math>d_1</math> is not prime then there exists a positive divisor <math>d_2</math> of <math>d_1</math>, so <math>1 < d_2 < d_1 < n</math>. But this implies that <math>d_2 \in S \setminus \{ 1 \}</math> and <math>d_2 < d_1</math>, which is a contradiction since <math>d_1</math> is the least element in <math>S \setminus \{ 1 \}</math>. Therefore <math>d_1</math> is actually prime and again, <math>d_1 \mid n</math>.</li>
</ul>
<ul>
<li>Therefore if <math>n > 1</math> then there exists a prime <math>p \in \mathbb{P}</math> such that <math>p \mid n</math>. <math>\blacksquare</math></li>
</ul>
===Composite Numbers===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>n \geq 1</math> is said to be Composite if it is not prime, that is, there exists a <math>d \in \mathbb{Z}</math>, <math>1 < d < n</math> such that <math>d \mid n</math> OR <math>n = 1</math>.</td>
</blockquote>
For example, the numbers in the following set are composite:
<div style="text-align: center;"><math>\begin{align} \quad \{4, 6, 8, 9, 10, 12, ... \} \end{align}</math></div>
One important aspect of composite numbers is that every nonnegative even number (except <math>2</math>) is composite since all of such numbers has <math>2</math> as a nontrivial divisor.

===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T01:40:54Z

Lila: /* Algebraic properties */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\Z</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T01:20:51Z

Lila: /* Greatest Common Divisor */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\mathbb{Z}</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a | b</math> then it's not hard to verify that <math>\text{gcd} (a, b) = | a |</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T01:09:15Z

Lila: /* Bézout's Identity */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\mathbb{Z}</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a \mid b</math> then it's not hard to verify that <math>\gcd (a, b) = \mid a \mid</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

====Proof====
Given any nonzero integers {{mvar|a}} and {{mvar|b}}, let <math>S=\{ax+by \mid x,y\in\mathbb{Z} \text{ and } ax+by>0\}.</math> The set {{mvar|S}} is nonempty since it contains either {{mvar|a}} or {{math|–''a''}} (with {{math|1=''x'' = ±1}} and {{math|1=''y'' = 0}}). Since {{mvar|S}} is a nonempty set of positive integers, it has a minimum element <math>d = as + bt</math>, by the Well-ordering principle. To prove that {{mvar|d}} is the greatest common divisor of {{mvar|a}} and {{mvar|b}}, it must be proven that {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}, and that for any other common divisor {{mvar|c}}, one has {{math|''c'' ≤ ''d''}}.

The Euclidean division of {{mvar|a}} by {{mvar|d}} may be written
:<math>a=dq+r\quad\text{with}\quad 0\le r<d.</math>
The remainder {{mvar|r}} is in <math>S\cup \{0\}</math>, because
:<math>
\begin{align}
r & = a - qd \\
& = a - q(as+bt)\\
& = a(1-qs) - bqt.
\end{align}
</math>
Thus {{mvar|r}} is of the form <math>ax+by</math>, and hence <math>r\in S\cup \{0\}</math>. However, {{math|0 ≤ ''r'' < ''d''}}, and {{mvar|d}} is the smallest positive integer in {{mvar|S}}: the remainder {{mvar|r}} can therefore not be in {{mvar|S}}, making {{mvar|r}} necessarily 0. This implies that {{mvar|d}} is a divisor of {{mvar|a}}. Similarly {{mvar|d}} is also a divisor of {{mvar|b}}, and {{mvar|d}} is a common divisor of {{mvar|a}} and {{mvar|b}}.

Now, let {{mvar|c}} be any common divisor of {{mvar|a}} and {{mvar|b}}; that is, there exist {{mvar|u}} and {{mvar|v}} such that
{{math|1=''a'' = ''cu''}} and {{math|1=''b'' = ''cv''}}. One has thus
:<math>\begin{align}d&=as + bt\\
& =cus+cvt\\
&=c(us+vt).\end{align} </math>
That is {{mvar|c}} is a divisor of {{mvar|d}}, and, therefore {{math|''c'' ≤ ''d.''}}

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T01:04:49Z

Lila: /* Finding GCD and LCM with Prime Decomposition */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\mathbb{Z}</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a \mid b</math> then it's not hard to verify that <math>\gcd (a, b) = \mid a \mid</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

===Bézout's Identity===
Let ''a'' and ''b'' be integers with greatest common divisor ''d''. Then there exist integers ''x'' and ''y'' such that ''ax'' + ''by'' = ''d''. More generally, the integers of the form ''ax'' + ''by'' are exactly the multiples of ''d''.

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T00:21:35Z

Lila: /* Greatest Common Divisor */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\mathbb{Z}</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>

Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a \mid b</math> then it's not hard to verify that <math>\gcd (a, b) = \mid a \mid</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T00:18:42Z

Lila: /* Finding GCD and LCM with Prime Decomposition */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\mathbb{Z}</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>
Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a \mid b</math> then it's not hard to verify that <math>\gcd (a, b) = \mid a \mid</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \{m, n\} = p_1^{\min \{k_1, l_1\} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \{m, n\} = p_1^{\max \{k_1, l_1\} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license

Abstract Algebra: Preliminaries

2021-11-19T00:15:34Z

Lila: /* Finding GCD and LCM with Prime Decomposition */

==Natural numbers==
The '''natural numbers''' are those numbers used for counting (as in "there are ''six'' coins on the table") and ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers", and words used for ordering are "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"), that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.

Some definitions begin the natural numbers with 0, corresponding to the non-negative integers {0, 1, 2, 3, ...}, whereas others start with 1, corresponding to the positive integers {1, 2, 3, ...}. Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, while in other writings, that term is used instead for the integers (including negative integers).

The natural numbers are a basis from which many other number sets may be built by extension: the integers, by including (if not yet in) the neutral element 0 and an additive inverse ({{math|−''n''}}) for each nonzero natural number {{mvar|n}}; the rational numbers, by including a multiplicative inverse (<math>\tfrac 1n</math>) for each nonzero integer {{mvar|n}} (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on. This chain of extensions make the natural numbers canonically embedded (identified) in the other number systems.

In common language, particularly in primary school education, natural numbers may be called '''counting numbers''' to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement — a hallmark characteristic of real numbers.

===Well-ordering principle===
Consider the following set which we define to be the set of natural numbers:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{N} = \{ 1, 2, 3, ... \} \end{align}</math></div>
Now consider any subset <math>A</math> of <math>\mathbb{N}</math>. For example, let us consider the subsets <math>A_1 = \{ 7, 29 \}</math>, <math>A_2 = \{1, 3, 5, ... \}</math>, and <math>A_3 = \{2, 4, 6, ... \}</math>. The subset <math>A_1</math> describes a finite set containing two integers. The subset <math>A_2</math> describes the infinite set of all odd natural numbers, and the subset <math>A_3</math> describes the infinite set of all even natural numbers. Notice that in each of these cases we can identify a least element of the subset. For example, the least element in <math>A_1</math> is <math>7</math>, the least element in <math>A_2</math> is <math>1</math>, and the least element in <math>A_3</math> is <math>2</math>.
In fact, it is impossible to construct a nonempty subset <math>A</math> of <math>\mathbb{N}</math> that does not contain a least element. We describe this very important result below in the following theorem.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let <math>A</math> be a nonempty subset of the natural numbers <math>\mathbb{N} = \{1, 2, 3, ...\}</math>. Then there exists a least element <math>x \in A</math>, that is, <math>x \leq y</math> for all <math>y \in A</math>.</td>
</blockquote>
We do not have the tools to prove Theorem 1 although it is a rather intuitively obvious result. Nevertheless, we should note that while Theorem 1 holds true for the set of natural numbers, it does not hold true for all sets of real numbers. For example, consider the set of rational numbers <math>\mathbb{Q}</math>:
<div style="text-align: center;"><math>\begin{align} \quad \mathbb{Q} = \left \{ \frac{a}{b} : a, b \in \mathbb{Z}, b \neq 0 \right \} \end{align}</math></div>
Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between <math>0</math> and <math>1</math> noninclusive, that is, <math>A = \{ x \in \mathbb{Q} : 0 < x < 1 \}</math>. We claim that <math>A</math> has no least element. To prove this, assume that <math>A</math> does have a least element, say <math>x \in A</math> is such that <math>x \leq y</math> for all <math>y \in A</math>. Since <math>x \in A</math> we have that <math>x = \frac{a}{b}</math> where <math>a, b \in \mathbb{Z}</math> Since <math>x \in A</math> we must have that <math>x > 0</math> too since <math>0 < x < 1</math> and so:
<div style="text-align: center;"><math>\begin{align} \quad 0 < \frac{a}{b} < 1 \end{align}</math></div>
Consider the number <math>y</math> in the middle of <math>0</math> and <math>\frac{a}{b}</math>. This number is <math>y = \frac{a}{2b}</math>. Since <math>b \in \mathbb{Z}</math> and <math>b \neq 0</math> we have that <math>2b \in \mathbb{Z}</math> and <math>2b \neq 0</math>, so indeed, <math>y = \frac{a}{2b} \in A</math> and <math>0 < y < x < 1</math> which is a contradiction to our assumption that a least element exists in <math>A</math>.
It's important to note that while not every subset of the rational numbers <math>\mathbb{Q}</math> contains a least element, there are still subsets of <math>\mathbb{Q}</math> that do contain a least element.

===Induction===
'''Mathematical induction''' is a mathematical proof technique. It is essentially used to prove that a statement ''P''(''n'') holds for every natural number ''n'' = 0, 1, 2, 3, . . . ; that is, the overall statement is a sequence of infinitely many cases ''P''(0), ''P''(1), ''P''(2), ''P''(3), . . . . Informal metaphors help to explain this technique, such as falling dominoes or climbing a ladder:

<blockquote style="background: white; border: 1px solid black; padding: 1em;">
"Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the '''basis''') and that from each rung we can climb up to the next one (the '''step''')."
::: — Concrete Mathematics, page 3 margins.
</blockquote>

A '''proof by induction''' consists of two cases. The first, the '''base case''' (or '''basis'''), proves the statement for ''n ='' 0 without assuming any knowledge of other cases. The second case, the '''induction step''', proves that ''if'' the statement holds for any given case ''n = k'', ''then'' it must also hold for the next case ''n'' = ''k'' + 1. These two steps establish that the statement holds for every natural number ''n''. The base case does not necessarily begin with ''n ='' 0, but often with ''n ='' 1, and possibly with any fixed natural number ''n = N'', establishing the truth of the statement for all natural numbers ''n ≥ N''.

====Description====
The simplest and most common form of mathematical induction infers that a statement involving a natural number {{mvar|n}} (that is, an integer {{math|''n'' ≥ 0}} or 1) holds for all values of {{mvar|n}}. The proof consists of two steps:
# The '''initial''' or '''base case''': prove that the statement holds for 0, or 1.
# The '''induction step''', '''inductive step''', or '''step case''': prove that for every {{mvar|n}}, if the statement holds for {{mvar|n}}, then it holds for {{math|''n'' + 1}}. In other words, assume that the statement holds for some arbitrary natural number {{mvar|n}}, and prove that the statement holds for {{math|''n'' + 1}}.

The hypothesis in the inductive step, that the statement holds for a particular {{mvar|n}}, is called the '''induction hypothesis''' or '''inductive hypothesis'''. To prove the inductive step, one assumes the induction hypothesis for {{mvar|n}} and then uses this assumption to prove that the statement holds for {{math|''n'' + 1}}.

Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value.

====Example====

Mathematical induction can be used to prove the following statement ''P''(''n'') for all natural numbers ''n''.

:<math>P(n)\!:\ \ 0 + 1 + 2 + \cdots + n \,=\, \frac{n(n + 1)}{2}.</math>

This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: <math>0 = \tfrac{(0)(0+1)}2</math>, <math>0+1 = \tfrac{(1)(1+1)}2</math>, <math>0+1+2 = \tfrac{(2)(2+1)}2</math>, etc.

'''Proposition.''' For any <math>n\in\mathbb{N}</math>, <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math>

'''Proof.''' Let ''P''(''n'') be the statement <math>0 + 1 + 2 + \cdots + n = \tfrac{n(n + 1)}{2}.</math> We give a proof by induction on ''n''.

''Base case:'' Show that the statement holds for the smallest natural number ''n'' = 0.

''P''(0) is clearly true: <math>0 = \tfrac{0(0 + 1)}{2}\,.</math>

''Inductive step:'' Show that for any ''k ≥'' 0, if ''P''(''k'') holds, then ''P''(''k''+1) also holds.

Assume the induction hypothesis that for a particular ''k'', the single case ''n = k'' holds, meaning ''P''(''k'') is true:<blockquote><math>0 + 1 + \cdots + k \ =\ \frac{k(k{+}1)}2.</math></blockquote>It follows that:

:<math>(0 + 1 + 2 + \cdots + k )+ (k{+}1) \ =\ \frac{k(k{+}1)}2 + (k{+}1).</math>

Algebraically, the right hand side simplifies as:

:<math>
\begin{align}
\frac{k(k{+}1)}{2} + (k{+}1) & \ =\ \frac {k(k{+}1)+2(k{+}1)} 2 \\
& \ =\ \frac{(k{+}1)(k{+}2)}{2} \\
& \ =\ \frac{(k{+}1)((k{+}1) + 1)}{2}.
\end{align}
</math>

Equating the extreme left hand and right hand sides, we deduce that:<blockquote><math>0 + 1 + 2 + \cdots + k + (k{+}1) \ =\ \frac{(k{+}1)((k{+}1)+1)}2.</math></blockquote>That is, the statement ''P''(''k+''1) also holds true, establishing the inductive step.

''Conclusion'': Since both the base case and the inductive step have been proved as true, by mathematical induction the statement ''P''(''n'') holds for every natural number ''n''. ∎

==Integers==
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called ''whole numbers'' or ''counting numbers'', and their additive inverses (the '''negative integers''', i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface ({{math|'''Z'''}}) or blackboard bold <math>(\mathbb{Z})</math> letter "Z"—standing originally for the German word ''Zahlen'' ("numbers").

<math>\mathbb{Z}</math> is a subset of the set of all rational numbers <math>\mathbb{Q}</math>, which in turn is a subset of the real numbers <math>\mathbb{R}</math>. Like the natural numbers, <math>\mathbb{Z}</math> is countably infinite.

=== Algebraic properties ===

Like the natural numbers, <math>\mathbb{Z}</math> is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, 0), <math>\mathbb{Z}</math>, unlike the natural numbers, is also closed under subtraction.

The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring <math>\mathbb{Z}</math>.

<math>\mathbb{Z}</math> is not closed under division, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).

The following table lists some of the basic properties of addition and multiplication for any integers {{math|''a''}}, {{math|''b''}} and {{math|''c''}}:
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+Properties of addition and multiplication on integers
!
!scope="col" |Addition
!scope="col" |Multiplication
|-
!scope="row" |Closure:
|{{math|''a'' + ''b''}} is an integer
|{{math|''a'' × ''b''}} is an integer
|-
!scope="row"|Associativity:
|{{math|''a'' + (''b'' + ''c'') <math>=</math> (''a'' + ''b'') + ''c''}}
|{{math|''a'' × (''b'' × ''c'') <math>=</math> (''a'' × ''b'') × ''c''}}
|-
!scope="row" |Commutativity:
|{{math|''a'' + ''b'' <math>=</math> ''b'' + ''a''}}
|{{math|''a'' × ''b'' <math>=</math> ''b'' × ''a''}}
|-
!scope="row" |Existence of an identity element:
|{{math|''a'' + 0 <math>=</math> ''a''}}
|{{math|''a'' × 1 <math>=</math> ''a''}}
|-
!scope="row" |Existence of inverse elements:
|{{math|''a'' + (−''a'') <math>=</math> 0}}
|The only invertible integers (called units) are {{math|−1}} and {{math|1}}.
|-
!scope="row" |Distributivity:
|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math>=</math> (''a'' × ''b'') + (''a'' × ''c'')}} and {{math|(''a'' + ''b'') × ''c'' <math>=</math> (''a'' × ''c'') + (''b'' × ''c'')}}
|-
!scope="row" |No zero divisors:
| || | If {{math|''a'' × ''b'' <math>=</math> 0}}, then {{math|''a'' <math>=</math> 0}} or {{math|''b'' <math>=</math> 0}} (or both)
|}

The first five properties listed above for addition say that <math>\mathbb{Z}</math>, under addition, is an abelian group. It is also a cyclic group, since every non-zero integer can be written as a finite sum 1 + 1 + ... + 1 or (−1) + (−1) + ... + (−1). In fact, <math>\mathbb{Z}</math> under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is isomorphic to <math>\mathbb{Z}</math>.

The first four properties listed above for multiplication say that <math>\mathbb{Z}</math> under multiplication is a commutative monoid. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that <math>\mathbb{Z}</math> under multiplication is not a group.

All the rules from the above property table (except for the last), when taken together, say that <math>\mathbb{Z}</math> together with addition and multiplication is a commutative ring with unity. It is the prototype of all objects of such algebraic structure. Only those equalities of expressions are true in <math>\mathbb{Z}</math> for all values of variables, which are true in any unital commutative ring. Certain non-zero integers map to zero in certain rings.

The lack of zero divisors in the integers (last property in the table) means that the commutative ring <math>\mathbb{Z}</math> is an integral domain.

The lack of multiplicative inverses, which is equivalent to the fact that <math>\mathbb{Z}</math> is not closed under division, means that <math>\mathbb{Z}</math> is ''not'' a field. The smallest field containing the integers as a subring is the field of rational numbers. The process of constructing the rationals from the integers can be mimicked to form the field of fractions of any integral domain. And back, starting from an algebraic number field (an extension of rational numbers), its ring of integers can be extracted, which includes <math>\mathbb{Z}</math> as its subring.

Although ordinary division is not defined on <math>\mathbb{Z}</math>, the division "with remainder" is defined on them. It is called Euclidean division, and possesses the following important property: given two integers {{math|''a''}} and {{math|''b''}} with {{math|''b'' ≠ 0}}, there exist unique integers {{math|''q''}} and {{math|''r''}} such that {{math|''a'' <math>=</math> ''q'' × ''b'' + ''r''}} and {{math|0 ≤ ''r'' < {{!}}''b''{{!}}}}, where {{math|{{!}}''b''{{!}}}} denotes the absolute value of {{math|''b''}}. The integer {{math|''q''}} is called the ''quotient'' and {{math|''r''}} is called the ''remainder'' of the division of {{math|''a''}} by {{math|''b''}}. The Euclidean algorithm for computing greatest common divisors works by a sequence of Euclidean divisions.

The above says that <math>\mathbb{Z}</math> is a Euclidean domain. This implies that <math>\mathbb{Z}</math> is a principal ideal domain, and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.

===Construction===
[[File:Relative numbers representation.svg|thumb|alt=Representation of equivalence classes for the numbers −5 to 5
|Red points represent ordered pairs of natural numbers. Linked red points are equivalence classes representing the blue integers at the end of the line.|upright=1.5]]
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers, zero, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers {{math|(''a'',''b'')}}.

The intuition is that {{math|(''a'',''b'')}} stands for the result of subtracting {{math|''b''}} from {{math|''a''}}. To confirm our expectation that 1 − 2 and 4 − 5 denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
:<math>(a,b) \sim (c,d) </math>
precisely when
:<math>a + d = b + c. </math>

Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using {{math|[(''a'',''b'')]}} to denote the equivalence class having {{math|(''a'',''b'')}} as a member, one has:
:<math>[(a,b)] + [(c,d)] := [(a+c,b+d)].</math>
:<math>[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].</math>

The negation (or additive inverse) of an integer is obtained by reversing the order of the pair:
:<math>-[(a,b)] := [(b,a)].</math>

Hence subtraction can be defined as the addition of the additive inverse:
:<math>[(a,b)] - [(c,d)] := [(a+d,b+c)].</math>

The standard ordering on the integers is given by:
:<math>[(a,b)] < [(c,d)]</math> if and only if <math>a+d < b+c.</math>

It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes.

Every equivalence class has a unique member that is of the form {{math|(''n'',0)}} or {{math|(0,''n'')}} (or both at once). The natural number {{math|''n''}} is identified with the class {{math|[(''n'',0)]}} (i.e., the natural numbers are embedded into the integers by map sending {{math|''n''}} to {{math|[(''n'',0)]}}), and the class {{math|[(0,''n'')]}} is denoted {{math|−''n''}} (this covers all remaining classes, and gives the class {{math|[(0,0)]}} a second time since {{math|−0 <math>=</math> 0.}}

Thus, {{math|[(''a'',''b'')]}} is denoted by
:<math>\begin{cases} a - b, & \mbox{if } a \ge b \\ -(b - a), & \mbox{if } a < b. \end{cases}</math>

If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.

This notation recovers the familiar representation of the integers as {{math|{..., −2, −1, 0, 1, 2, ...} }}.

Some examples are:
:<math>\begin{align}
0 &= [(0,0)] &= [(1,1)] &= \cdots & &= [(k,k)] \\
1 &= [(1,0)] &= [(2,1)] &= \cdots & &= [(k+1,k)] \\
-1 &= [(0,1)] &= [(1,2)] &= \cdots & &= [(k,k+1)] \\
2 &= [(2,0)] &= [(3,1)] &= \cdots & &= [(k+2,k)] \\
-2 &= [(0,2)] &= [(1,3)] &= \cdots & &= [(k,k+2)].
\end{align}</math>

==Division algorithm==
One rather important aspect of the divisibility of integers is that if <math>a, b \in \mathbb{Z}</math> then <math>a</math> can be written as the product of some quotient <math>q</math> with <math>b</math> plus a remainder <math>r</math>. For example, if <math>a = 11</math> and <math>b = 3</math>, then <math>a = 3(b) + 2</math> where <math>q = 3</math> and <math>r = 2</math>.
The following theorem known as the Division Algorithm shows us that for any pair of integers <math>a, b \in \mathbb{Z}</math> where <math>b > 0</math> that the form <math>a = bq + r</math> exists and is unique.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Division Algorithm): Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math>. Then there exists unique <math>q, r \in \mathbb{Z}</math> where <math>0 \leq r < b</math> and such that <math>a = bq + r</math>.</td>
</blockquote>
The value of <math>r</math> is often called the Remainder when <math>a</math> is divided by <math>b</math>. That is, <math>a</math> is equal to some multiple <math>q</math>, known as the Quotient of <math>b</math>, plus a remainder <math>r</math> for which <math>0 \leq r < b</math>.
<ul>
<li>Proof: Let <math>a, b \in \mathbb{Z}</math> with <math>b > 0</math> and consider the following set <math>A</math>:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad A = \{ a - bq : q \in \mathbb{N} \} = \{ a, a - b, a - 2b, ... \}. \end{align}</math></div>
<ul>
<li>Consider <math>B \subset A</math> of nonnegative elements in <math>A</math>. If <math>a > 0</math> then <math>a \in B</math> and if <math>a < 0</math> then <math>a - ba = a(1 - b) \in B</math>, so <math>B</math> is a nonempty set. Moreover, <math>B</math> is a nonempty subset of integers and so by the Well-Ordering principle there exists a least element of <math>B</math>, say <math>r \in B</math> is this least element. Since <math>r \in B</math> we have that <math>r = a - bn</math> for some <math>n \in \mathbb{N}</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r \end{align}</math></div>
<ul>
<li>So <math>a</math> if of the form we desire. Now since <math>r \in B</math> and every element in <math>B</math> is nonnegative then we have that <math>0 \leq r</math>. Suppose that <math>r \geq b</math>. Then <math>r = b + p</math> where <math>p \geq 0</math> and so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad a = bq + r = bq + (b + p) = b(q + 1) + p \end{align}</math></div>
<ul>
<li>Therefore <math>p = a - b(q + 1) \geq 0</math> and so <math>p \in B</math>. Since <math>b > 0</math> and <math>q+1 > q</math> we have that <math>0 \leq p < r</math>. But this implies that <math>r</math> is not the least element of <math>B</math> which is a contradiction, so our assumption that <math>r \geq b</math> is false, and so <math>0 \leq r < b</math>.</li>
</ul>
<ul>
<li>We lastly show that <math>q</math> and <math>r</math> are unique for <math>a = bq + r</math> and <math>0 \leq r < b</math>. Suppose that <math>a = bq_1 + r_1</math> and <math>a = bq_2 + r_2</math> where <math>0 \leq r_1, r_2 < b</math>. Then by subtracting these two equations we get:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad 0 = bq_1 - bq_2 + r_1 - r_2 \\ \quad 0 =b(q_1 - q_2) + r_1 - r_2 \\ \quad r_2 - r_1 = b(q_1 - q_2) \end{align}</math></div>
<ul>
<li>Therefore <math>r_2 - r_1</math> is a multiple of <math>b</math>. Since <math>0 \leq r_1 < b</math> and <math>0 \leq r_2 < b</math> we must have that <math>-b < r_2 - r_1 < b</math>. The only multiple of <math>b</math> such that <math>-b < r_2 - r_1 < b</math> is <math>0</math>, so:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad r_2 - r_1 = 0 \\ \quad r_1 = r_2 \end{align}</math></div>
<ul>
<li>Since <math>r_1 = r_2 = r</math>, we have that <math>0 = b(q_1 - q_2)</math>. Since <math>b > 0</math> this implies that <math>q_1 - q_2 = 0</math> and so <math>q_1 = q_2</math>. Therefore <math>a</math> can be uniquely expressed in the form <math>a = bq + r</math> where <math>0 \leq r < b</math>. <math>\blacksquare</math></li>
</ul>

==Congruence modulo m==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td>Definition: If <math>a, b, m \in \mathbb{Z}</math> then we say that <math>a</math> is Congruent to <math>b</math> modulo <math>m</math> denoted <math>a \equiv b \pmod m</math> if <math>m \mid (a - b)</math>.</td>
</blockquote>
The congruence of two integers is an equivalence relation which we can describe in other words. We say that <math>a</math> and <math>b</math> are said to be congruent modulo <math>m</math> if <math>a</math> divided by <math>m</math> leaves the same remaining as <math>b</math> divided by <math>m</math>.
For example, <math>3 \equiv 7 \pmod 2</math> since <math>2 \mid (3 - 7) = -4</math>. Notice also that when <math>3</math> and <math>7</math> are divided by <math>2</math> they both leave the same remainder, <math>1</math>.
We can denote the remainder of an integer <math>a</math> when divided by <math>m</math> as simply <math>a \pmod m</math>. For example, <math>1 = 3 \pmod 2</math> since the remaining of <math>3</math> when divided by <math>2</math> is <math>1</math>. We should note that from <a href="/the-division-algorithm">The Division Algorithm</a> that the remainder <math>r</math> when an integer <math>a</math> is divided by <math>m</math> satisfies the inequality <math>0 \leq r < m</math>. Therefore we note that for all nonnegative integers <math>a</math> and positive integers <math>m</math> that:
<div style="text-align: center;"><math>\begin{align} \quad 0 \leq a \pmod m < m \end{align}</math></div>
In other words, for all nonnegative integers <math>a</math> and positive integers <math>m</math>, the set of possible remainders when <math>a</math> is divided by <math>m</math> is <math>\{ 0, 1, 2, ..., m - 1 \}</math>.
Consequentially, we can define the set of odd integers and the set of even numbers in terms of these elements in the set of even integers being congruent to one another and elements in the set of odd numbers being congruent to one another.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: An integer <math>z \in \mathbb{Z}</math> is said to be Odd if <math>z \equiv 1 \pmod 2</math> or Even if <math>z \equiv 0 \pmod 2</math>.</td>
</blockquote>
We will now look at some basic theorems regarding integer congruence modulo <math>m</math>.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: Let <math>a, b, c, d, m \in \mathbb{Z}</math>. Then: 
a) If <math>a \equiv b \pmod m</math> then <math>b \equiv a \pmod m</math>. 
b) <math>a \equiv a \pmod m</math>. 
c) If <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math> then <math>a \equiv c \pmod m</math>. 
d) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>(a + c) \equiv (b + d) \pmod m</math>. 
e) If <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math> then <math>ac \equiv bd \pmod m</math>.</td>
</blockquote>
<ul>
<li>Proof of a) Suppose that <math>a \equiv b \pmod m</math>. Then <math>m \mid (a - b)</math> so there exists an integer <math>k</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk = a - b \\ \quad -mk = b - a \\ \quad m(-k) = b - a \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (b - a)</math> so <math>b \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>We have that <math>m \mid 0 = (a - a)</math> for any integer <math>a</math>. Therefore <math>a \equiv a \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of c) Suppose that <math>a \equiv b \pmod m</math> and <math>b \equiv c \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (b - c)</math> and there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = b - c \end{align}</math></div>
<ul>
<li>From the second equation we have that <math>b = mk_2 + c</math>. Substituting this into the first equation gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - (mk_2 + c) \\ \quad mk_1 + mk_2 = a - c \\ \quad m(k_1 + k_2) = a - c \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a - c)</math> so <math>a \equiv c \pmod m</math>. <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of d): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \end{align}</math></div>
<ul>
<li>Adding these equations together gives us:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 + mk_2 = (a - b) + (c - d) \\ \quad m(k_1 + k_2) = (a + c) - (b + d) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (a + c) - (b + d)</math> so <math>(a + c) \equiv (b + d) \pmod m</math> <math>\blacksquare</math></li>
</ul>
<ul>
<li>Proof of e): Suppose that <math>a \equiv b \pmod m</math> and <math>c \equiv d \pmod m</math>. Then <math>m \mid (a - b)</math> and <math>m \mid (c - d)</math>, so there exists integers <math>k_1</math> and <math>k_2</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad mk_1 = a - b \quad \mathrm{and} \quad mk_2 = c - d \\ \quad a = b + mk_1 \quad \mathrm{and} \quad c = d +mk_2 \end{align}</math></div>
<ul>
<li>Multiplying these equations together yields:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad ac = (b + mk_1)(d + mk_2) \\ \quad ac = bd + bmk_2 + dmk_1 + m^2k_1k_2 \\ \quad ac - bd = m(bk_2 + dk_1 + mk_1k_2) \end{align}</math></div>
<ul>
<li>Therefore <math>m \mid (ac - bd)</math> so <math>ac \equiv bd \pmod m</math>. <math>\blacksquare</math></li>
</ul>

==Algebra on <math>\Z_m</math>==

==GCD, LCM, and Bézout's identity==
===Greatest Common Divisor===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> then the Greatest Common Divisor of <math>a</math> and <math>b</math> denoted <math>\gcd (a, b)</math> or <math>(a, b)</math> is an integer <math>d \in \mathbb{Z}</math> such that: 
1. <math>d \mid a</math> and <math>d \mid b</math>. 
2. If <math>c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \leq d</math>.</td>
</blockquote>
Note that the number <math>1 \in \mathbb{Z}</math> has the property such that <math>1 \mid a</math> and <math>1 \mid b</math> for all <math>a, b \in \mathbb{Z}</math> and so we see that the greatest common divisor of any two integers is always at least <math>1</math>. As a result, we only need to compare positive divisors of <math>a</math> and <math>b</math>. Furthermore, if <math>a \mid b</math> then it's not hard to verify that <math>\gcd (a, b) = \mid a \mid</math>.
Let's look at an example of finding the greatest common divisor among two integers <math>a, b \in \mathbb{Z}</math>.
Consider the integers <math>a = 14</math> and <math>b = 35</math>. The factors of <math>a</math> are <math>1, 2, 7, 14</math> and the factors of <math>b</math> are <math>1, 5, 7, 35</math>. Both <math>a</math> and <math>b</math> share the factor <math>7</math> and it is the largest factor they share, and so:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (14, 35) = 7 \end{align}</math></div>
As we elaborated on just moments ago, the greatest common divisor of any two integers is always at least <math>1</math>. We give a special name to pairs of integers for which their greatest common divisor is <math>1</math> which we define below.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Definition: If <math>a, b \in \mathbb{Z}</math> and <math>\gcd (a, b) = 1</math> then <math>a</math> and <math>b</math> are said to be Relatively Prime to each other.</td>
</blockquote>
For example, consider the numbers <math>a = 5</math> and <math>b = 8</math>. The greatest common divisor between these numbers is:
<div style="text-align: center;"><math>\begin{align} \quad \gcd (5, 8) = 1 \end{align}</math></div>
Therefore <math>5</math> and <math>8</math> are relatively prime to each other.
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1: If <math>a, b, c \in \mathbb{Z}</math>, <math>c \mid a</math>, and <math>c \mid b</math> then <math>c \mid \gcd (a, b)</math>.</td>
</blockquote>

===Lowest Common Multiple===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
For all <math>a, b \in \Z : a b \ne 0</math>, there exists a smallest <math>m \in \Z : m > 0</math> such that <math>a | m</math> and <math>b | m</math>.

This <math>m</math> is called the '''lowest common multiple of <math>a</math> and <math>b</math>''', and denoted <math>\text{lcm} \{a, b\}</math>.
</blockquote>

===Finding GCD and LCM with Prime Decomposition===
Let <math>m, n \in \Z</math>.

Let:
:<math>m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}</math>
:<math>n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}</math>
:<math>p_i | m \lor p_i | n, 1 \le i \le r</math>.

That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers <math>m</math> or <math>n</math>.

Note that if one of the primes <math>p_i</math> does not appear in the prime decompositions of either one of <math>m</math> or <math>n</math>, then its corresponding index <math>k_i</math> or <math>l_i</math> will be zero.

Then the following results apply:

:<math>\text{gcd} \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \{k_2, l_2\} } \ldots p_r^{\min \{k_r, l_r\} }</math>

:<math>\text{lcm} \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \{k_2, l_2\} } \ldots p_r^{\max \{k_r, l_r\} }</math>

==Primes==
===Euclid's Lemma===

==Fundamental Theorem of Arithmetic==

== Licensing ==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Natural_number Natural number, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Integer Integer, Wikipedia] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Mathematical_induction Mathematical induction, Wikipedia] under a CC BY-SA license
* [http://mathonline.wikidot.com/abstract-algebra Abstract Algebra, mathonline.wikidot.com] under a CC BY-SA license