The Integers - Revision history

Lila: /* Licensing */

2021-11-16T21:48:30Z

Licensing

Lila at 21:31, 16 November 2021

2021-11-16T21:31:30Z

Lila at 21:30, 16 November 2021

2021-11-16T21:30:44Z

Lila: /* Licensing */

2021-11-16T21:29:06Z

Licensing

Lila at 21:24, 16 November 2021

2021-11-16T21:24:31Z

Lila at 21:23, 16 November 2021

2021-11-16T21:23:22Z

Lila at 21:22, 16 November 2021

2021-11-16T21:22:25Z

Lila at 21:21, 16 November 2021

2021-11-16T21:21:51Z

Lila at 21:20, 16 November 2021

2021-11-16T21:20:50Z

Lila at 21:20, 16 November 2021

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@@ Line 173: / Line 173: @@
 <ul>
 <li><strong>Proof of c)</strong> From (b) if we set <span class="math-inline"><math>x = 1</math></span> and <span class="math-inline"><math>y = 1</math></span> we get that <span class="math-inline"><math>a \mid (b + c)</math></span> and if we set <span class="math-inline"><math>x = 1</math></span> and <span class="math-inline"><math>y = -1</math></span> we get that <span class="math-inline"><math>a \mid (b - c)</math></span> as desired. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>

@@ Line 147: / Line 147: @@
 <li><strong>Proof of a)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>b \mid c</math></span> then there exists <span class="math-inline"><math>q_1, q_2 \in \mathbb{Z}</math></span> such that:</li>
 </ul>
-<div style="text-align: center;"><math>\begin{align} \quad aq_1 = b \quad \text{and} \quad bq_2 = c. \end{align}<math></div>
+<div style="text-align: center;"><math>\begin{align} \quad aq_1 = b \quad \text{and} \quad bq_2 = c. \end{align}</math></div>
 <ul>
 <li>Substituting the first equation into the second gives us:</li>

@@ Line 147: / Line 147: @@
 <li><strong>Proof of a)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>b \mid c</math></span> then there exists <span class="math-inline"><math>q_1, q_2 \in \mathbb{Z}</math></span> such that:</li>
 </ul>
-<div style="text-align: center;"><math>\begin{align} \quad aq_1 = b \quad \mathrm{and} \quad bq_2 = c. \end{align}<math></div>
+<div style="text-align: center;"><math>\begin{align} \quad aq_1 = b \quad \text{and} \quad bq_2 = c. \end{align}<math></div>
 <ul>
 <li>Substituting the first equation into the second gives us:</li>

← Older revision		Revision as of 21:29, 16 November 2021
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	The technique for the construction of integers presented above in this section corresponds to the particular case where there is a single basic operation '''pair'''<math>(x,y)</math> that takes as arguments two natural numbers <math>x</math> and <math>y</math>, and returns an integer (equal to <math>x-y</math>). This operation is not free since the integer 0 can be written '''pair'''(0,0), or '''pair'''(1,1), or '''pair'''(2,2), etc. This technique of construction is used by the proof assistant Isabelle; however, many other tools use alternative construction techniques, notable those based upon free constructors, which are simpler and can be implemented more efficiently in computers.		The technique for the construction of integers presented above in this section corresponds to the particular case where there is a single basic operation '''pair'''<math>(x,y)</math> that takes as arguments two natural numbers <math>x</math> and <math>y</math>, and returns an integer (equal to <math>x-y</math>). This operation is not free since the integer 0 can be written '''pair'''(0,0), or '''pair'''(1,1), or '''pair'''(2,2), etc. This technique of construction is used by the proof assistant Isabelle; however, many other tools use alternative construction techniques, notable those based upon free constructors, which are simpler and can be implemented more efficiently in computers.
		+
		+
		+	==Integer Divisibility==
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	<td><strong>Definition:</strong> Let <span class="math-inline"><math>a, b \in \mathbb{Z}</math></span>. Then <span class="math-inline"><math>b</math></span> is said to be <strong>Divisible</strong> by <span class="math-inline"><math>a</math></span>, or, <span class="math-inline"><math>a</math></span> is said to <strong>Divide</strong> <span class="math-inline"><math>b</math></span> written <span class="math-inline"><math>a \mid b</math></span> if there exists a <span class="math-inline"><math>q \in \mathbb{Z}</math></span> such that <span class="math-inline"><math>aq = b</math></span>. The number <span class="math-inline"><math>a</math></span> is said to be a <strong>Divisor</strong> of <span class="math-inline"><math>b</math></span> and <span class="math-inline"><math>b</math></span> is said to be a <strong>Multiple</strong> of <span class="math-inline"><math>a</math></span>. If there does not exist an <span class="math-inline"><math>q \in \mathbb{Z}</math></span> such that <span class="math-inline"><math>aq = b</math></span> then we say that <span class="math-inline"><math>a</math></span> does not divide <span class="math-inline"><math>b</math></span>, or, <span class="math-inline"><math>b</math></span> is not divisible by <span class="math-inline"><math>a</math></span> denoted <span class="math-inline"><math>a \not \mid b</math></span>.</td>
		+	</blockquote>
		+	<p>For example, the number <span class="math-inline"><math>a = 2</math></span> divides <span class="math-inline"><math>b = 6</math></span> since for <span class="math-inline"><math>q = 3</math></span> we have that <span class="math-inline"><math>aq = b</math></span>, so we write <span class="math-inline"><math>a \mid b</math></span>. However, the number <span class="math-inline"><math>a = 2</math></span> does not divide any odd number <span class="math-inline"><math>b = 2n + 1</math></span> for <span class="math-inline"><math>n \in \mathbb{Z}</math></span>, so in this case we have that <span class="math-inline"><math>a \not \mid b</math></span>, i.e., <span class="math-inline"><math>2 \not \mid (2n + 1)</math></span> for all integers <span class="math-inline"><math>n</math></span>.</p>
		+	<p>We will now look at some rather elementary proofs regarding the divisibility of integers.</p>
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	<td><strong>Theorem 1:</strong> Let <span class="math-inline"><math>a, b, c \in \mathbb{Z}</math></span>.<br />
		+	<strong>a)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>b \mid c</math></span> then <span class="math-inline"><math>a \mid c</math></span>.<br />
		+	<strong>b)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>a \mid c</math></span> then for all <span class="math-inline"><math>x, y \in \mathbb{Z}</math></span>, <span class="math-inline"><math>a \mid (bx + cy)</math></span>.<br />
		+	<strong>c)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>a \mid c</math></span> then <span class="math-inline"><math>a \mid (b + c)</math></span> and <span class="math-inline"><math>a \mid (b - c)</math></span>.</td>
		+	</blockquote>
		+	<p><em>For the following proofs we utilize the fact that the set of integers is closed under standard addition and standard multiplication - that is, if <span class="math-inline"><math>x, y \in \mathbb{Z}</math></span> then <span class="math-inline"><math>(x + y) \in \mathbb{Z}</math></span> and <span class="math-inline"><math>x \cdot y = xy \in \mathbb{Z}</math></span>.</em></p>
		+	<ul>
		+	<li><strong>Proof of a)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>b \mid c</math></span> then there exists <span class="math-inline"><math>q_1, q_2 \in \mathbb{Z}</math></span> such that:</li>
		+	</ul>
		+	<div style="text-align: center;"><math>\begin{align} \quad aq_1 = b \quad \mathrm{and} \quad bq_2 = c. \end{align}<math></div>
		+	<ul>
		+	<li>Substituting the first equation into the second gives us:</li>
		+	</ul>
		+	<div style="text-align: center;"><math>\begin{align} \quad bq_2 = c \\ \quad a(q_1q_2) = c \end{align}</math></div>
		+	<ul>
		+	<li>Since <span class="math-inline"><math>q_1, q_2 \in \mathbb{Z}</math></span> we have that their product <span class="math-inline"><math>q_1q_2 \in \mathbb{Z}</math></span>. Therefore <span class="math-inline"><math>a \mid c</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
		+	</ul>
		+	<ul>
		+	<li><strong>Proof of b)</strong> If <span class="math-inline"><math>a \mid b</math></span> and <span class="math-inline"><math>a \mid c</math></span> then there exists <span class="math-inline"><math>q_1, q_2 \in \mathbb{Z}</math></span> such that:</li>
		+	</ul>
		+	<span class="equation-number">(3)</span>
		+	<div style="text-align: center;"><math>\begin{align} \quad aq_1 = b \quad \mathrm{and} \quad aq_2 = c. \end{align}</math></div>
		+	<ul>
		+	<li>Multiply the first equation by <span class="math-inline"><math>x \in \mathbb{Z}</math></span> and the second equation by <span class="math-inline"><math>y \in \mathbb{Z}</math></span> to get:</li>
		+	</ul>
		+	<div style="text-align: center;"><math>\begin{align} \quad axq_1 = bx \quad \mathrm{and} \quad ayq_2 = cy. \end{align}</math></div>
		+	<ul>
		+	<li>We now add both of these equations:</li>
		+	</ul>
		+	<div style="text-align: center;"><math>\begin{align} \quad axq_1 + ayq_2 = bx + cy \\ \quad a(xq_1 + yq_2) = bx + cy \end{align}</math></div>
		+	<ul>
		+	<li>Since <span class="math-inline"><math>q_1, q_2, x, y \in \mathbb{Z}</math></span> we have that <span class="math-inline"><math>(xq_1 + yq_2) \in \mathbb{Z}</math></span> and so <span class="math-inline"><math>a \mid (bx + cy)</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
		+	</ul>
		+	<ul>
		+	<li><strong>Proof of c)</strong> From (b) if we set <span class="math-inline"><math>x = 1</math></span> and <span class="math-inline"><math>y = 1</math></span> we get that <span class="math-inline"><math>a \mid (b + c)</math></span> and if we set <span class="math-inline"><math>x = 1</math></span> and <span class="math-inline"><math>y = -1</math></span> we get that <span class="math-inline"><math>a \mid (b - c)</math></span> as desired. <span class="math-inline"><math>\blacksquare</math></span></li>
		+	</ul>

@@ Line 63: / Line 63: @@
 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'' = ''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.
+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.
@@ Line 105: / Line 105: @@
 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 = 0.}}
+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

@@ Line 53: / Line 53: @@
 |}
-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 {{nowrap|1 + 1 + ... + 1}} or {{nowrap|(−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 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.
@@ Line 85: / Line 85: @@
 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 {{nowrap|1 − 2}} and {{nowrap|4 − 5}} denote the same number, we define an equivalence relation {{math|~}} on these pairs with the following rule:
+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

@@ Line 47: / Line 47: @@
 |-
 !scope="row" |Distributivity:
-|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') = (''a'' × ''b'') + (''a'' × ''c'')}} &nbsp;&nbsp; and &nbsp;&nbsp; {{math|(''a'' + ''b'') × ''c'' = (''a'' × ''c'') + (''b'' × ''c'')}}
+|colspan=2 align=center |{{math|''a'' × (''b'' + ''c'') <math> = </math> (''a'' × ''b'') + (''a'' × ''c'')}} &nbsp;&nbsp; and &nbsp;&nbsp; {{math|(''a'' + ''b'') × ''c'' <math> = </math> (''a'' × ''c'') + (''b'' × ''c'')}}
 |-
 !scope="row" |No zero divisors:

@@ Line 50: / Line 50: @@
 |-
 !scope="row" |No zero divisors:
-| || | If {{math|''a'' × ''b'' = 0}}, then {{math|''a'' = 0}} or {{math|''b'' = 0}} (or both)
+| || | If {{math|''a'' × ''b'' <math> = </math> 0}}, then {{math|''a'' <math> = </math> 0}} or {{math|''b'' <math> = </math> 0}} (or both)
 |}

@@ Line 27: / Line 27: @@
 |-
 !scope="row" |Closure:
-|{{math|''a'' + ''b''}}{{pad|1em}}is an integer
+|{{math|''a'' + ''b''}} &nbsp;&nbsp; is an integer
-|{{math|''a'' × ''b''}}{{pad|1em}}is an integer
+|{{math|''a'' × ''b''}} &nbsp;&nbsp; is an integer
 |-
 !scope="row"|Associativity: