Abstract Algebra: Preliminaries

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 (−n) for each nonzero natural number n; the rational numbers, by including a multiplicative inverse (${\tfrac {1}{n}}$) for each nonzero integer 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:

${\begin{aligned}\quad \mathbb {N} =\{1,2,3,...\}\end{aligned}}$

Now consider any subset $A$ of $\mathbb {N}$. For example, let us consider the subsets $A_{1}=\{7,29\}$, $A_{2}=\{1,3,5,...\}$, and $A_{3}=\{2,4,6,...\}$. The subset $A_{1}$ describes a finite set containing two integers. The subset $A_{2}$ describes the infinite set of all odd natural numbers, and the subset $A_{3}$ 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 $A_{1}$ is $7$, the least element in $A_{2}$ is $1$, and the least element in $A_{3}$ is $2$.

In fact, it is impossible to construct a nonempty subset $A$ of $\mathbb {N}$ that does not contain a least element. We describe this very important result below in the following theorem.

Theorem 1 (The Well-Ordering Principle of the Natural Numbers): Let $A$ be a nonempty subset of the natural numbers $\mathbb {N} =\{1,2,3,...\}$. Then there exists a least element $x\in A$, that is, $x\leq y$ for all $y\in A$.

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 $\mathbb {Q}$:

${\begin{aligned}\quad \mathbb {Q} =\left\{{\frac {a}{b}}:a,b\in \mathbb {Z} ,b\neq 0\right\}\end{aligned}}$

Any subset of the rational numbers need not have a least element. For example, consider the following subset of rational numbers between $0$ and $1$ noninclusive, that is, $A=\{x\in \mathbb {Q} :0<x<1\}$. We claim that $A$ has no least element. To prove this, assume that $A$ does have a least element, say $x\in A$ is such that $x\leq y$ for all $y\in A$. Since $x\in A$ we have that $x={\frac {a}{b}}$ where $a,b\in \mathbb {Z}$ Since $x\in A$ we must have that $x>0$ too since $0<x<1$ and so:

${\begin{aligned}\quad 0<{\frac {a}{b}}<1\end{aligned}}$

Consider the number $y$ in the middle of $0$ and ${\frac {a}{b}}$. This number is $y={\frac {a}{2b}}$. Since $b\in \mathbb {Z}$ and $b\neq 0$ we have that $2b\in \mathbb {Z}$ and $2b\neq 0$, so indeed, $y={\frac {a}{2b}}\in A$ and $0<y<x<1$ which is a contradiction to our assumption that a least element exists in $A$.

It's important to note that while not every subset of the rational numbers $\mathbb {Q}$ contains a least element, there are still subsets of $\mathbb {Q}$ that do contain a least element.

Licensing

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