Axioms of the Real Numbers

Donovan D Chang

17 Axioms of the Real Numbers

The set of real numbers can be described as a “complete ordered field” [12]. In this section and in the following, we shall discuss

What a field is
What an ordered field is
What completeness is, and thus what a complete ordered field is

To begin with, we first need to define a field.

Definition. (Field)
A field is a commutative division ring. [9]

However, we have not introduced what a ring is. Therefore, we formally define a ring as follows:

Definition. (Ring)
A ring $\langle R,+,\cdot \rangle$ is a set $R$ , on which two binary operations, addition $(+)$ and multiplication $(\cdot)$ , are defined, with the following properties:

1. For all $a,b,c\in R$ ,
(a) Closure

$(a+b),(a\cdot b)\in R$

(b)Associativity

$(a+b)+c=a+(b+c) \text{ and } (a\cdot b)\cdot c=a \cdot (b \cdot c)$

(c) The existence of an identity element
$0\in R$ such that $0+a=a+0=a$
$1\in R$ such that $1\cdot a=a\cdot 1=a$

(d) The existence of an inverse
$a^{-1}=-a\in R$ such that $a+(-a)=(-a)+a=0$
$a^{-1}=\frac{1}{a}\in R$ such that $a\cdot \frac{1}{a}=\frac{1}{a}\cdot a=1$ , except when $a=0$

2. Addition $(+)$ is commutative, i.e. $a+b=b+a$

3. The left and right distributive law holds, i.e.

$a\cdot (b+c)=(a\cdot b)+(a\cdot c)\text{ and }(a+b)\cdot c=(a\cdot c)+(b\cdot c)$

Therefore, a field, a commutative division ring, refers to a ring, where multiplication $(\cdot)$ is also commutative.

Now we turn our focus to orderedness. To begin with, let us formally define well-ordering.

Definition. (Well-Ordering Property of $\mathbb{N}$ )
Let $S\subseteq \mathbb{N}$ . Then, there exists $m\in S$ such that $m \leq k$ for all $k\in S$ .

In other words, this can be restated as every nonempty subset of $\mathbb{N}$ has a smallest element. Also, note that this well-ordering property is in fact equivalent to the principles of mathematical induction. That is, one implies the other, and vice versa.

Let us illustrate with an example.

Example. Show that $\mathbb{Q}^c$ defined by $\mathbb{Q}^c=\mathbb{R}\setminus\mathbb{Q}$ is not an ordered field.

Let us first examine whether $\mathbb{Q}^c$ is a field.

In fact, showing one counterexample for any of the axioms that have to be satisfied for $\mathbb{Q}^c$ to be a field would suffice to prove $\mathbb{Q}^c$ is not a field.

Such counterexample is $\sqrt{2}\in \mathbb{Q}^c$ .

$\sqrt{2}\cdot \sqrt{2} = 2 \notin \mathbb{Q}^c$

One of important properties of a field, of any binary operation, is a closure property, and we can see $\mathbb{Q}^c$ is not closed under multiplication.

Therefore, $\mathbb{Q}^c$ is not a field.

For a field to become an ordered field, we need 4 more conditions or axioms as follows:

Let us define a relation “ $<$ ” on $\mathbb{R}$ . Then, for all $x,y,z\in \mathbb{R}$ ,