Axioms of the Real Numbers

Abigail Huettig

18 Axioms of the Real Numbers

Axioms of the real numbers are statements that describe the qualities and properties that the real numbers possess. We are going to cover the field axioms and then the triangle inequality.

Field Axioms

A field is a set that satisfies the field axioms given in the definition below, which comes from Abbott [1].

Definition IV.1

A set $F$ is a field if there exist two operations, addition $(x+y)$ and multiplication $(xy)$ , that satisfy the following list of conditions:

(Commutativity) $x+y=y+x$ and $xy=yx$ for all $x,y\in F$ .
(Associativity) $(x+y)+z = x+(y+z)$ and $(xy)z = x(yz)$ for all $x,y,z \in F$ .
(Identities exist) There exist two special elements 0 and 1 with $0\neq 1$ such that $x+0=x$ and $1x=x$ for all $x\in F$ .
(Inverses exist) Given $x\in F$ , there exists an element $-x\in F$ such that $x+(-x)=0$ . If $x\neq 0$ , there exists an element $x^{-1}$ such that $xx^{-1}=1.$
(Distributive property) $x(y+z)=xy+xz$ for all $x,y,z \in F$ .

Basically, a set is a field if and only if the commutative and associative properties hold for all of the elements under addition and multiplication, the distributive property of multiplication holds, the set contains identity elements for addition and multiplication, and the inverses of each element under addition and multiplication are also elements of the set. We have already been working with fields without even realizing it. The set of rational numbers and the set of real numbers are both fields.They are also ordered fields, which are fields that must satisfy the following additional properties [1].

For any two elements $x,y$ in a set $F$ , either $x\leq y$ or $y\leq x$ .
If $x\leq y$ and $y\leq x$ , then $x$ must equal $y$ .
If $x\leq y$ and $y\leq z$ , then $x\leq z$ .
If $y\leq z$ , then $x+y\leq x+z$ .
If $x\geq 0$ and $y\geq 0$ , then $xy\geq 0$ .

Looking at these properties, it becomes clear why we refer to fields satisfying them as ordered fields. These properties ensure that there is a specific order preserved throughout the field.

Let’s now look at a few well known sets and determine whether they are fields. The following exercise is a homework problem from Abbott that I completed for this paper [1].

Example 37
Determine whether the sets $\mathbb{N},\mathbb{Z}$ , and $\mathbb{Q}$ are fields. For any that are fields, determine whether they are ordered fields.

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Solution
First off, we know that the commutative, associative, and distributive properties all hold for these sets.Regarding the natural numbers, for $x\in \mathbb{N}$ , $x+0=x$ . That is, the identity element for addition is 0. However, additive inverses do not exist within the natural numbers. For example,

$5+(-5)=0,$

but -5 is not a natural number. Thus, $\mathbb{N}$ is not a field.Regarding the integers, for $x\in \mathbb{Z}$ , $x\cdot 1=x$ . That is, the multiplicative identity is 1. However, multiplicative inverses do not exist within the integers. For example,

$5\cdot \frac{1}{5}=1,$

but $\frac{1}{5}$ is not an integer. Thus, $\mathbb{Z}$ is not a field.Regarding the rational numbers, for $x\in \mathbb{Q}$ ,

$x+0=x, x\cdot 1=x, x+(-x)=0 \text{ and } x\cdot\frac{1}{x}=1 \text{ (where $x\neq 0$) }.$

Since 0, 1, $-x$ , and $\frac{1}{x}$ for all rational numbers $x$ are themselves elements of the rational numbers, $\mathbb{Q}$ is a field. Furthermore, $\mathbb{Q}$ is an ordered field because the five properties listed above are satisfied for all $x,y,z \in \mathbb{Q}$ .

Later on, we will revisit the idea of fields in Part V: Chapter 24 when we are discussing Modern Algebra. But for now, we are going to move on to a simple but powerful axiom of the real numbers known as the triangle inequality.

Triangle Inequality

The triangle inequality is stated by Abbott as follows [1]:

$|a+b|\leq |a|+|b|$

where $a$ and $b$ are real numbers.

Much like the Pigeonhole Principle (see Part III: Chapter 16), the triangle inequality is a simple statement that helps us prove many more complex ideas.

Recall that in Part I: Chapter 1, we discussed some properties of limits that made it easier to solve limits analytically. We will now present the proof for the property

$\lim_{x\rightarrow c}(f(x)-g(x))=L-K$

where $\lim_{x\rightarrow c} f(x)=L$ and $\lim_{x\rightarrow c} g(x)=K$ . To prove this, we will be using definition IV.20 as given in chapter 21.

The following proof is a homework problem from Abbott that I completed for Introduction to Analysis [1].

Example 38
Show that

$\lim_{x\rightarrow c}(f(x)-g(x))=L-K$

where $\lim_{x\rightarrow c} f(x)=L$ and $\lim_{x\rightarrow c} g(x)=K$ .

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Proof.
Using the triangle inequality, we have

$\begin{align*} |(f(x)-g(x))-(L-K)|&=|(f(x)-L)+(-g(x)+K)|\\ &\leq |f(x)-L|+|-g(x)+K| \end{align*}$

Let $\epsilon>0$ . By definition IV.20, there exists $\delta_1>0$ such that if $0<|x-c|<\delta_1$ ,

$|f(x)-L|<\frac{\epsilon}{2}.$

By definition IV.20, there also exists $\delta_2>0$ such that if $0<|x-c|<\delta_2$ ,

$\begin{align*} |g(x)-K|&=|-g(x)+K|\\ &<\frac{\epsilon}{2} \end{align*}$

Let $\delta=\text{min}\{\delta_1,\delta_2\}.$ If $0<|x-c|<\delta$ ,

$\begin{align*} |(f(x)-g(x))-(L-K)|&\leq |f(x)-L| + |-g(x)+K|\\ &< \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align*}$

Therefore,

$\lim_{x\rightarrow c}(f(x)-g(x))=L-K$

by definition IV.20.

This is just one example of how useful the triangle inequality is. There is so much that we would not be able to prove without it. But time is precious, and we must move on to the topology of the real numbers, which contains some of the most interesting ideas covered in analysis.

18 Axioms of the Real Numbers

Field Axioms

Triangle Inequality

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