3.1 Why do we Need the Real Numbers?

The rational numbers have many desirable properties by themselves which allow us to do a lot of mathematics with them. One of these is that they are closed under regular arithmetic operations of addition and multiplication. In fact, $\mathbb{Q}$ forms an ordered field when paired with the total order $\leq$ . The rational numbers, along with the integers and natural numbers, satisfy the Archimedean property,

\forall q\in\mathbb{Q},\ \exists n\in\mathbb{N}\ \mathrm{s.t.}\ n>q,

and they also possess the property of density, which states that between any two rational numbers there exists another rational number.

Theorem 3.1

$\mathbb{Q}$ is dense.

Proof.

Let $p,q\in\mathbb{Q}$ with $p<q$ . Then let $r=\frac{p+q}{2}$ , so $r\in\mathbb{Q}$ and $p<r<q$ . ∎

This process can be repeated indefinitely, which shows that there are actually infinitely many rational numbers between each rational number. This being the case, there are still many numbers that we need that are not in $\mathbb{Q}$ . Specifically, since limits are one of the central topics of study in analysis, it would be nice if we could guarantee the existence of limits for convergent sequences. It turns out that this is equivalent to the completeness property, which we will now define.

Definition 3.1.

A partially ordered set $X$ is complete¹¹ 1 This property is sometimes called Dedekind completeness, or simply the least-upper-bound property. if every non-empty subset of $X$ which is bounded above (has an upper bound) has a least upper bound (supremum) in $X$ .

It can be shown wih a counterexample that the rational numbers are not complete.

Theorem 3.2

Let $S=\{q\in\mathbb{Q}\,:\,q^{2}<2\}$ . Then there does not exist a rational number $r$ such that $r=\sup(S)$ .

Proof.

$S$ is clearly bounded from above, for example $\frac{3}{2}$ is an upper bound $\left(\left(\frac{3}{2}\right)^{2}=\frac{9}{4}>2\right)$ , so by the completeness property we would expect a supremum to exist. Let us assume by way of contradiction that $r=\sup(S)$ and that $r\in\mathbb{Q}$ . Then, by the trichotomy law, we have $r^{2}<2$ , $r^{2}>2$ , or $r^{2}=2$ .
If $r^{2}<2$ , let $\varepsilon>0$ , then

$\displaystyle(r+\varepsilon)^{2}$	$\displaystyle=r^{2}+\varepsilon(2r+\varepsilon)$	(3.1)
	$\displaystyle<r^{2}+\varepsilon(2r+1)$	( $\varepsilon<1$ )
	$\displaystyle<r^{2}+2-r^{2}$	( $\varepsilon<\frac{2-r^{2}}{2r+1}$ )
	$\displaystyle=2.$	(3.2)

So there exists a rational number greater than $r$ which is an element of $S$ , which means that $r$ is not an upper bound for $S$ , contradicting our assumption that $r=\sup(S)$ .
If $r^{2}<2$ , let $\varepsilon>0$ , then

$\displaystyle(r-\varepsilon)^{2}$	$\displaystyle=r^{2}-\varepsilon(2r+\varepsilon)$	(3.3)
	$\displaystyle>r^{2}-\varepsilon(2r+1)$	( $\varepsilon<1$ )
	$\displaystyle>r^{2}-r^{2}+2$	( $\varepsilon<\frac{r^{2}-2}{2r+1}$ )
	$\displaystyle=2.$	(3.4)

So there exists a rational number less than $r$ which is an upper bound for $S$ , which contradicts our assumption that $r=\sup(S)$ .

Thus $r^{2}=2$ , which we have already shown is impossible if $r\in\mathbb{Q}$ , meaning that no supremum exists in $\mathbb{Q}$ for $S$ . ∎