Between Every Two Reals Exists a Rational

Theorem

Let $a, b \in \R$ be real numbers such that $a < b$.

Then $\exists r \in \Q: a < r < b$.

Corollary

$\exists r \in \R: a < r < b$.

Proof

As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.

Thus $\dfrac 1 {b - a} \in \R$.

By the Archimedean Principle, $\exists n \in \N: n > \dfrac 1 {b - a}$.

Now let $m \in \N$ be the smallest such that $m > a n$.

It follows that $a < \dfrac m n$.

It also follows that $m - 1 \le a n$.

As $n > \dfrac 1 {b - a}$ it follows that $\dfrac 1 n < b - a$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle m - 1\)	\(\le\)	\(\displaystyle a n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle m\)	\(\le\)	\(\displaystyle a n + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac m n\)	\(\le\)	\(\displaystyle a + \frac 1 n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle a + \left({b - a}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus we have shown that $a < \dfrac m n < b$.

$\blacksquare$

Corollary

Follows trivially from the main result, as $\Q$ is a subset of $\R$.

$\blacksquare$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Corollary $1.1.7$