Between two Real Numbers exists Rational Number

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:

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

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