Between two Real Numbers exists Rational Number

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

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

Proof
Suppose that $a \ge 0$.

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:
 * $\exists n \in \N: n > \dfrac 1 {b - a}$

Let $M := \left\{{x \in \N: \dfrac x n > a}\right\}$.

By the Well-Ordering Principle, there exists $m \in \N$ such that $m$ is the smallest element of $M$.

That is:
 * $m > a n$

and, by definition of smallest element:
 * $m - 1 \le a n$

As $n > \dfrac 1 {b - a}$, from Ordering of Reciprocals, it follows that $\dfrac 1 n < b - a$

Thus:

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

That is:
 * $\exists r \in \Q: a < r < b$

such that $r = \dfrac m n$.

Now suppose $a < 0$.

If $b > 0$ then $0 = r$ is a rational number such that $a < r < b$.

Otherwise we have $a < b \le 0$.

Then $0 \le -b < -a$ and there exists $r \in \Q$ such that:
 * $-b < r < -a$

where $r$ can be found as above.

That is:
 * $a < -r < b$

All cases have been covered, and the result follows.