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

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

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

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

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

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

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

Thus:

$$ $$ $$ $$ $$

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