Between two Real Numbers exists Rational Number/Proof 2

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

Let $M := \set {x \in \Z: x > a n}$.

By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ 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 $n > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals 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 $r = \dfrac m n$.