Related to Real Analysis

As $r < s$, it follows that $r \neq s$ and so $s - r \neq 0$. Thus:
\[ \frac{1}{s - r} \in \mathbb{R} \]
By the Axiom of Archimedes:
\[ \exists n \in \mathbb{N}: n > \frac{1}{s - r} \]
Let $M := \{x \in \mathbb{Z}: x > \frac{r}{n}\}$. Since any set of integers that is bounded below has smallest element, there exists $m \in \mathbb{Z}$ such that $m$ is the smallest element of $M$. That is:
\[ m > rn  \Rightarrow  r< \frac{m}{n}      (1)\]
and, by definition of smallest element:
\[ m - 1 \leq rn \]

As $n > \frac{1}{s - r}$, it follows from Ordering of Reciprocals that:
\[ \frac{1}{n} < s - r \]
Thus:
\[ m - 1 \leq rn    \Rightarrow   m \leq rn+ 1 \]
\[ \frac{m}{n} \leq r+\frac{1}{n} < r + (s - r) = s.      (2)\]

Thus it follows from (1) and (2) that $r < \frac{m}{n} < s$. That is:
\[ \exists q \in \mathbb{Q}: r < q < s \]
such that $q = \frac{m}{n}$.