Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$ Let $m$ and $n$ be natural numbers. Prove that if $m/n < \sqrt{2}$, then there is another rational number $m'/n'$ with $m/n < m'/n' < \sqrt{2}$
I interpreted this statement as $$\frac{m}{n} < \sqrt{2} \rightarrow \exists \frac{m'}{n'} \in \mathbb{Q} \left( \frac{m}{n} < \frac{m'}{n'} \wedge \frac{m'}{n'} < \sqrt{2} \right)$$ The contrapositive would then be $$\forall \frac{m'}{n'} \in \mathbb{Q} \left(\frac{m'}{n'} < \sqrt{2} \rightarrow \frac{m}{n} \geq \frac{m'}{n'}\right) \rightarrow m/n \geq \sqrt{2}$$
Partial Proof: Suppose that $\forall \frac{m'}{n'} \in \mathbb{Q} \left(\frac{m'}{n'} < \sqrt{2} \rightarrow \frac{m}{n} \geq \frac{m'}{n'}\right)$. Now suppose that $\frac{m}{n} < \sqrt{2}$. Since $\frac{m}{n} \in \mathbb{Q}$ and  $\frac{m}{n} < \sqrt{2}$, we have $\frac{m}{n} \leq \frac{m}{n}$...
I'm just wondering why I was not able to draw a contradiction at the end. Is there something wrong with how I interpreted the statement?

Here is a somewhat different approach: assume that $\dfrac mn < \sqrt 2$.  Then
\begin{align*}
m^2 &< 2n^2\\
2m^2 &< 4n^2 \\
2m^2 + 3mn &< 3mn + 4n^2 \\
m(2m + 3n) &< n (3m + 4n) \\
\frac mn &< \frac{3m + 4n}{2m + 3n}
\end{align*}
and
\begin{align*}
m^2 &< 2n^2\\
9 m^2 - 8m^2 &< 18 n^2 - 16n^2 \\
9m^2 + 16n^2 &< 8m^2 + 18n^2 \\
9m^2 + 24 mn + 16 n^2 &< 8m^2 + 24mn + 18n^2 \\
(3m + 4n)^2 &< 2 (2m + 3n)^2 \\
\frac{(3m+4n)^2}{(2m+3n)^2} &< 2 \\
\frac{3m+4n}{2m+3n} &< \sqrt{2}
\end{align*}
so that
$$ \frac mn < \frac{3m+4n}{2m+3n} < \sqrt 2.$$
