Prove that if $\frac{m}{n}<\sqrt{2}$, then there is a $\frac{m}{n}<\frac{m'}{n'}<\sqrt{2}$.

Question

I found this question on "Calculus" by Michael Spivak and the "Answer Book for Calculus" is unclear to me.

Note:
$$\frac{m^2}{n^2}<2\iff \frac{(m+2n)^2}{(m+n)^2}>2$$
and
$$\frac{m^2}{n^2}>2\iff \frac{(m+2n)^2}{(m+n)^2}<2$$
Have already been proven.

Edit:
$$m,n\in\mathbb{N}$$

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"Calculus 3rd edition" by Michael Spivak — Chapter 2 question 16 (c):

Prove that if $\frac{m}{n}<\sqrt{2}$, then there is another rational number $\frac{m'}{n'}$ with $\frac{m}{n}<\frac{m'}{n'}<\sqrt{2}$.

"Answer Book for Calculus" states:

Let $m_1=m+2n$ and $n_1=m+n$, and then choose
$$m'=m_1+2n_1=3m+4n,$$
$$n'=m_1+n_1=2m+3n.$$

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I sort of understand the solution but I'd like it if someone could explain to make sure. Thanks in advance.

Answer 1

If we have $\frac{m}{n} \le \sqrt{2}$ then $\frac{m+2n}{m+n} \ge \sqrt{2}$ & ...

If we have $\frac{m+2n}{m+n} \ge \sqrt{2}$ then $\frac{3m+4n}{2m+3n} \le \sqrt{2}$

We just need to show that \begin{eqnarray*} \frac{m}{n} \le \frac{3m+4n}{2m+3n} \le \sqrt{2} \\ \end{eqnarray*}