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

"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
$$m'=m_1+2n_1=3m+4n,$$
$$n'=m_1+n_1=2m+3n.$$


I sort of understand the solution but I'd like it if someone could explain to make sure. Thanks in advance.
 A: 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*}
A: I guess he did not want to include continued fractions or quadratic forms.  
Taking $m,n$ positive integers. The hypothesis tells us that $m^2 - 2 n^2 < 0.$ So, let  $m^2 - 2 n^2 = -k.$ 
The given answer is
$$  (m,n) \mapsto (3m+4n, 2m+3n)   $$
Note that we get 
$$ (3m+4n) + (2m+3n)\sqrt 2 > m + n \sqrt 2  $$
Next,
$$ (3m+4n)^2 - 2 (2m+3n)^2 =  m^2 - 2 n^2 = -k  $$
A: It's just mechanic and consequence.
Mechanics:
$\frac{(m+2n)^2}{(m+n)^2} >,=,< 2 \iff (m+2n)^2 >,=,< 2(m+n)^2 \iff m^2 + 4nm + 4n^2 >,=,< 2m^2 + 4mn + 2n^2 \iff 2n^2 >,=,< m^2 \iff \frac {m^2}{n^2} <,=, > 2$.
Consequence:
If $\frac {m^2}{n^2} < 2 \implies \frac {(m+2n)^2}{(m+n)^2} > 2 \implies \frac {([m+2n]+2[m+n])^2}{([m+2n]+[m+n])^2} = \frac {(3m+4n)^2}{(2m+3n)^2} < 2$
Mechanics.
If $m,n \in \mathbb N$ then if we compare $\frac mn$ to $\frac {3m+4n}{2m+3n}$ are result will be the same as comparing $m(2m+3n)=2m^2 + 3mn$ to $n(3m+4n) = 3mn +4n^2$ which is the same as comparing $2m^2$ to $4n^2$ which is the same as comparing $\frac {m^2}{n^2}$ to $\frac 42 = 2$.... which is ... less.
So if we are given $m,n\in \mathbb N$ and $\frac {m^2}{n^2} < 2$ then $\frac mn < \frac{3m+4n}{2m+3n}$ and $(\frac mn)^2 < (\frac {3m+4n}{2m+3n})^2 < 2$.
Consequence:
If $r\in \mathbb Q$ so that $r^2 < 2$ there exists a larger rational $q$ so that $r < q$ and $r^2 < q^2 < 2$.
MAJOR and VERY important CONSEQUENCE:

There is no largest rational number whose square is less than $2$

