Calculus Spivak Chapter 2 problem 16(c) The question asks to prove that if $\frac mn \lt \sqrt{2}$, then there is another rational number $\frac {m'}{n'}$ with $\frac mn \lt \frac {m'}{n'} \lt \sqrt{2}$.
Intuitively, it's clear that such a number exists, but I don't understand the solution to this problem. It states: let $m_1 = m + 2n$ and $n_1 = m + n$, and choose $m' = m_1 + 2n_1 = 3m + 4n$, and $n' = m_1 + n_1 = 2m + 3n$.  
Apparently $\frac {(m + 2n)^2}{(m+n)^2} \gt 2$, but can someone explain why and how plugging in those equations for $m'$ and $n'$ ensures that $\frac {m'}{n'}$ lies between $ \frac mn$ and $\sqrt {2}$?

$m/n < m'/n'$ if and only if $0 < m'n -m n' = (3m+4n)n - (2m+3n) m = 2(2n^2-m^2)$. It is clear from $m/n < \sqrt{2}$ that $2n^2 - m^2 > 0$.
$m'/n' < \sqrt{2}$ if and only if $0 < 2 n'^2 -m'^2 = 2(2m+3n)^2 - (3m+4n)^2 = 2n^2 - m^2$. As before, $2n^2 - m^2 > 0$ follows from $m/n < \sqrt{2}$.
