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 $\sqrt {2}$?
 A: I actually have a slightly different answer to the above, which I think is closer to the book as it relies directly on parts (a) and (b).
If anyone spots anything wrong I'd appreciate if you comment below and point out any mistakes:
We have proven in part (a) that:
$\frac{m^2}{n^2} < 2 \implies \frac{(m + 2n)^2}{(m + n)^2}>2$
or equivalently since $m,n \in \mathbb{N}$ and therefore all terms are positive:
$\frac{m}{n} < \sqrt{2} \implies \frac{m + 2n}{m + n}>\sqrt{2}$
Similarly in part (b) we have proven that:
$\frac{m}{n} > \sqrt{2} \implies \frac{m + 2n}{m + n}<\sqrt{2}$
Therefore if we start with a ratio $\frac{m}{n} < \sqrt{2}$ we can use (a) to get a ratio $\frac{m + 2n}{m + n}>\sqrt{2}$ and then use (b) in sequence to get a ratio $\frac{3m + 4n}{2m + 3n}<\sqrt{2}$.
We just need to prove that $\frac{m}{n} < \frac{3m + 4n}{2m + 3n}$.
We have:
$
\begin{aligned}
\frac{m}{n}<\frac{3m+4n}{2m+3n} &\iff \\
m(2m+3n)<n(3m+4n) &\iff \\
2m^2 +3mn < 3mn + 4n^2 &\iff \\
2m^2 < 4n^2 & \iff \\
\frac{m^2}{n^2}  < 2 & \iff \\
\frac{m}{n}  < \sqrt{2} &
\end{aligned}
$
Please do let me know if this is a proper solution.
I also should note that the idea to apply (a) and (b) in sequence and thus "try" the ratio $\frac{3m+4n}{2m+3n}$ originated from the second item proven in (a) and (b). Specifically we had proven:
(a): $\quad \frac{m^2}{n^2} < 2 \implies \frac{(m + 2n)^2}{(m+n)^2} - 2 < 2 - \frac{m^2}{n^2}$
(b): $\quad \frac{m^2}{n^2} > 2 \implies \frac{(m + 2n)^2}{(m+n)^2} - 2 > 2 - \frac{m^2}{n^2}$
Notice that the expressions in the inequalities to the right of the "implies" sign can be considered as "measures of the distance". For example $\frac{(m + 2n)^2}{(m+n)^2} - 2$ is a measure of the distance of point $\frac{m+2n}{m+n}$ from the point $\sqrt{2}$. The same can be said of $2 - \frac{m^2}{n^2}$ which can be considered as a measure of the distance of point $\frac{m}{n}$ from point $\sqrt{2}$.
This means that starting from any ratio $r_1=\frac{m}{n}<\sqrt{2}$ and creating a series of ratios $r_2=\frac{m+2n}{m+n}$, $r_3=\frac{3m+4n}{2m+3n}$, ... you would be "hopping" from one side of $\sqrt{2}$ to the other (on the number line). Hopping to the right of it will take you a bit closer to it, then hopping to the left in succession will get you slightly further away. In part (c) I think we are essentially proving that the decrease when hopping to the right is more than the increase when hopping back to the left, and thus we are getting "ever closer" to the point $\sqrt{2}$ with every pair of successive "hops".
A: $m/n < m'/n'$ 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 $m/n < \sqrt{2}$.
A: We check: $\dfrac{m'}{n'}=\dfrac{3m+4n}{2m+3n}>\dfrac{m}{n}\iff3mn+4n^2 > 2m^2+3mn\iff 2 > \dfrac{m^2}{n^2} \iff \dfrac{m}{n} < \sqrt{2}$ which is true. The other inequality is done similarly.
A: Actually we have the following estimates:
\begin{gather*}
\frac{m}{n}<\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
provided $m,n$ are positive integers with $\frac{m}{n}<\sqrt{2}.$ 
Verification:  Since $0<\frac{m}{n}<\sqrt{2},$ we have $\frac{n}{m}>\frac{1}{\sqrt{2}},$ and so
\begin{gather*}
\frac{2(m+n)}{m+2n}=\frac{m}{n}\cdot\frac{2\left(1+\frac{n}{m}\right)}{2+\frac{m}{n}}>\frac{m}{n}\cdot\frac{2\left(1+\frac{1}{\sqrt{2}}\right)}{2+\sqrt{2}}=\frac{m}{n}.
\end{gather*}
In the other direction, we have
\begin{align*}
&\frac{2(m+n)}{m+2n}=\frac{2}{\frac{m+2n}{m+n}}=\frac{2}{1+\frac{n}{m+n}}\\
=&\frac{2}{1+\frac{1}{\frac{m}{n}+1}}<\frac{2}{1+\frac{1}{\sqrt{2}+1}}=\frac{2}{1+(\sqrt{2}-1)}=\sqrt{2}.
\end{align*}
Therefore, we have 
\begin{gather*}
\frac{m}{n}<\frac{m'}{n'}=\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
if $0<m/n<\sqrt{2}.$
