Proving that if $\frac{m}{n}<\sqrt{2}$ then there exists $\frac{m'}{n'}$ such that $\frac{m}{n}< \frac{m'}{n'}<\sqrt{2}$ This is from Spivak. The problem is as follows:
Prove that if $\frac{m}{n}<\sqrt{2}$ then there exists $\frac{m'}{n'}$ such that $\frac{m}{n}< \frac{m'}{n'}<\sqrt{2}$
So far, I have that since $\frac{m^2}{n^2}<2$ it follows that:
$$\frac{(m+2n)^2}{(m+n)^2}-2<2-\frac{m^2}{n^2}<2$$
and: $$\frac{2n^2-m^2}{(m+n)^2}<\frac{2n^2-m^2}{n^2}<2$$
I suppose at this point, I would need to prove that both the terms $\sqrt{\frac{2n^2-m^2}{(m+n)^2}}$ and $\sqrt{\frac{2n^2-m^2}{n^2}}$ are rational. This is about where I'm stuck... Any hints would be much appreciated.
Edit:
I should probably add that at this point, I'm mostly interested in a solution that uses the techniques given in Spivak so far. Basically the only things that have been covered are the properties of $\mathbb{R}$ and induction...
 A: I am wondering why the following steps cannot be considered as a legitimate proof:


*

*Let $K$ be the number of digits in the representation of $\frac{m}{n}$ on base $n$.

*Calculate the first $K+1$ digits in the representation of $\sqrt{2}$ on base $n$.

*Use the result above in order to calculate the values of $m'$ and $n'$.
Note: base $n$ is in order to ensure a finite number of digits.
A: Here's a simple proof. Since $m/n<\sqrt{2}$, there is an $\varepsilon>0$ such that
$m/n+\varepsilon<\sqrt{2}$. But $\lim_k\frac{1}{k}=0$, therefore there is some $N \in \mathbb{N}$ such that $1/k \le \varepsilon$ for every $k \ge N$. Hence, for every $k \ge N$ we have
$$
\frac{m}{n}<\frac{km+n}{kn}=\frac{m}{n}+\frac{1}{k}\le \frac{m}{n}+\varepsilon<\sqrt{2}.
$$
Added to the proof: 
If you don't want to use limits, you may proceed as follows. 
Choose a sufficiently small number $\varepsilon>0$ such that
$m/n+\varepsilon<\sqrt{2}$. This is possible because $\sqrt{2}-m/n>0$, in fact you may choose e.g. $\varepsilon=\frac{\sqrt{2}-m/n}{10^{100}}$.
Setting $N_\varepsilon=\lfloor \varepsilon^{-1}\rfloor+1 \in \mathbb{N}$, we have
$$
N_\varepsilon-1 \le \varepsilon^{-1}<N_\varepsilon.
$$
Then for every $k \ge N_\varepsilon$ we have
$$
\frac{m}{n}+\frac{1}{k}\le \frac{m}{n}+\frac{1}{N_\varepsilon}<\frac{m}{n}+\varepsilon<\sqrt{2}.
$$
Hence
$$
\frac{m}{n}<\frac{km+n}{kn}=\frac{m}{n}+\frac{1}{k}<\sqrt{2} \quad \forall k \ge N_\varepsilon.
$$
A: The easiest way to do this is probably just to construct $m'$ and $n'$ and show that they satisfy the requirement. Assume wlog $m > n > 0$. (If $m, n \leq 0$ then try again with $-m$ and $-n$. If $m \leq n$ then $4/3$ is the answer.) Since $m/n < \sqrt{2}$, we have $m^2 < 2n^2$.
Now consider the number $x = \frac{m^2 + 2n^2}{2mn} = \frac{1}{2}(\frac{m}{n} + \frac{2n}{m}) = \frac{m}{2n} + \frac{n}{m}$. (I got this by using Newton's method to approximate $\sqrt{2}$, starting with $m/n$.) Because $2n^2 > m^2$, we have $$x = \frac{m^2 + 2n^2}{2mn} < \frac{2n^2 + 2n^2}{2mn} = \frac{2n}{m}.$$ However, we also have $$x^2 - 2 = \left(\frac{m}{2n} + \frac{n}{m}\right)^2 - 2 = \frac{m^2}{4n^2} + 1 + \frac{n^2}{m^2} - 2 = \frac{m^2}{4n^2} - 1 + \frac{n^2}{m^2} = \left(\frac{m}{2n} - \frac{n}{m}\right)^2 > 0.$$ Therefore $x^2 > 2$ and $x > \sqrt{2}$. 
In conclusion, $\sqrt{2} < x < \frac{2n}{m}$, so $2/x = \frac{4mn}{m^2 + 2n^2}$ satisfies the requirements.
