Need help with proof for Dedekind cuts on $\mathbb{Q}^+$ I am working on a proof about Dedekind cuts on the positive rational numbers. I have been stuck for a while on the following point and would appreciate any help.
Given $x\in \mathbb{Q}^+$ such that $x^2<2$, how to find $y\in \mathbb{Q}^+$ such $y>x$ and $y^2<2$?

Here is an exercise (which is not too far from what @HanDeBruijn is doing) from Hardy's "A Course of Pure Mathematics", 3rd edition, page 12

The book is old and should be freely available online.
Basically


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*$\frac{m}{n}<\sqrt{2} \Rightarrow \sqrt{2} < \frac{m+2n}{m+n}$.
$$\frac{m}{n}<\sqrt{2}=\frac{2-\sqrt{2}}{\sqrt{2}-1}\Rightarrow \\ m(\sqrt{2}-1)<n(2-\sqrt{2})\Rightarrow \\ \sqrt{2}(m+n)<m+2n$$

*Similarly if $\frac{m}{n}>\sqrt{2} \Rightarrow \sqrt{2} > \frac{m+2n}{m+n}$.

*$\left|\frac{m+2n}{m+n} - \sqrt{2}\right|< \left|\frac{m}{n} - \sqrt{2}\right|$. $$\left|\frac{\frac{m+2n}{m+n} - \sqrt{2}}{\frac{m}{n} - \sqrt{2}}\right|=\left|\frac{\frac{m+2n}{m+n} - \sqrt{2}}{\frac{m^2}{n^2} - 2}\cdot \left(\frac{m}{n} + \sqrt{2}\right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left(\frac{m+2n}{m+n} - \sqrt{2}\right)\cdot \left(\frac{m}{n} + \sqrt{2}\right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left(  \frac{m^2+2mn}{nm+n^2} +\sqrt{2}\frac{m+2n}{m+n} -\sqrt{2}\frac{m}{n} -2 \right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left(  \frac{m^2+2mn}{nm+n^2} - 2 +\sqrt{2}\left(\frac{m+2n}{m+n} -\frac{m}{n}\right) \right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left(  \frac{m^2-2n^2}{nm+n^2} +\sqrt{2}\frac{2n^2-m^2}{nm+n^2} \right)\right|\\=\left|\frac{n^2}{nm+n^2}\cdot \left(  1-\sqrt{2} \right)\right|=\left|\frac{n^2}{nm+n^2}\right| \cdot \left| 1-\sqrt{2} \right|<1$$


And this process can be repeated, by taking $$m_1=m+2n \\n_1=m+n \\...\\m_{k+1}=m_k+2n_k \\n_{k+1}=m_k+n_k$$ considering alterations of course, jumping around $\sqrt{2}$ and becoming closer to $\sqrt{2}$ each step. 
This will lead to $\frac{m}{n} < \frac{m_2}{n_2} < \sqrt{2} < \frac{m_1}{n_1}$ or $\frac{m^2}{n^2} < \frac{m_2^2}{n_2^2} < 2< \frac{m_1^2}{n_1^2}$. So if $x=\frac{m}{n}$ then the first suitable $y$ is $y=\frac{m_2}{n_2}$.
