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 $y^2<2$?
 A: Note that if $0<\epsilon<1$, then $$(x+\epsilon)^2=x^2+2x\epsilon+\epsilon^2<x^2+(2x+1)\epsilon.$$  If you want $(x+\epsilon)^2$ to be less than $2$, you can choose $\epsilon$ such that $x^2+(2x+1)\epsilon\leq 2$, which is equivalent to $\epsilon\leq\frac{2-x^2}{2x+1}$.
So to find your $y$, just pick $\epsilon>0$ such that $\epsilon<\min\left(\frac{2-x^2}{2x+1},1\right)$ and take $y=x+\epsilon$.
A: The Stern-Brocot tree is one of my favorite tools. It is assumed to be well-known in what follows. Let $m$ and $n$ be positive integers. Initialize (zeroth iteration):
$$
x = \frac{m}{n} < \sqrt{2} \quad ; \quad y = 2/x = \frac{2n}{m} > \frac{2}{\sqrt{2}} \quad \Longrightarrow \quad y > \sqrt{2}
$$
Now walk through the Stern-Brocot tree iteratively until $\,y < \sqrt{2}$ . From the properties of the tree, we know that always will be $\,y > x$ .
First iteration:
$$
y := \frac{m+2n}{n+m} = \frac{x+2}{x+1} < \sqrt{2} \quad \mbox{?}
$$
No, because a contradiction with $\,x<\sqrt{2}\,$ is derived:
$$
(x+2)^2 < 2(x+1)^2 \quad \Longleftrightarrow \quad x^2+4x+4 < 2x^2+4x+2 \quad \Longleftrightarrow \quad x^2 > 2
$$
Second iteration:
$$
y := \frac{m+(m+2n)}{n+(n+m)} = \frac{2m+2n}{m+2n} = \frac{2x+2}{x+2} < \sqrt{2} \quad \mbox{?}
$$
Yes, because:
$$
(2x+2)^2 < 2(x+2)^2 \quad \Longleftrightarrow \quad 4x^2+8x+4 < 2x^2+8x+8 \quad \Longleftrightarrow \quad x^2 < 2
$$
So the outcome is:
$$
y(x) = 2\frac{x+1}{x+2}
$$
Check, check, double check .. If we are allowed to have an embedding of the rationals in the reals, then the derivative is:
$$
y'(x) = \frac{2(x+2)-(2x+2)}{(x+2)^2} = \frac{2}{(x+2)^2} > 0
$$
And some function values are:
$$
y(0) = 1 \quad ; \quad y(\sqrt{2}) = 2\frac{\sqrt{2}+1}{2+\sqrt{2}} = \sqrt{2}
$$
But $y(x)$ is monotonically increasing, so $\sqrt{2}$ is the maximum at the interval $\left[0,\sqrt{2}\right]$ and all other values of $y(x)$ are smaller than this, but greater than $x$, as requested.
BONUS. Let $N$ be any positive integer. Now generalize the question as follows.
Given $x\in \mathbb{Q}^+$ such that $x^2 < N$, how to find $y\in \mathbb{Q}^+$ such that $y>x$ and $y^2 < N$ ?
It's left as an exercise for the reader to prove that this is a solution:
$$
y(x) = N\frac{x+1}{x+N}
$$
Can't resist to display some members of the $y$ - family in a $[0,3]\times[0,3]$ picture:

A: 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


*

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

*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 $y=\frac{m_2}{n_2}$.
