Constructing the proof that $\sup \{x:x\in\mathbb Q \wedge x<\sqrt 2 \}$ doesn't exist. I know there is a very well known proof that for any rational such that
$$x=\frac p q < \sqrt 2$$
there exists another rational $y=\dfrac mn$ such that
$$x=\frac p q < \frac m n <\sqrt 2$$
Happily I've forgotten most of the proof, which lets me try and build it myself. 
Could someone provide a hint to produce a proof? I guess I should start with $2p<3q$ to produce a number larger than $p/q$ but smaller than $\sqrt 2$. 
In parenthesis, would this prove that $\sqrt 2 \notin \mathbb Q$?
 A: Following my comment above, if $p,q>0$ and $\frac p q < \sqrt{2}$, then define $p'=3p+4q$ and $q'=2p+3q$.  Then you can show that $\frac{p}{q}<\frac{p'}{q'} < \sqrt{2}$.
This comes from noting that $0<3-2\sqrt{2}<1$, so if $p-q\sqrt{2}<0$ then $$p-q\sqrt{2}<(p-q\sqrt{2})(3-2\sqrt{2})=p'-q'\sqrt{2} <0$$.
But dividing by $-q$, and noting $q'>q$, we get $$\sqrt{2}-\frac{p}{q}>  \frac{1}{q}(q'\sqrt{2}-p') > \frac{1}{q'}(q'\sqrt{2}-p') = \sqrt{2}-\frac{p'}{q'}>0$$
As noted by Chaz, this doesn't prove that $\sqrt{2}$ is not rational.
More generally, if $n$ is not a perfect square, let $r=\lceil\sqrt{n}\rceil$.  Then $0<r - \sqrt{n}<1$, so if $p/q<\sqrt{n}$, set $p'=rp+nq$ and $q'=p+rq$.  Then $p/q<p'/q'<\sqrt{n}$.
A: Hint: Let $x > 0 \in Q$ such that $x^2 < 2$.
Let $q = 2 \dfrac{x + 1}{x + 2} = x - \dfrac{x^2 - 2}{x + 2}$
Clearly $q \in Q$.
Notice that $q^2 - 2 = 2 \dfrac{(x^2 - 2)}{(x + 2)^2}$
Proceed to show that $2 > q^2 > x^2 > 0$
A: This would most certainly not prove that $\sqrt2\notin\mathbb{Q}$.
In fact, for every $x = p/q<2$, exists $m/n;\,p/q<m/n<2$ and certainly $2\in\mathbb{Q}$.
The proof as I remember is to suppose that $\sqrt2$ is rational, take the smallest $m/n$ which describes $\sqrt2$ (that is, cannot be simplified) and show that both are even (edit), which contradicts them being the smallest.
