TITLE: certain proofs of the irrationality of $\sqrt{2}$
QUESTION [6 upvotes]: I had the impression that there might be proofs of the irrationality of $\sqrt{2}$ that showed that
$$
\left|\frac a b - \sqrt{2} \right| \ge (\text{something possibly depending on $a$ or $b$}) >0
$$
where $a,b\in\mathbb{Z}$.  But the one I saw in Wikipedia's article on the square root of $2$ spoke of whether the multiplicity of $2$ as a factor of $a$ or $b$ is even or odd, and that makes it seem not all that different from the old-fashioned proof that we all learned in childhood (you know, in 500 BC when we were children).  Is there a short and simple proof of that kind to which this last criticism will not apply?

REPLY [6 votes]: You might like the geometric proof posted here:
http://blog.plover.com/math/sqrt-2-new.html
I think it can be converted into the form you presented.
Edit:
20 proofs of the irrationality of $\sqrt{2}$ can be found here, but none with the exact form you mentioned.

REPLY [6 votes]: You can use that if $a^2=2b^2$ then $(2b-a)^2=2(a-b)^2$ to show that if $\frac ab $ is a square root of 2 then so is $\frac {2b-a}{a-b}$. So there is not a fraction with smallest denominator.
Also if $\frac ab$ is an approximation to $\sqrt 2$, then $\frac {a+2b}{a+b}$ is in general a better one, which may be what you are remembering.
If $$2-\frac{a^2}{b^2}=\epsilon$$ then
$$\frac{(a+2b)^2}{(a+b)^2}-2 = \frac{2b^2-a^2}{(a+b)^2}=\epsilon \left(\frac b{a+b}\right)^2$$