What is the most unusual proof you know that $\sqrt{2}$ is irrational?

Here is my favorite:

> 
> Theorem: $\sqrt{2}$ is irrational.
> 
> 
> Proof:
>  $3^2-2\cdot 2^2 = 1$.
> 
> 
> 

(That's it)

That is a corollary of
this result:

> 
> Theorem:
>  If $n$ is a positive integer
>  and there are positive integers
>  $x$ and $y$ such that
>  $x^2-ny^2 = 1$,
>  then
>  $\sqrt{n}$ is irrational.
> 
> 
> 

The proof is in two parts,
each of which
has a one line proof.

### Part 1:

> 
> Lemma: If 
>  $x^2-ny^2 = 1$,
>  then there are arbitrarily large integers
>  $u$ and $v$ such that
>  $u^2-nv^2 = 1$.
> 
> 
> 

Proof of part 1:

> 
> Apply the identity
>  $(x^2+ny^2)^2-n(2xy)^2
> =(x^2-ny^2)^2
> $
>  as many times as needed.
> 
> 
> 

### Part 2:

> 
> Lemma: If
>  $x^2-ny^2 = 1$
>  and
>  $\sqrt{n} = \frac{a}{b}$
>  then
>  $x < b$.
> 
> 
> 

Proof of part 2:

> 
> $1
> = x^2-ny^2
> = x^2-\frac{a^2}{b^2}y^2
> = \frac{x^2b^2-y^2a^2}{b^2}
> $
>  or
>  $b^2
> = x^2b^2-y^2a^2
> = (xb-ya)(xb+ya)
> \ge xb+ya
> > xb
> $
>  so
>  $x < b$.
> 
> 
> 

These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.

Two things to note about
this proof.

First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.

Second,
the key property of
positive integers needed
 is that
if $n > 0$
then
$n \ge 1$.