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

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
$n \ge 1$.

Answer

Suppose that $\sqrt{2} = a/b$, with $a,b$ positive integers. Meaning $a = b\sqrt{2}$. Consider $$A = \{ m \in \Bbb Z \mid m > 0 \text{ and }m\sqrt{2} \in \Bbb Z \}.$$

Well, $A \neq \varnothing$, because $b \in A$. By the well-ordering principle, $A$ has a least element, $s$. And $s,s\sqrt{2} \in \Bbb Z_{>0}$. Then consider the integer: $$r= s\sqrt{2}-s.$$
We have $r =s(\sqrt{2}-1) < s$, and $r > 0$. But $r\sqrt{2} = 2s-s\sqrt{2}$ is again an integer. Hence $r \in A$ and $r < s$, contradiction.

Attribution
Source : Link , Question Author : marty cohen , Answer Author : Ivo Terek

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