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*