What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Solution 1:
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.
Solution 2:
Here is something that I just came up with.
If $\sqrt2$ were rational, it would have been in every field of characteristics $0$.
It is also well-known that there are infinitely many prime numbers $p$, such that $x^2-2$ has no root over $\Bbb F_p$. Let $P$ be the set of these primes, and let $U$ be a free ultrafilter over $P$. Now consider $F=\prod_{p\in P}\Bbb F_p/U$.
Using Los theorem we have that:
- $F$ is a field.
- $F$ has characteristics $0$.
- $\lnot\exists x(x^2-2=0)$.
This means exactly that we found a field which extends the rational numbers, but has no roots for $x^2-2$, which in other words means $\sqrt2\notin F$ and therefore $\sqrt2\notin\Bbb Q$.
Solution 3:
Consider the linear application $A:\mathbb{R}^2\to \mathbb{R}^2$ given by $$A=\begin{pmatrix} -1&2 \\ 1&-1 \end{pmatrix} .$$ $A$ maps $\mathbb{Z}^2$ into itself and $V=\{y=\sqrt 2 x\}$ is an eigenspace relative to the eigenvalue $\sqrt 2-1$. But $A\mid_V$ is a contraction mapping, so $\mathbb{Z}^2\cap V=\emptyset$.