The proof of the day is: $\sqrt{2}$ is irrational
Theorem: $\sqrt{2} \not\in \mathbb{Q}$
Proof 1:
Define the p-adic valuation of n as $\nu_{p}(n) = max\{k \in \mathbb{N} : p^k \mid n \}$, where
p is prime.
Suppose, by contradiction that $\sqrt{2}=\frac{p}{q},\,p,q\in\mathbb{Z}$ and $q\neq0$.
Then $2=\frac{p^2}{q^2} \iff 2q^2 = p^2$, so we have that $\nu_{2}(p^2)=2\nu_{2}(p)$,
but $\nu_{2}(2q^2) = 2\nu_{2}(q)+1$.
By the fundamental theorem of arithmetic, the p-adic valuation of integers is unique. Contradiction.
Proof 2:
Suppose $\sqrt{2}=\frac{p}{q},\,p,q\in\mathbb{Z}$ and $q>0$.
Let $A=\{m\in\mathbb{Z}^{+} : m\sqrt{2}\in\mathbb{Z}\}$. Then $p=q\sqrt{2} \in\mathbb{Z}^{+}
\implies q\in A$.
Since $A\subset\mathbb{Z}^{+}$ and $A\neq\emptyset$, then $\exists s = min\{A\}$ (well-ordering).
Let $r=s\sqrt{2}-s=s(\sqrt{2}-1)$. Then $r\sqrt{2}=2s-s\sqrt{2}$, and since $2s\in\mathbb{Z}^{+}$ and
$s\sqrt{2}\in\mathbb{Z}^{+} \implies r\in\mathbb{Z}^{+} \implies r\in A.$
But $r=s(\sqrt{2}-1)<s$, contradiction, since $s=min\{A\}$.