The Pythagoreans, a wacky bunch of integer-worshipping Greeks, were so against the notion of irrational numbers that they supposedly murdered the poor soul who leaked their existence. In honor of Hippasus, we’ll prove that $$\sqrt{2}$$ is irrational, ie $$\sqrt{2}$$ can’t be expressed $$\frac{p}{q}$$ for some $$p, q \in \mathbb{Z}$$. We’ll do a proof by contradiction.

Let’s assume $$\sqrt{2}$$ is rational. Then $$\exists a, b, \in \mathbb{Z}: \sqrt{2} = \frac{a}{b}$$. $$a$$ and $$b$$ have a greatest common divisor and by dividing each by the gcd, we obtain an equivalent fraction $$\frac{p}{q}$$ that’s in lowest terms, i.e. $$p, q \in \mathbb{Z}$$, $$q \neq 0$$, $$\gcd{p, q} = 1$$.

$$
\begin{aligned}
\sqrt{2} &= \frac{p}{q}\
2 &= \frac{p^{2}}{q^{2}}\
2q^{2} &= p^{2}
\end{aligned}
$$

$$2q^{2}$$ is even so $$p^{2}$$ is even. Since only even numbers squared are even, this means $$p$$
is even. So $$\exists r \in \mathbb{Z}: p = 2r$$.

$$
\begin{aligned}
2q^{2} &= (2r)^{2}\
2q^{2} &= 4r^{2}\
q^{2} &= 2r^{2}
\end{aligned}
$$

But this contradicts $$\gcd{p, q} = 1$$. So $$\sqrt{2}$$ is irrational.