## Proof of the Irrationality of √2

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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{p2}{q2}\ 2q2 &= p2 \end{aligned}

$$2q2$$ is even so $$p2$$ is even. Since only even numbers squared are even, this means $$p$$ is even. So $$\exists r \in \mathbb{Z}: p = 2r$$.

\begin{aligned} 2q2 &= (2r)2\ 2q2 &= 4r2\ q2 &= 2r2 \end{aligned}

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