## The square root of 2 is irrational

Let $x$ be a positive number such that $x^2 = 2$. Assume for the sake of contradiction that $x$ is rational. Then there exist coprime non-negative integers $a,b$, with $b \ne 0$, such that $x = \frac{a}{b}$. Then we have
$2 = x^2 = \frac{a^2}{b^2}$,
so $a^2 = 2b^2$.
Since $2$ is prime and $2$ divides $a^2$, $2$ must divide $a$. Say $a = 2c$ for some integer $c$. Then
$4c^2 = a^2 = 2b^2$,
so $2c^2 = b^2$, and therefore $2$ divides $b$. But then $a$ and $b$ are not coprime, which is a contradiction.
Therefore $x$ is irrational.
QED