# Maths

Proof that $\sqrt{2}$ is irrational

Let $x$ be $\sqrt{2}$. Then

${x}^{2}+=2$

If $x$ were a rational number it would be expressible in the form

$x=\frac{m}{n}$

Where $m$ and $n$ are rational numbers with no common divisor (other than 1). It follows that

${m}^{2}=2{n}^{2}$

So ${m}^{2}$ is even. This implies that $m$ is even. So we can write

$m=2k$

Hence

$4{k}^{2}=2{n}^{2}$

So

$2{k}^{2}={n}^{2}$

Thus $n$ is even. We have therefore shown that $m$ and $n$ are divisible by 2. This is a contradiction. It therefore follows that $x$ cannot be rational, i.e. $\sqrt{2}$ must be an irrational real number.