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

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