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

$$m^2=2n^2$$

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

$$m=2k$$

Hence

$$4k^2=2n^2$$

So

$$2k^2=n^2$$

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.