Assume $\sqrt 2$ is rational.
$$\sqrt 2= \frac{P}{Q}$$
$$2 = \biggr(\frac{P}{Q}\biggr)^2$$
$$2 = \frac{P^2}{Q^2}$$
$P^2 = 2Q^2$ which is impossible because there doesn't exist a number whose square, nor its double, is equal to any other squares. The conclusion produces a contradiction.
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Assume on the contrary that $\sqrt{2}$ is a rational number.
$\Rightarrow \exists $ integers $a$ and $b$ such that $\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are relatively prime.
$\Rightarrow a=\sqrt{2}b\Rightarrow a^2=2b^2$
Clearly, $2|2b^2$ $\Rightarrow$ $2$ must divide $a^2$.
$2$ is a prime number that divides $a^2$, so it must divide $a$.
Therefore, $a=2m$, where $m\in \mathbb{N}$.
Recall that $a^2=2b^2$, replacing $a$ with $2m$,
we get $(2m)^2=2b^2$ $\Rightarrow$ $4m^2=2b^2$$\Rightarrow$$2m^2=b^2$.
Clearly, $2|2m^2$ $\Rightarrow$ $2$ must divide $b^2$.
$2$ is a prime number that divides $b^2$, so it must divide $b$.
Therefore, $b=2n$, where $n\in \mathbb{N}$.
Now we have two results; $a=2m$ and $b=2n$, which means $a$ and $b$ have $2$ as a common factor.
This contradicts the original hypothesis that $a$ and $b$ are relatively prime.
$\Rightarrow$ The original assumption that $\sqrt{2}$ is a rational number must be wrong.
$\Rightarrow$ $\sqrt{2}$ is an irrational number.
This is an incomplete proof at best, since you only claim, but don't prove, that there is no (natural) number $n$ such that $2n^2$ is the square of another (natural) number.