The standard way we prove that the square root of $2$ is irrational is the following:
Let us assume the square root of $2$ is rational and is equal to $\frac{a}{b}$ where $a$ and $b$ are co-prime.
∴$\sqrt{2} = \frac{a}{b} \Rightarrow 2=\frac{a^2}{b^2}\Rightarrow b^2 = \frac{a^2}{2}$
∴ $a^2$ is divisible by $2$
∴ $a$ is divisible by $2$
I want to stop here. How does the statement $a^2$ is divisible by $2$ imply $a$ is divisible by $2$? For example, $4$ is divisible by $4$, but that does not mean $2$ is divisible by $4$.
An even square must be even. To see this note that $(2k)^2=2(2k^2)$ which is even. An odd number squared is $(2k+1)^2 = 4k^2+4k +1 = 2(2k^2+2k) + 1$ which is odd. Since all integers are either even or odd this accounts for all squares which can be verified by considering their remainder when divided by $2$.