Prove that 3 is an irrational number.

Let us suppose that $\sqrt{3}$ is a rational number.
Then there are positive integers a and b such that $\sqrt{3}=\frac{a}{b}$, where a and b are co-prime, meaning their HCF is 1.
$3=\frac{a}{b}\Rightarrow 3=\frac{a^2}{b^2}\Rightarrow 3b^2=a^2\Rightarrow 3divides{a}^2[\because 3divides3b^2]\Rightarrow 3dividesa.....................\left(i\right)\Rightarrow a=3cforsomeintegerc\Rightarrow a^2=9c^2\Rightarrow 3b^2=9c^2[\because a^2=3b^2]\Rightarrow b^2=3c^2\Rightarrow 3divides{b}^2[\because 3divides3c^2]\Rightarrow 3dividesb..............................\left(ii\right)$
We can see that a and b share at least 3 as a common factor from (1) and (2).
The fact that a and b are co-prime, on the other hand, contradicts this and indicates that our hypothesis is incorrect.
Therefore, $\sqrt{3}$​ is an irrational number.