Prove that √3 is an irrational number.

Suppose that $\sqrt{3}$ is a rational number. Then, it can be written in $\frac{p}{q}$ form where integers $p$ and $q$ have no common prime factor.

$\sqrt{3}=\frac{p}{q}$

$\text{Where, HCF}(p,q)=1\text{ and }q\ne 0.$

$\sqrt{3}q=p$                $(1)$

Squaring both sides, we get

$(\sqrt{3}q{)}^2=p^2$

$(\sqrt{3}{)}^2(q{)}^2=p^2$

$3q^2=p^2$

So, $3$ is a factor of $p^2$.

$3$ is also a factor of $p$.

$⟹3r=p$

From $(1)$, we get

$3r=\sqrt{3}q$               $(2)$

Squaring both sides, we get

$(3r{)}^2=(\sqrt{3}q{)}^2$

$3^2{r}^2=3q^2$

$3r^2=q^2$

So, $3$ is a factor of $q^2$.

$3$ is also a factor of $q$.

It shows that $3$ is a common prime factor of both $p$ and $q$ which cannot be possible. So, our assumption is wrong. Hence, $\sqrt{3}$ is an irrational number.