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.

$\hspace{1.75 cm}\sqrt{3}=\large\frac{p}{q}$

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

$\hspace{1.3 cm}\sqrt{3}\; q = p$                $(1)$

Squaring both sides, we get

$\hspace{0.6 cm} (\sqrt{3}\; q)^2 = p^2$

$(\sqrt{3})^2\; (q)^2 = p^2$

$\hspace{1.3 cm} 3\; q^2 = p^2$

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

$3$ is also a factor of $p$.

$\implies\hspace{0.1 cm} 3\; r = p$

From $(1)$, we get

$\hspace{1.3 cm}3\; r = \sqrt{3}\; q$               $(2)$

Squaring both sides, we get

$\hspace{0.7 cm}(3\; r)^2 = (\sqrt{3}\; q)^2$

$\hspace{0.8 cm}3^{2}\; r^{2} = 3\; q^{2}$

$\hspace{1 cm}3\; r^2 = q^{2}$

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.