Prove that $\displaystyle \sqrt{3}$ is an irrational number.

Let \(\displaystyle \sqrt{3}\) be a rational number.

\(\displaystyle \therefore \sqrt{3}=\frac{\mathbf{p}}{\mathbf{q}}\), where \(\displaystyle q \neq 0\) and \(\displaystyle p \& q\) are coprime.

\(\displaystyle 3 q^{2}=p^{2} \Rightarrow p^{2}\) is divisible by \(\displaystyle 3 \Rightarrow p\) is divisible by $\displaystyle 3$ ---- (i)

Let \(\displaystyle p=3 a\), where ' \(\displaystyle a\) ' is some integer

\(\displaystyle 9 a^{2}=3 q^{2} \Rightarrow q^{2}=3 a^{2} \Rightarrow q^{2}\) is divisible by $\displaystyle 3$

\(\displaystyle \Rightarrow q\) is divisible by $\displaystyle 3$ ----- (ii)

and (ii) leads to a contradiction as ' \(\displaystyle p\) ' and ' \(\displaystyle q\) ' are coprime.

\(\displaystyle \therefore \sqrt{3}\) is an irrational number.