Prove that $\displaystyle \sqrt{5}$ is an irrational number.

Let \(\displaystyle \sqrt{5}\) be a rational number. \(\displaystyle \therefore \sqrt{5}=\frac{\mathbf{p}}{\mathbf{q}}\), where \(\displaystyle \mathrm{q} \neq 0\) and p & q are coprime. \(\displaystyle 5 \mathrm{q}^{2}=\mathrm{p}^{2} \Rightarrow \mathrm{p}^{2}\) is divisible by \(\displaystyle 5 \Rightarrow \mathrm{p}\) is divisible by $\displaystyle 5$ ----- (i) Let \(\displaystyle \mathrm{p}=5 \mathrm{a}\), where 'a' is some integer \(\displaystyle 25 \mathrm{a}^{2}=5 \mathrm{q}^{2} \Rightarrow \mathrm{q}^{2}=5 \mathrm{a}^{2} \Rightarrow \mathrm{q}^{2}\) is divisible by \(\displaystyle 5 \Rightarrow \mathrm{q}\) is divisible by $\displaystyle 5$ ----- (ii) (i) and (ii) leads to contradiction as 'p' and 'q' are coprime. \(\displaystyle \therefore \sqrt{5}\) is an irrational number.