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

Let $\displaystyle \sqrt{5}$ be a rational number.

$\displaystyle \therefore \sqrt{5}=\frac{\mathrm{p}}{\mathrm{q}}$, 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)

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

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