Prove that √ 5 is an irrational number.

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