Prove that √ 5 is an irrational number.

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