Prove that √5 is an irrational number.

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