Prove that √p + √q is irrational,

Let us suppose that $\sqrt{p}+\sqrt{q}$ is rational.

Again, let $\sqrt{p}+\sqrt{q}=a$, where $a$ is rational.

Therefore, $\sqrt{q}=a-\sqrt{p}$

On squaring both sides, we get

$q=a^{2}+p-2 a \sqrt{p}$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

Therefore, $\sqrt{p}=\frac{a^{2}+p-q}{2 a}$, which is a contradiction as the right hand side is rational

number while $\sqrt{p}$ is irrational, since $\rho$ and $q$ are prime numbers. Hence, $\sqrt{p}+\sqrt{q}$ is irrational.