If p, q are prime positive integers,

Let us assume that $\sqrt{p}+\sqrt{q}$ is rational. Then, there exist positive co primes $a$ and $b$ such that

$\sqrt{p}+\sqrt{q}=\frac{a}{b}$

$\sqrt{p}=\frac{a}{b}-\sqrt{q}$

$(\sqrt{p})^{2}=\left(\frac{a}{b}-\sqrt{q}\right)^{2}$

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

$p-q=\left(\frac{a}{b}\right)^{2}-\frac{2 a \sqrt{q}}{b}$

$p-q=\left(\frac{a}{b}\right)^{2}-\frac{2 a \sqrt{q}}{b}$

$\left(\frac{a}{b}\right)^{2}-(p-q)=\frac{2 a \sqrt{q}}{b}$

$\frac{a^{2}-b^{2}(p-q)}{b^{2}}=\frac{2 a \sqrt{q}}{b}$

$\left(\frac{a^{2}-b^{2}(p-q)}{b^{2}}\right)\left(\frac{b}{2 a}\right)=\sqrt{q}$

$\sqrt{q}=\left(\frac{a^{2}-b^{2}(p-q)}{2 a b}\right)$

Here we see that $\sqrt{q}$ is a rational number which is a contradiction as we know that $\sqrt{q}$ is an irrational number

Hence $\sqrt{p}+\sqrt{q}$ is irrational