Question
Prove that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is irrational, where p, q are primes.

Answer

Let us suppose that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is rational,
again, let $\sqrt{\text{p}}+\sqrt{\text{q}}=\text{a},$ where a is rational.
$\therefore \sqrt{\text{q}}=\text{a}-\sqrt{\text{p}}$
On squaring both sides, we get
$\text{q}=\text{a}^2+\text{p}-2\text{a}\sqrt{\text{p}}\ \big[\because(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab}\big]$
$\therefore\ \sqrt{\text{p}}=\frac{\text{a}^2+\text{p}-\text{q}}{2\text{a}},$ which is a contradiction as the right hand side is rational number while $\sqrt{\text{p}}$ is irrational, since p and q are prime number.
Hence, $\sqrt{\text{p}}+\sqrt{\text{q}}$ is irrational.

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