Question

Question 14
Prove that $\sqrt{p}+\sqrt{q}$ is irrational, where p and q are primes.

Answer 1

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
$q=a^2+p-2a\sqrt{p}$ $[\because (a-b{)}^2=a^2+b^2-2ab]$
Therefore, $\sqrt{p}=\frac{a^2+p-q}{2a}$.

This contradicts our assumption, as the right-hand side is a rational number whereas $\sqrt{p}$ is irrational.
Hence, $\sqrt{p}+\sqrt{q}$ is irrational.