Prove that $\sqrt{p}+\sqrt{q}$ is irrational,where $p$ and $q$ are distinct prime numbers.

(N/A) Let us assume that $\sqrt{p}+\sqrt{q}$ is a rational number.
Let $\sqrt{p}+\sqrt{q} = a$,where $a$ is a non-zero rational number.
Then,$\sqrt{q} = a - \sqrt{p}$.
Squaring both sides,we get:
$(\sqrt{q})^2 = (a - \sqrt{p})^2$
$q = a^2 + p - 2a\sqrt{p}$.
Rearranging the terms to isolate $\sqrt{p}$:
$2a\sqrt{p} = a^2 + p - q$
$\sqrt{p} = \frac{a^2 + p - q}{2a}$.
Since $a$ is a rational number,the expression $\frac{a^2 + p - q}{2a}$ must also be a rational number.
However,we know that $\sqrt{p}$ is an irrational number because $p$ is a prime number.
This leads to a contradiction because an irrational number cannot be equal to a rational number.
Therefore,our initial assumption is false,and $\sqrt{p} + \sqrt{q}$ must be an irrational number.