Question

prove that root p+root q is an irrational number

Answer 1

Let us suppose that √p + √q is rational. 
Let √p + √q = a, where a is rational. 
=> √q = a – √p 
Squaring on both sides, we get 
q = a2 + p - 2a√p

=> √p = (a2 + p - q)/2a, which is a contradiction as the right hand side is rational number, while√p is irrational. 
Hence, √p + √q is irrational.

Answer 2

First, we'll assume that √p + √q is rational, where p and q are distinct primes √p + √q = x, where x is rational 
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides. 
(√p + √q)² = x² p + 2√(pq) + q = x² 2√(pq) = x² - p - q 
√(pq) = (x² - p - q) / 2 
Now x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational. 
But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. Original assumption must be wrong. 
So √p + √q is irrational, where p and q are distinct primes 

:) Hope this Helps !!!