Question

If p, q are prime positive integers, prove that `sqrt(p) + sqrt(q)` is an irrational number.

Answer 1

Let us assume that  `sqrtp+sqrtq` is rational. Then, there exist positive co-primes a and b such that 

`sqrtp +sqrtq=a/b`

`sqrtp=a/b-sqrtq`

`(sqrtp)^2= (a/b-sqrtq)^2`

`p= (a/b)^2-(2asqrtq)/b+q`

`p-q=(a/b)^2-(2asqrtq)/b`

`(a/b)^2-(p-q)= (2asqrtq)/b`

`(a^2-b^2(p-q))/b^2 = (2asqrtq)/b`

`((a^2-b^2(p-q))/b^2)(b/(2a))=sqrtq`

`sqrtq=((a^2-b^2(p-q))/(2ab))`

Here we see that `sqrtq` is a rational number which is a contradiction as we know that `sqrtq` is an irrational number.

Hence `sqrtp+sqrtq` is irrational.