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`

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

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

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

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

`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