Question

Prove that for any prime positive integer p, `sqrt(p)` is an irrational number.

Answer 1

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

`sqrtq=a/b`

`p= (a/b)^2`

`⇒ p = a^2/b^2`

`⇒ pb^2=a^2`

`⇒ pb^2=a^2`

`⇒  p|a^2`

`⇒  p|a`

`⇒  a= pc` for some positive intger c

`⇒ b^2p=a^2`

`⇒ b^2 p = p^2c^2 (because a= pc)`

`⇒ p|b^2 ("since"  p|c^2p)`

`⇒ p|b`

`⇒ p|a and p|b`

This contradicts the fact that a and b are co-primes.

Hence `sqrtp` is irrational.