To prove √p is irrational, where p is a prime, we will need the following theorem: Theorem: If $$p$$ is prime then $$p|ab,$$ then $$p|a$$ or $$p|b.$$ Proof: If $$p|a,$$ we are done. So let us assume that $$pmid a.$$ Therefore, $$gcd(p,a)=1.$$ Hence,