How do I prove that an irrational number is indeed irrational?

[NOTE: An irreducible fraction (or fraction in lowest terms) is a fraction in which the numerator and denominator are integers that have no other common divisors than 1.]Given that p is a prime, positive integer and not a square number, we know that √p is irrational. Let's prove this:Proof: We shall use Proof by Contradiction;Let's suppose towards a contradiction that √p is in fact rational.This implies that there exists two non-negative integers, call them m and n, such that:√p = m / n , where n is not zero and m/n is an irreducible fraction.<=> p = m² / n²  (Squared both sides)<=> pn² = m²....(#)  (Multiplied through equation by n²)From eqn(#), it follows that, since p is prime, p | m² (i.e. p divides m²) which thus implies also that p | m. This means that there exists some natural number, call it k, such that:pk = m....(##).Now, sub eqn(##) into eqn(#), we thus obtain:pn² = p²k²<=> n² = pk²  (Divided through by p)This last implies that p | n², which further implies:p | n.BUT since p divides both m and n, this contradicts the fact that m and n were chosen to be irreducible, so our original assumption was incorrect.It thus follows that √p is in fact irrational, as required.