Have you ever caught yourself (assuming you like maths) proving that numbers like √ 2 or √ 5 etc. are irrationals. Here is, more or less, a general statement with the proof.
For ∀n ∈ N (natural numbers), either √ n ∈ N or √ n ∈ I (irrational numbers, i.e. real numbers minus rational numbers).
There are two possible cases:
1. If n = p2, p ∈ N ⇒ √ n = p ∈ N
2. If n ≠ p2 or there is no such a p ∈ N that would satisfy n = p2, then √ n is irrational.
Let's suppose the contrary, i.e. there ∃p,q ∈ N and (p,q) = 1 (greatest common divisor) so that
This means that n·q2 = p2. From (p,q) = 1 ⇒ (p2,q2) = 1.
From the Bézout’s theorem, there ∃z,t ∈ Z (integers) so that
z·p2 + t·q2 = 1 ⇔ z·n·q2 + t·q2 = 1 ⇔
q2·(z·n + t) = 1
which means q2 divides 1, but this is possible only if q = 1. As a result n = p2 - contradiction with our initial assumption.
This proves the statement.