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 = p^{2}, p ∈ N ⇒ √ n = p ∈ N

2. If n ≠ p^{2} or there is no such a p ∈ N that would satisfy n = p^{2}, 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·q^{2} = p^{2}. From (p,q) = 1 ⇒ (p^{2},q^{2}) = 1.

From the Bézout’s theorem, there ∃z,t ∈ Z (integers) so that

z·p^{2} + t·q^{2} = 1 ⇔ z·n·q^{2} + t·q^{2} = 1 ⇔

q^{2}·(z·n + t) = 1

which means q^{2} divides 1, but this is possible only if q = 1. As a result n = p^{2} - contradiction with our initial assumption.

This proves the statement.