Have you ever caught yourself (assuming you like maths) proving that numbers like \sqrt{2} or \sqrt{5} etc. are irrationals. Here is, more or less, a general statement with the proof.
For \forall n \in \mathbb{N} (natural numbers), either \sqrt{n} \in \mathbb{N} or \sqrt{n} \in \mathbb{I} (irrational numbers, i.e. real numbers minus rational numbers).
There are two possible cases:
1. If n = p^2, p\in \mathbb{N} ⇒ \sqrt{n}=p\in \mathbb{N}
2. If n \ne p^2 or there is no such a p \in \mathbb{N} that would satisfy n = p^2, then \sqrt{n} is irrational.
Let's suppose the contrary, i.e. there \exists p,q \in \mathbb{N} and \gcd(p,q) = 1 (greatest common divisor) so that
\sqrt{n}=\frac{p}{q}
This means that n·q^2=p^2. From \gcd(p,q) = 1 \Rightarrow \gcd(p^2,q^2) = 1.
From the Bézout’s theorem, there \exists z,t \in \mathbb{Z} (integers) so that
z\cdot p^2 + t\cdot q^2 = 1 \iff\\
z\cdot n\cdot q^2 + t\cdot q^2 = 1 \iff\\
q^2 \cdot (z\cdot 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.
