Here is an elementary proof of the fact that \( \sqrt{2} \) is irrational. But what does that mean? An irrational number is, well, a number that is not rational! So let's begin with a definition of the set of rational numbers \( \mathbb{Q} = \{ \frac{p}{q} | p, q \in \mathbb{Z}, q \neq 0 \}\). As we saw earlier, this is to be read: \( \mathbb{Q} \) is the set of all fractions \( \frac{p}{q} \) such that both \( p \) and \( q \) are integers, \( q \) not being \( 0 \), and with \( p \) and \( q \) in lowest terms. That simply means that they have no prime factor in common. (Let us recall that we can always write every natural number uniquely as a product of a number of prime factors.) What this definition means, at least for us right now, is that we're not going to be looking at numbers like \( \frac{12}{10} \); we're going to assume that all numbers have been reduced to their lowest form, like \( \frac{6}{5} \).
Another minor note: by our definition, a number like \( \frac{1.5}{3} \) would not count as a rational number. This is not true, of course, because we know we can simply rewrite \( 1.5 \) as \( \frac{3}{2} \), and then our number becomes \( \frac{\frac{3}{2}}{3} \), which is the same as \( \frac{1}{2} \), which is clearly a rational number.
So when we say a number is irrational, we're saying it's crazy in some really fundamental way: there really is no way to write it as the ratio of integers. (If you're wondering why we write \( \mathbb{Q} \) for the rational numbers, it's from the word "quotient".)
The proof of \( \sqrt{2} \) being irrational is a proof by contradiction, that is to say that we shall suppose that it is in fact rational, and show that this necessarily leads to a logical impossibility.