The rational numbers are the results of exact divisions of integers.

They are represented as fractions or, more exactly, by equivalence classes of fractions.

Namely, each rational number has an infinity of representations as fractions, deduced from a "canonical" one, with mutually prime numbers. The canonical representation of a rational number has a numerator and a denominator that do not share any common divisor. The other representations are so that their numerators and denominators are the same multiples of the numerator and denominator of the canonical representation.

Any rational number has a canonical representation because, if we have a common divisor, we may simplify the fraction by it, and we may continue to look for common divisors and simplify until we get the smaller number between the numerator and the denominator, that is a finite process.

The rational numbers are "dense" on the straight line of numbers, that is, we can always find a rational number between 2 rational numbers, e.g. there average. Thus, we could think that rational numbers are "all the numbers", couldn't we?

But this is not true. And we have a very simple example of a useful non-rational number, the square root of 2, that is the diagonal of a rectangle triangle of side 1 (or the ratio if the side is not 1).

This is because of the Pythagorean Theorem, that was already known by the ancient Greeks. And the ancient Greeks had already found the proof of the fact that the square root of 2 is non rational.

This is because if it was rational, then 2 would be the ratio of the squares of 2 mutually prime integers *a* and *b*, that is its canonical representation. Then the square of *a* should be the double of the square of *b*, being thus even, so that *a* would be even: *a=2k*. Then the square of *a*, that is the double of the square of *b*, should be the quadruple of the square of *k*. Thus the square of *b* should be the double of the square of *k*, being thus even, and *b* should so be even.

We arrive to the fact that both *a* and *b* are even, which is contradictory with the fact that *a* and *b* are mutually prime. Thus the hypothesis that the square root of 2 is a rational number is false.

Up to you to proove the same way that the square root of 3 isn't a rational number!

The key to prove this is the fact that 2 (as well as 3) is a prime number and, more generally, not a perfect square. It doesn't work with 4, happily enough, because the square root of 4 is the integer 2, that is a rational number!