Pythagoras of Samos, who lived during the sixth century BC, made significant advances in number theory (the study of integers, or whole numbers). He considered himself a philosopher (indeed, he coined the very word philosopher), and he founded the Pythagorean Brotherhood, which numbered some six hundred mathematicians at its peak. He is most famous for Pythagoras's Theorem which, as every schoolboy knows, is that "the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides" or, in other words, if the shorter sides are labelled a and b, and the hypotenuse is labelled c, then

Some scholars suggest that this fact was already well-known to those for whom it was a matter of practical importance (such as farmers and architects), but its association with Pythagoras's name is so strong that mathematicians will often say "Pythagoras" when they actually mean Pythagoras's Theorem.

Pythagoras's Theorem is perhaps the most-proved theorem ever. It is certainly true that there are more ways of proving it than there are days in a year. Many of these proofs are Byzantine geometrical tours de force, which seem to delight in adding construction lines to untidy gatherings of disparate triangles, apparently at random ("draw a line through P, perpendicular to AC, and then by similar triangles we know that BD is parallel to GH, so draw another line from..." until your head spins). I'm sure you know the sort of thing I mean. But there is one proof that is so delightfully simple that I cannot resist showing it to you.

We start off by drawing two squares, one inside the other. We rotate the inner square until it is in contact with the outer square at all four corners. The distances from these points of contact to the corners of the outer square, we call a and b respectively. The inner square has sides of length c.

If that sounds terribly complicated, don't worry -- I've drawn you a nice picture:

Proving Pythagoras

Before we use this diagram to prove Pythagoras's Theorem, let's make some simple observations and some elementary deductions. Firstly, it's clear that we can measure the area, A, of the large square by multiplying its side (a + b) by itself. Now, (a + b)(a + b) = a(a + b) + b(a + b) = aa + ab + ba + bb and of course this can be expressed as A = a² + 2ab + b².

There's another way we can measure the area of the big square, though. After all, it is made up of four triangles (I've placed a big red dot in each of these, so that you can easily see what I mean) and one (smaller) square -- which I've marked with a blue dot. We know that the area of a triangle is equal to half its base multiplied by its height. Each red-dot triangle therefore has an area of ab/2. We have four of these triangles, though; and the area of all four triangles put together is of course 4ab/2, or 2ab. The area of the inner square is cc, or c², so we can add that into the mix to get the total area of the large square: A = 2ab + c².

We now have two different expressions for the area of the large square. Since two things that are equal to the same thing are equal to each other, we can now say that:

The expression 2ab appears on both sides of the equation, so we can drop it altogether, which leaves us with...

And there it is. We have just proved -- indeed, we have derived -- Pythagoras's Theorem.