Considered to be another good introduction into how proofs are structured and to what number theory concerns, here is the elementary proof that the square root of 2 is irrational.

Pf. First suppose that √2 is rational. This means that √2 can be represented in a ratio of two integers, a very neat clean way of expressing the number. Let’s call those two integers a and b for conenvience’s sake and let’s note that the ratio of a to b is already in simplified form so that neither a or b and can be reduced by any factor. Now we have √2 = a / b since we can express ratios as one number divided by the other. If we square both sides we have 2 = a^2 / b^2 and if we multiply both sides by b^2 then we have the cleaner equation :

 2 * b^2 = a^2

Now consider the possibilities regarding this expression. Can the integer a ever be odd? No, because an odd number multiplied by another odd number (itself in this case) must be odd and the equation states that the number a multiplied by itself must be even (because it is equal to 2 times something either even or odd, no matter what something even). So we conclude that the integer a must be even. Now let’s see whether the integer b is even or odd. If we multiple both sides of the equation by a half and look at the equation from the vantage point of b^2 = a^2 / 2, then we have a nicer way to consider the possible forms of b.

We know that a is even so the number a^2 divided by 2 will be either even or odd but will always be an integer. In fact if we consider things a little closer we come to the conclusion that b must be even since the expression a^2 / 2 cannot be the square of any odd number, only an even. This is because a^2 with a being even implies that a^2 is also divisible by 4. So if b^2 is equal to a number that is divisible by 4 divided by 2, b^2 must be even and therefore b. Thus, we conclude that both a and b are evens.

However, this contradicts our initial statement that the ratio must be in lowest common terms; a and be cannot both be divisible 2 (in the way that even numbers are all divisible by 2) if they are in lowest common terms. Therefore we conclude that our initial statement was false. There is no way to state the number √2 as a ratio of two integers in lowest terms and so √2 must be irrational.

Q.E.D.

Supposedly, the Greek school of Pythagoras was so in love with ratios and rational numbers that when presented with a proof that there existed some numbers that cannot be expressed in a ratio, they called heresy and threw the man who proved that √2 was irrational overboard of a ship. Although that version of the proof depended on triangles and their sides, it was nonetheless true enough to upset the foundations for Greek mathematics at the time.

Such a revelationary proof opened up a new perspective for how numbers are to be treated and how they appear naturally in the universe.