The proof that frightened Pythagoras

I have been meaning to do this for a while …

Pythagoras was said to be shocked by the proof that \sqrt 2 is an irrational number (i.e., that it cannot be represented as a ratio of two whole numbers). The proof (by contradiction) is a simple one.

If \sqrt 2 is rational then \sqrt 2 = \frac{a}{b} and 2 = \frac{a^2}{b^2} where a and b are the smallest nominator and denominator possible, i.e, in the lowest terms.

Hence a^2 = 2b^2 and so a must be even (as two odd numbers multiplied always give an odd number – let l = 2x and m = 2y be even numbers then (l + 1)(m + 1) = lm + l + m + 1 ). Hence a = 2k .

But we also have 4k^2 = a^2 and thus 4k^2 = 2b^2 and b^2 = 2k^2 and so b is even.

So we have a and b sharing a common factor of 2, so they cannot be the nominator and denominator of the fraction in the lowest terms.

But if a and b are both even then they share a common factor of 2 and \frac{a}{b} = \frac{2k}{2p} = \frac{k}{p} : implying that a = k and a = 2k , an obvious contradiction: hence \sqrt 2 cannot be a rational.

Update: I have made the final step of this shorter and clearer.

Further update: I have been told (see comments below) I would have been better sticking with a clearer version of the original ending ie., we state that \frac{a}{b} are the lowest terms. Then, plainly as they have a factor in common (2) they cannot be the lowest terms and so we have a contradiction. Would be great if someone could explain why we cannot use the a = 2k = k contradiction.

And another update: Should have stuck with the original explanation – which I have now restored in a hopefully clearer way. The comments below are really interesting and from serious mathematicians, so please have a look!