## 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!