Projects {Notes} Now

He is unworthy of the name of man who does not know that the diagonal of a square is incommensurable with its sides

– Plato (maybe)

Apparently, Plato had a strong opinion on the merits of knowing that \sqrt{2} is irrational. I guess just memorizing the fact wouldn’t be enough. Let’s see if I can construct a proof to convince myself that \sqrt{2} is indeed irrational.

Strategy

I’ve read a proof online and this is a reconstruction in my own words. The high-level strategy is to first assume that \sqrt{2} is rational and then carefully examine what follows from that assumption. If one keeps in mind certain “facts” about squares of even numbers and ratios of whole numbers, the assumption \sqrt{2} is rational leads to an absurd conclusion.

Proof

Suppose \sqrt{2} is rational. Then it can be written as a ratio of two integers:

\sqrt{2} = \frac{p}{q}

where p, q \in \mathbb{Z} and q \not= 0, for we can’t divide by 0. Let’s see what happens when we square both sides:

2 = \frac{p^2}{q^2}

and then multiply them by q^2:

2q^2 = p^2

We can reason that p^2 is even, because any number multiplied by 2 is even. Moreover, p must also be even, for a square root of a whole even number must be even . Since p is even, we can write it as 2 times some whole number, e.g.:

It is not obvious that a square root of an even number must be even, but how can it not be? Suppose i=j^2 and j^2 is even. Could i be odd? If it was, we could write j^2 = (2z + 1)^2. Expanding the right hand side, we get j^2 = 4z^2 + 4z + 1, i.e., j^2 = 2(2z^2 + 2z) + 1. This implies that j^2 is odd, for it is written as some number times 2 plus 1. But we started from assuming j^2 is even! Thus, i cannot be an odd number.

p = 2m

Substituting this into 2q^2 = p^2 we get

2q^2 = 4m^2 \implies q^2 = 2m^2

I.e., q^2 is an even number. And again, since the square root of an even number is also even, we can conclude that q is even and can be written as 2n.

Now, let’s recall our original assumption that \sqrt{2} = \frac{p}{q}. We can substitute p and q with their counterparts and write:

\sqrt{2} = \frac{m}{n}

where both m and n are some whole numbers. Note that \frac{m}{n} is \frac{p}{q} simplified by factoring out 2. That is, \frac{m}{n} is a simpler fraction than \frac{p}{q}.

Nothing prevents us from applying the same steps to m and n as we did to p and q. We could do it forever, as each time we get a simpler ratio of two whole numbers. And here is the contradiction, because any ratio has a single simplest form – we can’t simplify forever!

Epilogue

Apparently that’s the gist of Euclid’s proof that \sqrt{2} is irrational. But isn’t something missing? Why can’t we simplify forever? Of course, I’ve experienced this first hand, when you simplify once or twice and then arrive at a ratio that you cannot simplify more. But I haven’t tried to simplify all possible ratios, so I don’t know for sure. The fact that we cannot simplify forever relies on the notion of a greatest common divisor, specifically that there always exists a greatest common divisor for a pair of two integers and that it is unique. The simplest form of a ratio of two integers is a form in which the greatest common divisor is 1. The existence and uniqueness of the greatest common divisor between two integers can be proved by noticing certain facts about sets of integers. I haven’t digested that proof yet, but I hope to do so in the future.