# Methods of Irrationality

Being a mathematician requires you to think in strange ways.

For starters, you might think that mathematicians spend all day pondering purely theoretical things that only exist in Mathematical abstraction, right? Well, it can get even stranger than that: sometimes we have to think about things that don’t exist, even in the abstraction! It’s called a proof by contradiction, and here’s an example:

Theorem: The number $$\sqrt{2}$$ is irrational.

Proof: To prove that $$\sqrt{2}$$ is irrational, we have to show that we can never write it as a ratio of integers, $$\sqrt{2}=a/b$$. So let’s assume we can find such a fraction and see where it leads us.

Let’s cancel common factors in the numerator and denominator so that $$a/b$$ is in lowest terms. Squaring our equation shows that $$a^2=2b^2$$, so $$a^2$$ is even and so $$a$$ itself is even. Since $$a/b$$ is in lowest terms, $$b$$ must be odd.

Since $$a$$ is even, $$a^2$$ is in fact divisible by 4. On the other hand, since $$b$$ is odd, $$2b^2$$ is not divisible by 4. But then the integer $$a^2=2b^2$$ is both divisible by 4 and not divisible by 4, which is absurd! The only explanation for this contradiction is that our original assumption—that $$\sqrt{2}$$ is rational—is false. So we’re done with the proof! Notice that we spent most of our effort reasoning about integers that never existed (specifically $$a$$ and $$b$$).

But that’s not the only twisted thing about Mathematical thinking. Here’s a delightfully short yet aggravatingly unsatisfying proof:

Theorem: There exist irrational numbers $$x$$ and $$y$$ such that $$x^y$$ is rational.

Proof: We already know that $$\sqrt{2}$$ is irrational, so maybe we can use $$\sqrt{2}^{\sqrt{2}}$$. Is this number rational? If it is, then we’re done: $$x=\sqrt{2},y=\sqrt{2}$$ solves the problem. But what if $$\sqrt{2}^{\sqrt{2}}$$ is irrational? In this case, I claim $$x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}$$ works: indeed, $$\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot\sqrt{2}} = \sqrt{2}^2 = 2,$$ which is rational. In either case the required numbers $$x$$ and $$y$$ exist, so this completes the proof.

But wait; which is it? The proof shows us that one of the pairs $$x=\sqrt{2},y=\sqrt{2}$$ or $$x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}$$ works, but it doesn’t tell us which one! So, have we proven the theorem? Yes, technically, but only nonconstructively.

Frustrating, isn’t it?