29.02.2016

There is a proof that I love, one that is very succinct and easy to understand to someone with only middle-school knowledge of mathematics. But it's remarkable in that inside of it lies a deep-seated idea, one that I find very difficult to wrap my head around sometimes. Let's have some fun and walk through it. I'm going to write this post in a mathematical language, but I want it to be accessible to a general audience, so I'm going to explain some terms as I go. (I generally won't do this).

Before we begin, we're going to need a small but general prior result. In mathematics, we call this a *lemma*, even though some things called lemmas turn out to be quite important indeed.

## Lemma

\(\sqrt{2}\) is irrational.

## Proof

A rational number takes the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers (e.g. whole numbers such as \(-2,-1,0,1,2...\)) which have no common factors. In other words, there is no other number that can be divided into both \(p\) and \(q\) and have integers come out both times. Any number that is not rational is called irrational. Now, suppose for a minute that \(\sqrt{2}\)

wererational. Then, there are some integers \(p\) and \(q\) such that \(\sqrt{2} = \frac{p}{q}\). Now, let's do some math for a bit.\[ \begin{align} \sqrt{2} &= \frac{p}{q}, \\\\ 2 &= \left(\frac{p}{q}\right)^2, \\\\ 2 &= \frac{p^2}{q^2}, \\\\ 2q^2 &= p^2. \end{align} \]

Now, suppose that \(p\) was an odd number. Then we can write \(p\) in the following form: \(p = 2k+1\) for some

otherinteger \(k\). For instance, \(5 = 2\times 2 + 1\). If \(p\) is odd, then\[ \begin{align} p^2 &= (2k+1)^2 \\\\ &= (2k+1)(2k+1) \\\\ &= 4k^2+4k+1. \end{align} \]

Since \(4k^2+4k\) is divisible by 2, it must be even; therefore, \(4k^2+4k+1\) is necessarily odd. This is a problem, because regardless of whether $q$ is odd or even, \(2q^2\)

mustbe even. Therefore, this means that \(p\)mustbe even. So lets say instead that \(p=2m\).Now, we have \(2q^2 = (2m)^2\) which is equivalent to \(2q^2 = 4m^2\), which is equivalent to \(q^2 = 2m^2\). But now let's apply our logic again: \(2m^2\) must be even, so by what we just did in the previous paragraph, \(q\) must be even. Therefore, \(p\) and \(q\) are both even, which means they have a common factor! This goes against our earlier assertion that a rational number has the form \(\frac{p}{q}\) where \(p\) and \(q\) have

no common factors. This absurdity establishes what's known as aproof by contrapositive: we assert the opposite of the conclusion we want to be true, and show that it must mean that our hypothesis is nonsense. Therefore, \(\sqrt{2}\)must be irrational.

Ok, now that we've spent some time on that proof, we can use the irrationality of \(\sqrt{2}\) as an unquestionable argument from here on out. So let's get to the meat of this remarkable little result. I will state a theorem. In mathematics, a theorem is a statement that can be proven true. Sometimes a theorem gives us important results. Other times, it establishes minor stepping stones along the way. In this case, because this is the focus of this blog post, I will call this teeny-tiny result a theorem.

## Theorem

There exists an irrational number that when raised to an irrational power, results in a rational number.

## Proof

Consider \(\sqrt{2}\), which is irrational. Then, \(\sqrt{2}^{\sqrt{2}}\) is either rational or irrational. If it is rational, then we are done. If it is irrational, then consider \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^2 = 2\).

That's it. Easy as can be. Theorems such as these are called *existence theorems*. They tell us facts about mathematics, but it is likely that we cannot use them to construct practical examples of what we want to see. Mathematicians deal with these all the time. In this case, this existence theorem tells us something remarkable: we can take two irrational numbers and construct a rational from them using exponentiation. This can be a surprising result, because there are *many, many* more irrational numbers than rational ones.

But something should be really uncomfortable here: We've never established whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational.

We've used what logicians call the *Law of the Excluded Middle*. In this case, the number \(\sqrt{2}^{\sqrt{2}}\) must be either irrational or not; there is no middle case. We don't immediately know what qualities \(\sqrt{2}^{\sqrt{2}}\) has. But it turns out *we don't care.* Regardless of which case is true, the pathway leads us to our desired result!

To me, this is very deep. It means that we can use truths we don't possess to prove things that are unquestionably true. It would be like mending a fence using either a screwdriver or a hammer without your ability to know which tool you're using. In either case the fence is mended, but you have no idea what truth kept your horses in the pasture.

This is not to say that we don't understand each of the possible truths individually--we know quite well what it means if \(\sqrt{2}^{\sqrt{2}}\) is rational or not. This knowledge allows us to see that both paths lead to the result.

To me, getting to a destination whilst having absolutely no idea which path is taken is a bit unnerving. But it is also liberating to know that such strict control is unnecessary to discover beautiful things in the world.

Posted: 29.02.2016

Built: 14.10.2024

Updated: 24.04.2023

Hash: 95f4054

Words: 901

Estimated Reading Time: 5 minutes