# An irrational problem

Investigating the power of thinking rationally

Image: Wikimedia commons

Can you find an $a$ and $b$ that are both irrational such that $a^b$ is a rational number? It’s a fairly easy question to understand, if not to answer. I was posed this question in middle school before I had even heard of logarithms. The only tool you will need to solve this is the following exponent rule: $(a^b)^c$ is equal to $a^{bc}$. If you want to answer this on your own, which I highly recommend, don’t read on yet.

## $e$ isn’t the answer?

If you’re reading this, you are probably more than familiar with $e^{i\pi} = -1$. This in itself isn’t quite a solution to the problem, though it might be somewhere to start. This is where I started when I first came across the problem. However, that $i$ is quite pesky to remove.

Euler’s identity is a specific case of the more general expression $x$, $e^{ix} = \cos{x} + i\sin{x}$, known as de Moivre’s formula. The $i$ would have to disappear to fit what the problem asks since $i$, not being real, is definitely not rational. The only way to turn $i$ rational would be to either multiply by $i$, which would not give a good $\sin$ or $\cos$, or to multiply by $0$, which is not exactly irrational, so we cannot make a valid solution from playing with $e^{i\pi}$.

I didn’t actually know this at the time, and don’t even have it memorised now (thank you internet), but I didn’t find anything down that path at the time and haven’t seen anyone come up with a solution using something similar to Euler’s identity.

## The roots of the problem

Sometimes the best way to look at a problem, however, is not to blindly push through in search of a solution, but to backtrack and look at something simpler. What’s the most basic irrational number, one you probably learned about before even $\pi$? Well, $\sqrt{2}$ is a simple irrational, probably the first discovered. By definition, squaring a rooted term gets rid of the root. (For example, $(\sqrt{2})^2 = 2$.) Now, to find a solution to the problem given, we just need a way to get the exponent $2$ that the $\sqrt{2}$ is raised to through irrational numbers. Recall that $(a^b)^c$ is equal to $a^{bc}$. So if we just took $\sqrt{2}$ and $\sqrt{2}$ for $b$ and $c$, then the final output will be
$$(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}},$$
which is 2. So how does that help us? Well, we know that $\sqrt{2}^{\sqrt{2}}$ is an irrational raised to an irrational power. It’s either rational, which means we found the answer, or it’s irrational, which means we need to keep going. Raise that irrational to the $\sqrt{2}$ and we know we get 2, so that’s an irrational to an irrational that turns out to be rational.

This means that $a$ can be any rooted irrational as long as $b$ and $c$ are $\sqrt{2}$, so we have just found infinitely many solutions!

The beauty of this is that we don’t actually need to know which of the two sets of $a^b$ gives the irrational to the irrational. We just know that one of them does, and that’s enough to get through the problem. A problem this hard to come up with an answer to can be solved using middle school maths. This is why I love mathematics; there is beauty in the simplicity sometimes.

Some of you are probably not satisfied with the beauty of not needing to know. I certainly was when I came across this problem in middle school. Looking back now though, it wasn’t enough that it worked. I wanted to know which one it was that worked. There is elegance in finding the complete solution to a problem, and, after some research, I even found another realm of solutions. But first, let’s figure out which of the two sets is the irrational to an irrational power.

Sometimes in maths you have to dig a little. Image: Pixabay

The first step requires that you have algebraic numbers, that is, numbers that can be found as the solution to an equation with integer coefficients. (Note that this is the same as saying rational coefficients.) Obviously $\sqrt{2}$ is one such number since $x^2-2=0$ gives $x=\pm \sqrt{2}$. There are mathematical proofs that an irrational algebraic number raised to the power of another gives a transcendental number. These numbers are in some sense the opposite of algebraic. They are always irrational, and cannot be defined as a solution to an equation with integer coefficients. Some examples of familiar transcendental numbers are $e$ and $\pi$. Therefore, it is the second set of $a^b$ that gives us the irrational to the power of the irrational, because $\sqrt{2}^{\sqrt{2}}$ must be transcendental.

## Oh. $e$ is an answer

Now for the other solution. Logarithms. The obvious way to go is with $e$. Since $e^{\ln{x}}$ outputs $x$, so as long as $x$ is rational, you have an irrational to an irrational outputting a rational number. There are other ways to go about this as well, though they require a little more mental manoeuvring. For example,
$$\sqrt{10}^{\log{4}} = \sqrt{10}^{\log{2^2}} = \sqrt{10}^{2\log{2}} = 10^{\log{2}} = 2.$$
Since $\sqrt{10}$ and $\log{2}$ are irrational, this gives us another of an infinite set of answers.

Problems like these are the reason I love maths so much. They’re so simple, yet so difficult. They require strange thinking that makes so much sense and that it’s impossible to forget the solution once you do it for yourself. Keep doing maths!

A junior at Poway High School in California, Jenny likes to spend her free time exploring recreational and competition maths, though you can also find her participating in FRC robotics, playing ukulele, making bad puns, or practicing for sports. She runs (and created) her school’s maths team: the Poway ?thagoreans, and participates in most of the academic clubs on her campus, as well as various STEM competitions.
@to_genyus    + More articles by Jennifer

• ### How to Make the World Add Up

We review the third of this year's nominees for the Book of the Year
• ### Top ten vote issue 11

Vote for your favourite maths-themed day out
• ### Page 3 model: Bees

You won't bee-lieve it
• ### The croissant equation

Each time you eat a croissant you might be biting more than 500 layers of dough!

Santa's sack of scientific surprises
• ### In conversation with Ian Stewart

We speak to one of Britain's most successful popularisers of maths