Irrational Thinking

The simplest questions to understand are sometimes the toughest to solve. Have a look at this one:

May an irrational (number) raised to the power of an irrational ever result in a rational?

The first time I laid my eyes on it, I sat and thought for a moment. All I had to do was provide an example which satisfied the premise, and I was done. Or prove mathematically that it would never occur, of course, which seemed infinitely tougher.

Gut instinct would tell us that the latter would be true. I mean, how can $irrational^{irrational} = rational$ ? Our basic knowledge (which is very basic) of the real number system leads us to guess so. However, as some of you may be knowing, in math, instinct can get the better of you when the answer is otherwise.

So I kept thinking, kept trying out trivial example to see whether I could find the easy way out. Unfortunately, my (relatively insignificant) efforts were in vain, and I couldn’t find an example as such. Then, almost convinced that my instinct was right, I moved on to proving that it would never happen. This was a more herculean task, as you would imagine, than providing a single example, and I had no idea where to start. And I gave up on that, too.

Having given my two cents to an attempt at answering the question, I reluctantly skimmed to the end of the book containing the solutions. I wasn’t worried that I couldn’t solve it myself, because you don’t get everything all the time, but there’s always that feeling of regret when you have to concede to that fact. As I read through the solution, I was impressed.

+Click for the solution!

First off, the answer is that it is possible. How? And what are those two irrational numbers? I may be able to answer the first question, but perhaps not the second. What do I mean? Well:

Consider $\sqrt{2}$ . We all know that it’s an irrational number. (the interwebs have many, many proofs)

Raise this to the power of $\sqrt{2}$ and let the resulting number be denoted by $x$ , i.e., we get

$x = \sqrt{2}^{\sqrt{2}}$

Any real number is either a rational or an irrational.

1. Let’s assume that $x$ is rational. Then we’re done here, because we’ve found a case of a rational equal to an irrational raised to an irrational power.

2. If, however, $x$ is irrational, raise x to the power $\sqrt{2}$ , and call it $y$ . That is,

$y = x^{\sqrt{2}} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$

This is now an irrational raised to an irrational power, since we considered $x$ to be irrational in the second case.

And this turns out to be the

$y = (\sqrt{2})^{\sqrt{2} * \sqrt{2}} = (\sqrt{2})^{2} = 2$

And this is obviously rational.

What do we conclude from this? In both cases (whether $x$ is rational or irrational), we can always find an example of an irrational to an irrational power resulting in a rational. And since one of the cases must be true, the answer is YES, it is possible, but we don’t know which case holds.

This kind of proof is called a non-constructive proof, as it proves the existence of something, yet doesn’t provide a concrete example. This is why I mentioned before that I probably wouldn’t be able to answer the question “And what are those two irrational numbers?”

But for the record, $\sqrt{2}^{\sqrt{2}}$ is an irrational number. In fact, it is a transcendental number.

Site June 7, 2016 at 1:50 pm

In other words, there are more irrational numbers than there are rational. Can you grasp that? It seems like something I accept as a proven fact but that is not tangible or easily illustrated in concrete terms which is often the case with irrational numbers. They seem to elude us, yet are fascinating to think about.

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