Irrational numbers


Visualisation of $\sqrt{2}$
Visualisation of $\sqrt{2}$

In antiquity it was believed that all numbers are either integers or fractions, what's called rational numbers. So when the Pythagorean Society discovered irrational numbers they were astonished. They didn't like the idea at all, and even denied it for a while.

$\sqrt{2}$ is irrational

We now know lots of numbers which are irrational. We know $\pi$ is, and so is $e$, but we don't have to look for such exotic numbers. $\sqrt{2}$ is also irrational (as well as $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{7}$, for instance).

The proof

There are several proofs which show that $\sqrt{2}$ is irrational, but the best known is the proof by contradiction. It's a kind of proof where you assume that your preposition is not true, and the logic that follows leads to a contradiction. So in the case of $\sqrt{2}$ we assume that it's indeed rational, then you can write the ration in its smallest terms: $$ \sqrt{2} = \dfrac{p}{q} $$ Smallest terms means that $p$ and $q$ have no common factors, so that the fraction can not be simplified any further. Then $$ 2 = \dfrac{p^2}{q^2} $$ and therefore $$ 2 \cdot q^2 = p^2 $$ This means that $p^2$ is even, and therefore $p$ must be even as well, and we can write it as $2\cdot p'$. Then $$ 2 \cdot q^2 = (2 \cdot p')^2 $$ $$ 2 \cdot q^2 = 4 \cdot p'^2 $$ $$ q^2 = 2 \cdot p'^2 $$ This time we conclude that $q^2$ is even, and therefore also $q$ must be even. But here's our contradiction: both $p$ and $q$ are even, so the fraction $\frac{p}{q}$ is not in it's lowest terms, which we assumed at the beginning. We can simplify $\frac{p}{q}$ by dividing numerator and denominator by 2, but the problem will remain: it will still not be in its lowest terms. We can still repeat the whole argument. It never will be in lowest terms, and for a rational number that's impossible; every rational has a lowest terms form.

Therefore we can conclude that $\sqrt{2}$ is not rational. It's irrational. QED.

Can an irrational number raised to an irrational power be rational?

That's an intriguing thought, and it may be surprising that the answer is yes. The proof below is a special kind which is called an existence proof. It means that you prove a solution exists, without necessarily finding the solution.

Let $r$ and $s$ be irrational numbers for which $r^s$ is rational. Above we proved that $\sqrt{2}$ is irrational, so let's use that value for both $r$ and $s$. Now $\sqrt{2}^{\sqrt{2}}$ is either rational or irrational, and it's not clear on sight which it is. Let's consider both possibilities.

Case 1: it's rational

Then we're done. We proved that an irrational number raised to an irrational power can be rational.

Case 2: it's irrational

If $\sqrt{2}^{\sqrt{2}}$ is irrational we can use it as a new value for $r$, so that $$ r^s = (\sqrt{2}^{\sqrt{2}})^\sqrt{2} = \sqrt{2}^{(\sqrt{2} \cdot \sqrt{2})} = \sqrt{2}^2 = 2 $$ So whether $\sqrt{2}^{\sqrt{2}}$ is rational or irrational, in any case there exists a solution, though we don't know what it is.

By choosing the right values for $r$ and $s$ you can construct a shorter, constructive proof, i.e. a proof which not only proves the existence of such a number, but also tells us what that number is. The following is due to Chris Reineke: let $r = \sqrt{10}$ and $s = log(4)$. Then $$ r^s = \sqrt{10}^{log(4)} = 10^{log(2)} = 2 $$