Case analysis

Proving by considering different cases is often used when we are not sure about the truth of some statement, and the trick is that, often, we do not really need the knowledge about its validity. Let us look at a concrete example.

Example 14

There exist irrational $x, y$ such that $x^y$ is rational.

Proof:

Let us start with $x = \sqrt{2}$ and $y = \sqrt{2}$. Now consider $x^y = \sqrt{2}^{\sqrt{2}}$. We do not know whether this number is rational or not, so let us try to consider both cases:

• In the first case, if $x^y$ is rational, we are already done since we explicitly present the values of $x$ and $y$.

• In the second case, if $x^y$ is irrational, we can define $z = x^y$ . Now notice that $z^y = \left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = 2$, which is rational. Therefore, $z$ and $y$ are the two numbers that validate the statement we want to prove.

Notice that, in either case, we establish the validity of the statement we are proving. Therefore, whether $\sqrt{2}^{\sqrt{2}}$ is rational or not does not really matter!