Here’s a cute little existence argument that I was exposed to as an undergraduate and have never forgotten. It shows that there must be irrational positive reals and for which is rational. Furthermore, it’s done by showing that one of two very specific pairs of numbers (either or ) satisfies the condition, but without establishing which of the two pairs satisfies the condition.
The argument is elementary. First look at the case If is rational, we have an example, and we’re done. If not then must be irrational, and so where (assumed to be irrational) and is an example of an irrational raised to an irrational. And it’s rational. In fact, it’s , because
It’s a great little elementary existence argument, and I think it’s worth being exposed to it. But, it’s also worth knowing that with (much) more advanced techniques we can actually say which of the two choices satisfies the condition (it’s the second choice ). The technique I’m thinking of is the Gelfond-Schneider theorem, which implies that if are positive algebraic irrationals, then is irrational: the actual theorem uses weaker conditions on and and reaches a stronger conclusion about And, of course, is an algebraic irrational, so the result applies.
And, it’s also worth knowing that there are different examples that can be established using elementary techniques: for example for which
Just for fun, here are elementary examples of all possible choices of rational and irrational choices in taking powers:
All that you need to prove the above results is that the three numbers are irrational, and there are elementary arguments for establishing each of these. For completeness, I’ll sketch proofs for the three results.
is irrational. Assume it’s not. Then there are integers for which i.e. for which (and, of course, ). We show that both have to be even numbers, and that’s impossible for all with satisfying because by removing common factors from any given we arrive at a simplified for which one of the two numbers is odd. To show that both are even, we first have that is even because is even (it’s ) and writing we have that and so which tells us that is even and so is even.
The above argument has been known for over two thousand years, and has been repeated in numerous places, including many popular mathematics books. I would recommend G.H. Hardy’s “A Mathematician’s Apology” if you want to find another source: Hardy calls it “Pythagoras’ proof”. With extremely minor modifications, the same argument can be used to prove that is irrational, as shown below.
is irrational. Assume it’s not. Then there are integers for which i.e. for which (and, of course, ). We show that both have to be divisible by , and that’s impossible for all with satisfying because by removing common factors from any given we arrive at a simplified for which one of the two numbers is not divisible by . To show that both are divisible by , we first have that is divisible by because is divisible by (it’s ) and writing we have that and so which tells us that is divisible by and so is divisible by .
is irrational. Assume it’s not. Then there are integers for which i.e. for which i.e. for which (and, of course, ). We show that both have to be zero, and that contradicts . If are both non-negative, then is an odd number and the only way that can be equal to an odd number is if and so , from which it immediately follows that and so and we’re done. The other three (partially overlapping) cases ( non-negative and non-positive, non-positive and non-negative, non-positive and non-positive) all can be ruled out using similar reasoning (on for non-negative integers in some cases).
And, while I’m on the subject, here are some elementary examples for and