Originally Posted by

**thomasmgill** Hi all. I was asked the following question in class and came up with a proof.

The question is as to whether there exists irrational x, y such that x^{y }is rational.

I know the classic proof to show that it exists is to find irrational x, y such that it works.

That involves a lot of guess and check, although it is obviously a valid proof.

The proof I came up with is as follows:

Let x=y=(1/2). Then for x^{x}, (1/2)^(1/2) is not rational. Therefore, there are rational numbers x that map to irrational numbers x^{x}.

There are some irrational x^{x} that are mapped to by rational x, and therefore, there cannot be a 1-1 correspondence between irrational x and irrational x^{x}, since the size of the set of irrational numbers x is the same size as the set of irrational numbers x^{x}.

Therefore, there must be some irrational x that map to non-irrational x^{x}. We are closed to the reals here, so numbers that are not irrational are rational.

The professor says this is wrong, but she cannot really explain why. She just says "it doesn't work because we are dealing with infinite sets". I am in my first year of undergraduate studies, so perhaps my understanding is not developed fully yet.

Is this proof valid? If not, can someone provide a better explanation as to why not?

Thanks.