# Proof that there exists irrational x,y such that x^y is rational. Valid or not?

• Sep 26th 2013, 07:41 PM
thomasmgill
Proof that there exists irrational x,y such that x^y is rational. Valid or not?
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 xy ​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 xx, (1/2)^(1/2) is not rational. Therefore, there are rational numbers x that map to irrational numbers xx.
There are some irrational xx that are mapped to by rational x, and therefore, there cannot be a 1-1 correspondence between irrational x and irrational xx, since the size of the set of irrational numbers x is the same size as the set of irrational numbers xx.
Therefore, there must be some irrational x that map to non-irrational xx. 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.
• Sep 26th 2013, 10:20 PM
SworD
Re: Proof that there exists irrational x,y such that x^y is rational. Valid or not?
No, it isn't valid, unless I'm missing something here. But by your logic, there exists an irrational number whose square root is rational, just because there exists a rational number whose square root is irrational. Think about it, the logic is analogous, yet it is clearly false.

You assumed that just because A is a proper subset of B, there cannot be a 1-1 correspondence between A and B, but this is only true for finite sets. For example, there IS a one-one correspondence between {all odd positive integers} and {all positive integers}. You can get it by listing out all the odd positive integers in order by size, and then each of them has an index, which is a positive integer.

Hilbert's paradox of the Grand Hotel - Wikipedia, the free encyclopedia
• Sep 30th 2013, 11:16 AM
ebaines
Re: Proof that there exists irrational x,y such that x^y is rational. Valid or not?
Why not simply use an existence proof - find an example of irrational values for x and y that satisfy the criteria that x^y is rational. One that comes to mind right away is

$e^{\ln (2)} = 2$

You would have to show that both $e$ and $\ln (2)$ are irrational, but that's not too difficult.
• Sep 30th 2013, 02:24 PM
SlipEternal
Re: Proof that there exists irrational x,y such that x^y is rational. Valid or not?
Quote:

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 xy ​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 xx, (1/2)^(1/2) is not rational. Therefore, there are rational numbers x that map to irrational numbers xx.
There are some irrational xx that are mapped to by rational x, and therefore, there cannot be a 1-1 correspondence between irrational x and irrational xx, since the size of the set of irrational numbers x is the same size as the set of irrational numbers xx.
Therefore, there must be some irrational x that map to non-irrational xx. 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.

You don't even have 1-1 correspondence among the rationals: $1^1 = 0^0 = 1$. See the wikipedia article $0^0$.
• Sep 30th 2013, 02:30 PM
HallsofIvy
Re: Proof that there exists irrational x,y such that x^y is rational. Valid or not?
You have been told why your proof is not valid. Consider, instead:

Look at $\sqrt{3}^{\sqrt{2}}$. IF it is rational we are done! If it is NOT rational then what about that to the $\sqrt{2}$ power?