Quote:
Originally Posted by bizaro86
Spoiler!
Thanks! I agree that to consider the problem complete a proof that those are both irrational is necessary, but I didn't feel like it, and probably wouldn't be able to do that any more 
A very slightly modified version of my solution works for your amended problem as well.
A=sqrt(2)
B= log-sqrt(2)(3) ie a logarithm of base sqrt(2) of 3
Which by the same rules of exponents/logarithms equals 3.
It is fairly trivial to prove that sqrt(2) is irrational - I haven't done that for 15 years either but recall the rough outline. If you assume it is rational in the form a/b and that a/b is fully reduced then a and b can't both be even, and by squaring both sides and moving the denominator you get 2b^2=a^2. That means a is even so b can't be even.
But if a^2 is even then a is even, so its a multiple of 2, and if you sub 2x for a (as its an even number) into the previous you get 2b^2=4x^2 which means that b must be even and is a contradiction.
I'm quite confident that logarithm is irrational as well but leave that as an exercise to the reader 
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Haha, the old "exercise to the reader" argument. I remember in undergrad someone put together a list of different proof types. Along with common ones like "proof by induction" or "proof by contradiction" were "back of the textbook proof" and "proof by intimidation".
I think your method works, but you do have to prove that that one item is irrational. It certainly looks irrational, but how do you know for sure?
Maybe, to help further narrow the scope, I'll amend the problem one last time to say there exists 2 irrational algebraic numbers a and b such that a^b is rational.
Algebraic numbers are numbers that are solutions to P(x)=0 where P(x) is any polynomial in x with rational coefficients.