Question: Given any two irrational numbers x and y, does x^y ever evaluate to a rational number?
Proof: Given the definitions:
r = 2^0.5 (square root of 2, known to be irrational)
z = r^r
- z is rational. Set x=r, y=r and you have a proof.
- z is irrational. Set x=z, y=r. x^y = z^r = (r^r)^r = r^(r*r) = r^2 = 2, which is rational.
This proof does not provide a particular value of x and y, but merely shows there must be one.