Answer to POW #3:

Consider the number x = sqrt(2)^sqrt(2).

If x is rational, then let a = sqrt(2) and b = sqrt(2), and we are done (it is well-known that sqrt(2) is irrational).

On the other hand, if x is irrational, then let a = x and b = sqrt(2). Then a^b = ((sqrt(2))^(sqrt(2)))^(sqrt(2)) = (sqrt(2))^2 = 2 is rational, so we are done.

Source: Paul Halmos