You should be able to solve this.

True or false: Are algebraic numbers countable?
You can't use axiom of choice.

>>8
NxN is countable (there is an injective function from NxN to N, in fact f(m,n) = 2^m 3^n works). There is a clear injective function from Q to NxN making Q countable. However there is no  bijection between UX and NxN (not even an injective function). You aren't gonna get any further without AC. Try it if you don't believe me.