You should be able to solve this.

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

>>7
Dumbass.  If X is a countable set whose members are all countable sets, theres a clear bijection between NxN and UX.  NxN is countable by a diagonalization argument, does not require AC at all.  Are you suggesting that the rationals cannot be proved countable if you assume the negation of AC?