Problem 1

1. Do there exist irrational numbers $r,s$ such that $r^s$ is rational? Do there exist irrational numbers $r,s$ such that $r^s$ is irrational?

Really, Wikipedia ruined this problem for me in advance, or at least part a. The answer to both these problems is yes.

Proof: (a) The easiest methods are nonconstructive; observe $α=\sqrt{2}^\sqrt{2}$. Either $α$ is rational, in which the proposition is established, or $α$ is irrational. If $α$ is irrational, then $α^\sqrt{2}=\sqrt{2}^(\sqrt{2}·\sqrt{2})=\sqrt{2}^2=2$ is a pair of irrationals $α,\sqrt{2}$ establishing the proposition.

(b) My method for this too is nonconstructive; assume the proposition is false. Then observe the map $\sqrt{2}^\_ : ℝ \setminus ℚ → ℚ$. Since the union of two countable sets would then be countable, and since $ℝ$ is uncountable, we have $ℝ \setminus ℚ$ is uncountable. Therefore this map is not injective and we have $\sqrt{2}^r=\sqrt{2}^{r'}$ for some distinct irrationals $r,r'$. But exponentiation is naturally injective, so we observe a contradiction.