The cardinality of the set Let $\mathbb{G} =\{ a^b + \sqrt{c}: a,b,c\in \mathbb Q \}$
I guess the set $\mathbb{G}$ is countable set, but I can't show it properly.
How to start the proof?

The proof is basically the same as for $\mathbb Q$ being countable. You start with an enumerations $a_j$, $b_k$ and $c_l$ (for $a$, $b$ and $c$) in your formula and iterate the triples by first taking those where $j+k+l=0$ then $j+k+l=1$, then $j+k+l=2$ and so on. For each level you have finitely many triples and by this construct all triples will be enumerated.
