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?
 A: You have a function $f \colon (a, b, c) \mapsto a^b + \sqrt c \colon \mathbb{Q}^3 \twoheadrightarrow \mathbb{G} \subseteq \mathbb{C}$, where $f$ maps $\mathbb{Q}^3$ onto $\mathbb{G}$. 
In general: if $f\colon X \twoheadrightarrow Y$ is a surjection, then $|Y| \le |X|$ i.e. cardinality of $Y$ $\le$  cardinality of $X$. 
Here, $\mathbb{G}$ is a surjective image of $\mathbb{Q}^3$ which is countable, so $\mathbb{G}$ is countable too.
I'm assuming you know/don't have to prove that (*) $\mathbb{Q}$ is countable, and that (*
*) if $X, Y$ are countably infinite, then so is $X \times Y$ (thus, given (*), $\mathbb{Q}^3$ is countably infinite).
A: 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.
