Let $\mathbb Z$ act on the additive group $\mathbb Z\left[\frac13\right] = \{a/3^k : a \in \mathbb Z, k \ge 0\}$ by $\varphi_n(r) = 3^n r$ for $n \in \mathbb Z$ and $r \in \mathbb Z\left[\frac13\right]$. Let $G = \mathbb Z\left[\frac13\right] \rtimes_\varphi \mathbb Z$, a semi-direct product.
(a) Compute the product $(r, m)(s, n)$ and the inverse $(r, m)^{-1}$ in the group $G$.
(b) Show $G$ is generated by $(1, 0)$ and $(0, 1)$.
I found part (a) to be easy; $(r, m)(s, n) = (3^ms + r, m + n)$ and $(r, m)^{-1} = (-r3^{-m}, -m)$. I'm struggling much more with part (b). In fact, any time I need to show that two elements generate a group, I tend to have trouble figuring out what to do. Any help on both solving this problem and any advice on how to show two elements generate a group would be greatly appreciated.
As noted in a comment, you can achieve $(a, 0)$ and $(0,n)$ for $a, n\in\mathbb Z$ as multiples of these generators. What if you want to get $(a/3^k,n)$? By definition of the semidirect product, $(0,k)(a, 0)(0,-k)=(a/3^k,0)$. Can you finish from here?