Why is the group of rational numbers with odd denominators residually finite?

Question

I want to prove that the additive group of rational numbers with odd denominators is residually finite. However, despite pondering for quite some time, I can't even find a single subgroup of finite index. Can someone give me a hint?

Answer 1

Denote this ring $\mathbb{Z}_{(2)}$. Then for $n \ge 0$, you can check that $2^n \mathbb{Z}_{(2)}$ is an ideal and $\mathbb{Z}_{(2)} / 2^n \mathbb{Z}_{(2)} \simeq \mathbb{Z}/2^n \mathbb{Z}$. If $x \in \mathbb{Z}_{(2)}$ is non zero, you can write $x= \frac{2^n a}{b}$ with $a, b$ odd, and check that $x$ is not in $2^{n+1} \mathbb{Z}_{(2)}$.