Clearly we may assume that $G\not=\{0\}$, so that $G\cap\mathbb{R}_{>0}\not=\emptyset$. We set $\delta=\inf(G\cap\mathbb{R}_{>0})$. If $\delta\in G\cap\mathbb{R}_{>0}$, then $G=\mathbb{Z}\delta$. Indeed, for $g\in G$, let $n$ be an integer such that $n\leq g/\delta<n+1$. Thus $0\leq g-n\delta<\delta$, and hence $g=n\delta$. Therefore, $G$ is discrete.

Next we assume that $\delta\not\in G\cap\mathbb{R}_{>0}$. Then there is a sequence $\{\delta_{n}\}_{n=1}^{\infty}$ in $G\cap\mathbb{R}_{>0}$ such that $\delta_{n}>\delta_{n+1}$ for all $n$ and $\lim_{n\to\infty}\delta_{n}=\delta$. If we set $a_{n}=\delta_{n}-\delta_{n+1}$, then $a_{n}\in G\cap\mathbb{R}_{>0}$ and $\lim_{n\to\infty}a_{n}=0$. For an open interval $(\alpha,\beta)$ of $\mathbb{R}$ ($\alpha<\beta$), we choose $a_{n}$ and an integer $m$ such that $a_{n}<\beta-\alpha$ and $m<\beta/a_{n}\leq m+1$. Then we have $ma_{n}<\beta$ and

so that $ma_{n}\in(\alpha,\beta)\cap G$. Thus $G$ is dense. ∎