Since $G$ is non empty, let $a \in \displaystyle{\mathbb{R}}$ be a real that belongs to $G$. But then, $aa$ must also belong to $G$, that is, $0 \in G. \\$ Now since $G$ is open, there must exist an open interval around $0$, say $(\delta,\delta) \subseteq G$ with $\delta >0$. Now pick any positive real $\theta. \\$ Choose a big enough $n \in \displaystyle{\mathbb{N}}$ such that $\theta / n < \delta$. This means that $\theta / n \in G$. By adding this number to $0 \in G, n$ times, you $\\$ get (every time) an element of $G$. Hence, $\theta \in G. \\$ If a negative real $t$ is, futhermore, selected, then $t$ is positive and according to the above, belongs to $G$. But then, $0  (t) = t$ must also belong to $G$. Hence, all positive and negative reals, and $0$, belong to $G$. Hence, $G = \mathbb{R}$
Efthymios Tsakaleris
