Let $H$ be a subgroup of $\Bbb R^*$. If $\Bbb R^+ \subseteq H \subseteq \Bbb R^*$, prove that $H = \Bbb R^+$ or $H = \Bbb R^*$. 
Let $H$ be a subgroup of $\Bbb R^*$, the group of nonzero real numbers under multiplication. If $\Bbb R^+ \subseteq H \subseteq \Bbb R^*$, prove that $H = \Bbb R^+$ or $H = \Bbb R^*$.

I think that $\Bbb R^+$ is supposed to mean the set of non-negative reals? I am a bit confused on what do I need to prove here? If I assume that $H \ne \Bbb R^+$ and get that $H=\Bbb R^*$, then would that prove the result?
Suppose that $H \ne \Bbb R^+$, then I only have that there exists $h \in H$ such that $h \notin \Bbb R^+$?

$\Bbb{R}^+=\{a\in\Bbb{R}:a>0\}$.
$\Bbb{R}^+, \Bbb{R}^{\star}$ both are multiplicative group.
If $\Bbb{R}^+\subset H\subset \Bbb{R}^{\star}$ and $H\le \Bbb{R}^{\star}$ then $H=\Bbb{R}^+$ or $H=\Bbb{R}^{\star}$
Proof: Suppose $\Bbb{R}^+\subsetneq H$.
Then $\exists a\in H\setminus \Bbb{R}^+$ i.e $a\le 0$ and since $0\notin H$ , $a<0$.
Claim : $-1\in H$.
$a<0$ implies $-a\in \Bbb{R}^+$ and hence $-\frac{1}{a}\in \Bbb{R}^+$.
Hence $-\frac{1}{a}\in H$.
Since $H\le \Bbb{R}^{\star}$ , $a\cdot-\frac{1}{a}=-1\in H$
Now it's easy to show that $\langle \Bbb{R}^+, -1\rangle =\Bbb{R}^{\star}$
$(b<0$ , then $b=-b\cdot -1)$
Notations:
$\le $: subgroup.
$\langle S \rangle $: subgroup generated by $S$
