Let $\mathbb{R}^*$ be the group of nonzero real numbers under multiplication and let
$$H = \{x \in \mathbb{R}^* : x^2 \text{ is rational}\}.$$
Prove that $H$ is a subgroup of $\mathbb{R}^*$.
I have so far that $H$ is nonempty because $e=1$ is in $H$ since $e^2$ is rational. Let $a$ and $b$ be elements of $H$. We know that $a^2$ and $b^2$ are rational.
We need to show that $ab^{-1}$ is in $\mathbb{R}^*$ and that $(ab^{-1})^2$ is rational.
I'm having trouble with this last part mainly because I'm not sure how to show that $b^{-1}$ is also an element of $H$.
Best Answer
HINT: $(ab^{-1})^2=a^2(b^{-1})^2={a^2\over b^2}.$ Now, $a^2$ and $b^2$ are both rational . . .
(To see that $b^{-1}\in H$, just think about $(b^{-1})^2={1\over b^2}$ . . .)