Are the additive group of rationals and the multiplicative group of positive rationals isomorphic?

HINT: For each $q\in\Bbb Q$ the equation $x+x=q$ has a solution; is it true that $x\cdot x=q$ has a solution for each $q\in\Bbb Q^+$?

So now you know they're not isomorphic, but you can go a bit further with what you already know. For example, $\mathbb{Q}^+$ has a subgroup whose elements are the powers $2^n$ of $2$, including $1$ and $1/2$, etc. Because the map $n \mapsto 2^n$ from $(\mathbb Z,+)$ is injective, this is a cyclic subgroup. You can do the same thing with the powers $3^n$ and see these two subgroups intersect only in $\{1\}$.

Now, you know every natural number has a unique factorization in terms of primes. Can you use that to say more about $(\mathbb Q^+,\cdot)$?

By comparison, you can check for any two elements $g$ and $h$ of $(\mathbb Q,+)$,there are integers $m$ and $n$ such that $mg = nh$. This means that, in contrast to the case of $(\mathbb Q^+,\cdot)$, any two cyclic subgroups have a nontrivial intersection. One says that $(\mathbb Q^+,\cdot)$ has rank equal to 1, meaning there's a $\mathbb Z$ subgroup, but only one independent such $\mathbb Z$ subgroup, in that all other isomorphic copies of $\mathbb Z$ meet it nontrivially.