$\mathbb R^{+}$ is isomorphic to $2^{\aleph}$ copies of $\mathbb Q$ Here is my homework problem.

Explain why the positive reals $\mathbb{R}^+$ is a divisible abelian group under multiplication. Show that the group $\mathbb{R}^+$ is isomorphic to the direct sum of $2^\aleph$ copies of $\mathbb{Q}$.

I think, unless my professor means something special by "divisible abelian group," that I am supposed to show that $(\mathbb{R}^+, \cdot)$ maps to $(\mathbb{R}, +)$ by the logarithm and that there's an invertible by the exponential. However, it seems as if he is asking me to use this fact in demonstrating that $2^\aleph$ copies of $\mathbb{Q}$ is isomorphic to $\mathbb{R}^+$.
 A: The positive reals are an  abelian group under multiplication, while the reals are an abelian group under addition. Working in the the latter case, the group is divisible since the equation $nx=r$ has a real solution for any positive integer $n$ and any real number $r.$ As a torsion-free divisible group, the additive group of reals must therefore be a direct sum of copies of $\mathbb{Q}.$ Since $\mathbb{R}$ is uncountable, the number of summands must be $2^{\aleph},$ $\aleph$ being the cardinality of the rationals.
This answers your original question because, as you correctly observed, logarithm provides an isomorphism between the multiplicative group of positive reals and the additive group of reals.
As @glofactics pointed out, $(\mathbb{R},+)$ is a $\mathbb{Q}$-vector space since the solution of $nx=r$ gives you a way to define $(1/n)r.$ You can proceed from this perspective easily.
Note that $n0=0$ for any positive integer $n.$ Indeed, we can choose $n$ to be any real number. Thus, we see that $0/0$ is indeterminate, because the expression does not have a unique value. It would be a good exercise to produce examples in which a limit has the indeterminate form $0/0,$ yet the value of the limit will be a specified real number $r.$
