How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module? How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module?
I guess there is only one way, and that is the  trivial case. i.e. 
$$\forall r\in \Bbb R,\, \forall a,b\in \Bbb Q \qquad r\cdot \frac{a}{b}=0$$
With an idea I can proof this:
$$\exists r\not=s\in \Bbb R \quad\text{s.t.}\quad r\cdot \frac{a}{b}=s\cdot \frac {a}{b}$$
but I think this idea is wrong.
 A: Here is another way to think about things.
First note that $\text{End}_\mathbf{Ab}(\mathbb{Q})$ (the abelian endomorphisms of $\mathbb{Q}$) is countable. Indeed if $g\in\text{End}_\mathbf{Ab}(\mathbb{Q})$ then for any $p\in\mathbb{Z}$ we have that $g(p)=pg(1)$ and thus if $\displaystyle qg\left(\frac{p}{q}\right)=g(p)=pg(1)$--in other words, $\displaystyle g\left(\frac{p}{q}\right)=\frac{p}{q}g(1)$. Thus, $\#\text{End}_\mathbf{Ab}(\mathbb{Q})=\#\mathbb{Q}$.
Now, an $\mathbb{R}$-module structure amounts to a ring map $\mathbb{R}\to\text{End}_\mathbf{Ab}(\mathbb{Q})$. This ring map clearly can't be unital, else it is an injection because $\mathbb{R}$ is a field which is impossible by cardinality issues. That said, note that if $1\mapsto g\in\text{End}_\mathbf{Ab}(\mathbb{Q})$ under our ring map, then $g^2=g$, and so $g(g-1)=0$. Thus, $g(g(1)-1)=g(g(1))-g(1)=0$, but as we have already observed $g(g(1))=g(1)g(1)=g(1)^2$, thus $g(1)^2=g(1)$. If $g(1)=1$ then $g=\text{id}$ and so our ring map was unital--bad. So, $g(1)=0$, and from the above we see that $g=0$, and so $1\mapsto 0$, and so the ring map is the zero map--this corresponds to the trivial module structure Zev alluded to.
A: There are no ways to consider $\mathbb{Q}$ as an $\mathbb{R}$-module (i.e. $\mathbb{R}$-vector space), if you follow the usual convention of requiring that modules be unital, i.e. $1\cdot q=q$ for any $q\in\mathbb{Q}$. This is because a vector space over $\mathbb{R}$ (or indeed, any field) is free; because $\mathbb{R}$ is uncountable, any $\mathbb{R}$-vector space must consist of either one element or uncountably many elements, neither of which is the case for $\mathbb{Q}$.
If you do not require your modules to be unital, then the trivial module structure (multiplication by any real number gives 0) is the only possible one. This is because, if there is some $r\in\mathbb{R}$ and some $q\in\mathbb{Q}$ such that $r\cdot q\neq 0$, then we reach the same problem as before, because for any non-zero real number $s$, we have that
$$0\neq rq=\left(\frac{r}{s}\right)\cdot sq$$
so that $sq\neq 0$ for any non-zero $s\in\mathbb{R}$, and we have $rq\neq sq$ if $r\neq s$ (because $(r-s)q\neq 0$), thereby making $\mathbb{Q}$ uncountable again (contradiction).
