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.

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).
