$(\mathbb R^2,+)$ as vector space over $\mathbb R$ It is not hard to see that $(\mathbb R^2,+)$ with this product

$
 {r\cdot(x,y)=(rx,ry) }
$

is vector space over field $\mathbb R$.
I'm looking  for another product that $(\mathbb R^2,+)$ is vector space over $\mathbb R$. I know 
$n*(x,y)=\underbrace{(x,y)\oplus (x,y) \oplus \cdots \oplus (x,y)}_n=(nx,ny)$ but I have no idea for arbitrary element of $\mathbb R$.
Any Suggestion.
Thanks 

If you are allowed to use the axiom of choice, then the abelian groups $(\mathbb R, +), (\mathbb R^2, +), (\mathbb R^3,+), \ldots$ are isomorphic. In fact they are all isomorphic to a direct sum of continuum-many copies of $(\mathbb Q,+)$.
So giving a vector space structure to any of them is the same thing. In particular, we know how to give $(\mathbb R^3,+)$ a structure of $\mathbb R$-vector space which is of dimension $3$, thus by transporting it to $\mathbb R^2$, we can give a vector space structure to $\mathbb R^2$ making it into a 3-dimensional $\mathbb R$-vector space (and similarly for any dimension you wish, as long as it's not more than the continuum cardinality).
In fact, even without the axiom of choice, giving a finite dimensional $\mathbb R$-vector space structure to any abelian group $G$ is the same as finding a group isomorphism from $G$ to $\mathbb R^n$ and transporting the natural vector space structure of $\mathbb R^n$ back to $G$. So the real question is how to find those group isomorphisms.
Sadly, I don't think it is possible to find any nontrivial one without the axiom of choice, so all the nontrivial vector space structure you can put on $\mathbb R^2$ need you to use it, and are not constructive.
