Space filling curve to simplify vector addition? Since points on a euclidean plane can be represented by one coordinate on a space-filling curve, is there any curve such that if two vectors $(x_0,y_0)$ and $(x_1,y_1)$ were represented by $a$ and $b$, the sum $a+b$ would represent the vector $(x_0+x_1,y_0+y_1)$? Could this curve be generalized to three dimensions?
EDIT: Even one with $ab$ representing $(x_0+x_1,y_0+y_1)$ would be a start, I can't find anything.
EDIT2: Never mind $a+b$, is there any way to do some sort of simple operation between $a$ and $b$ to represent vector addition?
 A: There can be no continuous space-filling curve respecting $+$ in the way you desire. By considering $0+b$, it follows that $0$ must represent $(0,0)$. Now, suppose that some real number $a$ represents $(1,1)$. It follows that $2a$ represents $(2,2)$ and $\frac{1}{2}$ represents $(\frac12,\frac12)$, and in general, a rational number $q$ must represent $(q,q)$. Extending by continuity, the whole curve will lie on the diagonal line, a contradiction.
A: In response to your edit 2: You may represent a pair of real numbers as a single real number by interweaving their decimal expansions.  For example, $(4.56, 1.23)$ is represented as $41.5263$.  A little care needs to be taken when dealing with non-terminating strings of 9s.
It turns out that arithmetic operations, like addition, in this encoding scheme are about as straightforward as un-interleaving and doing the ordinary arithmetic operations on each coordinate.  From a computational efficiency standpoint, it is difficult to beat plain old vector addition.
A: In other words, you want  $f:\mathbb{R}\to\mathbb{R}^2$ such that $f(a+b)=f(a)+f(b)$, that is a homomorphism of additive groups.
Such and $f$ is either $\mathbb{R}$-linear, thus its image is a line, or it is everywhere discontinuous. In the latter case, it may even be bijective (for instance, one can take $f$ an isomorphism between $\mathbb{R}$ and $\mathbb{R}^2$ as $\mathbb{Q}$-vector spaces).
