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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?

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closed as too localized by fedja, Andreas Blass, Dan Petersen, Ryan Budney, Andrés Caicedo Nov 16 '12 at 6:33

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Take small $t>0$ and think of what points will be represented by integer multiples of $t$. This will tell you what happens if you require continuity. If you don't, look up "Hamel basis" on Google. – fedja Nov 14 '12 at 22:41
For your $ab$ possibility, by considering $0\cdot b$ you will see that $0$ cannot represent any vector, unless all vectors represent $(0,0)$. – Joel David Hamkins Nov 14 '12 at 22:56
That also makes sense. Thank you all for your help. – Benjamin Hansel Nov 14 '12 at 23:01
...and if we agree to take positive $a$ and $b$ only, nothing really changes, because we are replacing the group $(\mathbb{R},+)$ with $(\mathbb{R} _ +,\cdot)$, isomorphic to the former. – Pietro Majer Nov 14 '12 at 23:01
up vote 4 down vote accepted

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.

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That makes sense; I ran into that problem when I tried to make one for integers. Do you think any sort of system could be developed to simplify vector operations by way of space-filling curve? – Benjamin Hansel Nov 14 '12 at 22:45
Do they need simplifying? It seems easier to add vectors than to compute a space-filling curve. – Joel David Hamkins Nov 14 '12 at 22:49
I suppose... If an elegant sort of system could be developed, though, vector operations could be done with simple scalar operations. It seems silly that we need to use two pieces of information to represent something where we only need one. – Benjamin Hansel Nov 14 '12 at 22:54
@Benjamin: you still need 2. The curve is the other piece of information. – Jason Polak Nov 14 '12 at 23:58
Well, by the same logic, you could say you need three pieces of information for normal vectors, because you have to specify what coordinate system you're using. – Benjamin Hansel Nov 15 '12 at 0:19

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.

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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).

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Did you just say what I was gonna say, but smarter? – Benjamin Hansel Nov 14 '12 at 22:58
In other words, the fact that $\mathbb{R}^2$ is isomorphic to $\mathbb{R}$ as additive group (or even $\mathbb{Q}$-linear space) is the more obvious meaning I can imagine to give to the idea of "simplifying the addition operation of $\mathbb{R}^2 $ ". Of course, this isomorphism would completely ignore the other structures (scalar multiplication, topology, order) – Pietro Majer Nov 14 '12 at 23:17

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