# Can 3D co-ordinates be transferred into 2D co-ordinates?

Is it possible to transform co-ordinates $$(a,b,c)$$ into $$(x,y)$$ such that $$(x,y)$$ is unique for each $$(a,b,c)$$ ? $$a, b, c, x, y$$ are in $$\Bbb{R}$$ .

• Just to clarify, do you want the map from $\Bbb R^3$ to $\Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection). – J.G. Mar 10 at 10:16
• I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks – BJKShah Mar 10 at 10:18

It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $$x=a$$, and $$y$$ is the alternating decimal expansion of $$b$$ and $$c$$. What that means is this:

Take the non-terminating decimal expansions of $$b$$ and $$c$$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $$b=1234.345$$ and $$c=76.987698798\ldots$$, we convert $$b$$ to its non-terminating form $$b=\color{red}{1234.344999999\ldots}$$, and we write $$c$$ as $$\color{blue}{0076.987698798\ldots}$$ to make the lengths before the decimal point equal.

Now we can write $$y$$ by taking digits from $$b$$ and $$c$$ in turn:

$$y=\color{red}{1}\color{blue}{0}\color{red}{2}\color{blue}{0}\color{red}{3}\color{blue}{7}\color{red}{4}\color{blue}{6}.\color{red}{3}\color{blue}{9}\color{red}{4}\color{blue}{8}\color{red}{4}\color{blue}{7}\color{red}{9}\color{blue}{6}\color{red}{9}\color{blue}{9}\color{red}{9}\color{blue}{8}\color{red}{9}\color{blue}{7}\color{red}{9}\color{blue}{9}\color{red}{9}\color{blue}{8}\ldots$$

This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.

Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.

Edited to add: As the OP points out in a comment, I have not define the mapping for $$b,c\le 0$$. My bad! But you can fix this by using the decimal expansions of $$e^b$$ and $$e^c$$, which are always strictly greater than zero.

• (1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example. – BJKShah Mar 10 at 10:34

It's is possible to have a bijection between $$\mathbb{R}^3$$ and $$\mathbb{R}^2$$, since it can be shown that $$\mathbb{R}$$ and $$\mathbb{R}^2$$, and hence every $$\mathbb{R}^n$$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.

• Can you please explain the last sentence a bit more? – BJKShah Mar 10 at 10:14
• Right. A map f:R^n\to R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n. – VENKITESH Mar 10 at 10:19
• Any good books for studying this? – BJKShah Mar 10 at 10:25
• Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra. – VENKITESH Mar 10 at 10:27