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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 $\Bbb{R}$ .

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  • $\begingroup$ 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). $\endgroup$ – J.G. Mar 10 at 10:16
  • $\begingroup$ I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks $\endgroup$ – BJKShah Mar 10 at 10:18
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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.

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

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  • $\begingroup$ Can you please explain the last sentence a bit more? $\endgroup$ – BJKShah Mar 10 at 10:14
  • $\begingroup$ 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. $\endgroup$ – VENKITESH Mar 10 at 10:19
  • $\begingroup$ Any good books for studying this? $\endgroup$ – BJKShah Mar 10 at 10:25
  • $\begingroup$ Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra. $\endgroup$ – VENKITESH Mar 10 at 10:27

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