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I know there are at least two questions on this site that ask about the negative dimensions. But I want to ask something more than that.

We have a number line. It contains all the real numbers we can imagine of. But there are also complex numbers like (1+5i) that extend the number line into a complex plane. So, the number line that contains all possible number sets, not only the real ones is really a 2d plane, not a line.

We can say the same for a 2d graph. It contains all the real numbers on both the $x$ and the $y$ axis. But if we make the 2d function so that it can hold not only the real number pairs, but also the imaginary number pairs like (1-2i, 9+14i), we can see that a 2d graph really is a 3d graph.

Same thing for 3d. If we have an 3d pair (1+i, -5, -3-9i), then we extend the function to 4d. It is still 3d function because it has three coordinates. It only has complex numbers as its coordinates, which implies that 3d function has 4 dimensions.

So as we can see, (theoritecally) an $n$ dimension graph has $n+1$ dimensions.

Is this thinking correct, and does that mean, that $0$th dimension is realy a point extended into an imaginary-only, not real-valued line, and the true $0$th dimension is the $-1$st dimension? Does my reasoning even make sense at all?

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    $\begingroup$ But if each complex number is two-dimensional, shouldn't an ordered pair of them $(a + bi, c + di)$ be four-dimensional? It takes four real numbers to specify such an ordered pair. $\endgroup$ – pjs36 Apr 20 '16 at 4:11
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    $\begingroup$ Following @pjs36, moving to complex numbers doubles the number of dimensions. $2 \times 0 = 0$, so taking a $0$ dimensional point (with no real coordinates) and making it a complex point (with no complex coordinates) leaves it $0$ dimensional. $\endgroup$ – Ross Millikan Apr 20 '16 at 4:16
  • $\begingroup$ Wait... does it really double? I was thinking wrong. I dont have time to edit the post now because I have to go something important, but I will edit it in couple of hours... $\endgroup$ – KKZiomek Apr 20 '16 at 4:19
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No, I'm afraid it's nonsensical / not correct.

The notation $\mathbb{R}$ stands for the set of real numbers, and $\mathbb{C}$ for the set of complex numbers.

You are correct that $\mathbb{R}$ is one-dimensional, and that $\mathbb{C}$ is two-dimensional. However there is no precise mathematical object that your phrase

"the number line that contains all possible number sets"

refers to; if you take it to mean $\mathbb{R}$ then the dimension of the "number line" is 1, and if you take it to mean $\mathbb{C}$ then the dimension of the "number line" is 2. These facts are not contradictory in any way to each other.

But your bigger mistake is that $\mathbb{C}^2$ (the space of pairs of complex numbers) is four-dimensional, not three-dimensional, and more generally,

  • $\mathbb{R}^n$ (the space of $n$-tuples of real numbers) is $n$-dimensional
  • $\mathbb{C}^n$ (the space of $n$-tuples of complex numbers) is $2n$-dimensional

That is because there are four "degrees of freedom" (dimensions) in choosing a pair of complex numbers: $$(a+bi,c+di)$$ (the numbers $a$, $b$, $c$, and $d$ can all be chosen independently).

Note that when $n=0$, these coincide; that is, each of $\mathbb{R}^0$ and $\mathbb{C}^0$ is a single point. You can verify that $\mathbb{R}^0$ consists of the single $0$-tuple of real numbers $()$, and $\mathbb{C}^0$ consists of the single $0$-tuple of complex numbers $()$.

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