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Basically what I'm asking is what set are complex numbers inside of? Surely there must be a set that encompasses complex numbers and so on.

In my pre-calculus book from senior year high school the most outer set taught was the complex number set in the form $a + bi$. This set contained integers, fractions, imaginaries, etc..

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    $\begingroup$ Well there's $\mathbb{H}$, en.wikipedia.org/wiki/Quaternion $\endgroup$ – Christopher Liu Dec 12 '13 at 13:54
  • $\begingroup$ ...or $\mathbb{R}^n$. $\endgroup$ – Lucian Dec 12 '13 at 13:56
  • $\begingroup$ Interesting. I've been using quaternions in AP physics mechanics this whole time without knowing it (as unit vector notation.) How are H and R^n related? Is one inside of the other? $\endgroup$ – Armend Veseli Dec 12 '13 at 14:00
  • $\begingroup$ Cross-products do, in a sense, apply the structure inherent to quaternions. After those, you have octonions, and I believe that there is a theorem stating that there are no "new" extensions beyond that. $\endgroup$ – Omnomnomnom Dec 12 '13 at 14:03
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    $\begingroup$ Ha! this is that theorem. $\mathbb{R} \subset \mathbb{C} \subset \mathbb{H}$ are the only associative normed division algebras over the real numbers. If you give up associativity, you have the octonions $\mathbb{O}$ and the sedenions $\mathbb{S}$. $\endgroup$ – Omnomnomnom Dec 12 '13 at 14:12
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Well, there's the quaternions, discovered by Hamilton in the mid-1800s. These are the numbers of form $a + bi + cj + dk$, where $i, j, k$ satisfy the following relations: $$ i^2 = j^2 = k^2 = ijk = -1 $$ from which we can derive $$ ij = k, jk = i, ki = j $$ with anticommutation of any two of $i, j, k$. So, multiplication is not commutative over these numbers. Similarly, the quaternions embed into the octonions, but beware that multiplication over these is not even associative.

If you want the extension to be a field (note that the quaternions and octonions are not fields, being noncommutative), there are certain constraints on what the field you're embedding $\mathbb{C}$ into can be. Because the complex numbers are algebraically closed, meaning that any polynomial with complex coefficients factors as a product of its roots and maybe another scaling term, you're not going to be able to find a finite field extension of the complex numbers (meaning, loosely, that you can never extend $\mathbb{C}$ to numbers of form $a_0 + a_1\alpha + \ldots + a_n\alpha^n$ with $\alpha$ satisfying any algebraic relations over $\mathbb{C}$. This is your representation of $\mathbb{C}$: numbers of form $a + b\alpha$ where $\alpha^2 + 1 = 0$). So, any thing $\alpha$ you adjoin to $\mathbb{C}$ to get something strictly larger than $\mathbb{C}$ can't satisfy any algebraic relations over $\mathbb{C}$. One such thing is $\alpha = t$, an indeterminate. This gives you $\mathbb{C}(t)$, the field of rational functions with complex coefficients.

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    $\begingroup$ discovered or invented ? :-) $\endgroup$ – lhf Dec 12 '13 at 14:09
  • $\begingroup$ @lhf summoned? $ $ $\endgroup$ – Julien Clancy Dec 12 '13 at 14:12
  • $\begingroup$ ^Seems like a Philosophical argument... $\endgroup$ – Armend Veseli Dec 12 '13 at 14:16
  • $\begingroup$ @ArmendVeseli, yeah, sorry for the noise. $\endgroup$ – lhf Dec 12 '13 at 14:21
  • $\begingroup$ @JulienClancy : I'd go with discovered. Gauss discovered them before Hamilton, but I don't know how much he did with them. Of course, Gauss also discovered the Fast Fourier Transform ages before Cooley and Tukey, but again, I don't know how much he did with the FFT. Same story with hyperbolic geometry. $\endgroup$ – Stefan Smith Dec 12 '13 at 18:22
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The complex numbers form one natural stopping point. To progress from naturals to integers you ask for solutions to certain linear equations which did not exist before ($x+n=0$). To go from integers to rationals you ask for solutions to all linear equations $px-q=0$. To go from rationals to reals you ask that you have all limits you could want ($3,3.1,3.14,3.141,3.1415,...$ should approach some kind of number). Another way to say this is that $\mathbb{R}$ is "complete" Finally for the complex numbers you ask that $x^2+1=0$ should have a solution. It turns out that for complex numbers, all polynomials equations have solutions, and it is complete. In fact, it is the only field (number system) containing the integers which is complete and has solutions to all polynomial equations.

In the comments someone mentions the quaternions as a possible extension, but it is up to personal taste whether you really consider them "numbers" or not. They are not commutative for instance (do not have $xy = yx$ for all $x$ and $y$).

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