What Number Set Contains The Subset of Complex Numbers? Is there even such a set? 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..
 A: 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.
A: 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 $y$).
