# Extending the set of complex numbers

Mathematics as a science became richer when Cantor invented the real numbers. Then scientists wanted to solve equations which were not solvable in the real numbers so they invented the complex numbers. My question is, will we ever need to extend the complex numbers to some other set which will contain all other existing sets and help us for something? Are complex numbers extendable? Thanks in advance.

• Complex numbers were discovered before Cantor came around...Real numbers were already known way before Cantor. However if you want to credit Cantor, read about set theory for example – imranfat May 16 '14 at 19:25
• There's the quaternions, polynomial rings, function fields, other algebras over the complex numbers... – bradhd May 16 '14 at 19:27
• In most formulations of set theory, there isn't a "set that contains all other existing sets", by the way. This is the content of Russell's paradox. – bradhd May 16 '14 at 19:29
• Just for the fun of it, we could add the hyperreals to the discussion... – abiessu May 16 '14 at 19:45

To solve polynomials, the answer is no. The Fundamental Theorem of Algebra says that every (non-constant) polynomial, of degree $n$, with complex coefficients has $n$ roots (including repeated roots) in $\mathbb{C}$.

As a result of this, the set of complex numbers, unlike the set of real numbers, is algebraically closed, which means that we cannot 'escape' $\mathbb{C}$ using any elementary operations, like $+, - , \times, \div, \sqrt{}, e^{...}$ etc. So, in this sense, we don't really 'need' to extend the complex numbers.

However, there do exist hypercomplex numbers, like the quaternions, $\mathbb{H}$, which, instead of using just one imaginary unit $i$, use three: $i, j, k$, each satisfying:$$i^2=j^2=k^2=ijk=-1$$

Quaternions are used in modelling 3D vectors, and have a lot of use in 3D mechanics. A useful property of a quaternion is that it does not satisfy commutativity: e.g. $i\times j \neq j \times i$.

N0w, if you still want to go further, you can explore the octonions, $\mathbb{O},$ which are an extension of the quaternions. Octonions have 7 imaginary units: $e_1, e_2, ...e_7$.

Octonions have far fewer (that I know of, at least) practical uses that quaternions. Octonions also have the interesting property that they lack associativity: e.g. $x\times(y\times z) \neq (x \times y) \times z$ (for $x,y,z \in \mathbb{O}$).

And, finally, at least as far as mathematical interest has gone, we have the sedenions, $\mathbb{S}$. These are an extension of the octonions and have 15 imaginary units: $e_1,e_2,e_3,...,e_{15}$. The special property about the sedenions is that they have zero-divisors (meaning that there exist non-zero sedenions $x$ and $y$ such that $xy=0$). Interesting, this property, isn't it? Not particularly, intuitive, to say the least.

Now, I'll leave as an exercise for you to find out about the applications of hypercomplex numbers, and how to multiply them (hint: Google 'Fano plane mnemonic', which explains how to multiply octonions- this same idea can be extended to the sedenions).

• Also, you 'can' extend even these (hypercomplex) number systems to ones with an arbitrary number of elements (see this- tony5m17h.net/Voudons.html which talks about so-coined 'voudons', $\mathbb{V}$, 'chingons', $\mathbb{X}$, routons, $\mathbb{U}$, voudons, $\mathbb{V}$, etc.), but whether or not these have any practical uses is a different story. Also, bear in mind that, as you'extend' a number system, you lose algebraic properties the higher up the hypercomplex ladder (e.g. commutativity, associativity, etc.), so a hypercomplex number system, with, say, $2^{20}$ elements,... – beep-boop May 16 '14 at 20:04
• ...will probably not have the algebraic properties to be considered even remotely useful. – beep-boop May 16 '14 at 20:04
• There exist a hierarchy for these number systems? I mean if is possibile to order of all these systems, the equations they allow us to solve and algebraic properties that lose validity? – MphLee May 16 '14 at 22:01
• @MphLee I'm not sure what you're asking, but if it's along the lines of 'why do equations lose their validity higher up the hypercomplex hierachy', then that's not true- what I'm saying is, as you go further up the ladder, you lose essential properties like associativity, and going from $\mathbb{O}$ to $\mathbb{S}$, you no longer have a division algebra, so, if you go even further up, you're likely to lose so many properties (e.g. power-associaitvity) that the algebra that you're left with won't be very practical, due to its strange algebraic properties. – beep-boop May 16 '14 at 22:10
• There is no reason to restrict to non-constant polynomials. A non-zero constant polynomial has degree $n=0$ and $n=0$ complex solutions, so the statement is just as true for those. Only for the zero polynomial you need an exception. – celtschk May 17 '14 at 10:37

As an account of history the question is nonsense. Real numbers and complex numbers were thought about before Cantor's birth (he was born in the 19th century), and solving algebraic equations with no real solutions was not the motive.

Complex numbers were introduced in the 16th century in order to find real solutions of third-degree algebraic equations with real coefficients.

That sort of process for solving algebraic equations will not take you beyond the complex numbers, because a theorem, often misleadingly called the "fundamental theorem of algebra" says that every algebraic equation with complex coefficients has a complex solution.

But, as others have already noted here, there are other motives for extending the system of complex numbers. The quaternions are the most widely known example.

You can extend the complex numbers and get quaternions. Which can be used in, for example, 3D computer graphics instead of matrices. It is possible to extend the quaternions as many times as you want, like you did with the complex numbers to get the quaternions.

Others have already mentioned the quaternions. I just wanted to mention the surcomplex numbers, which, while lacking the property of being a set, satisfy all the properties of an algebraicly closed field. You get these numbers by taking the surreal numbers (a gigantic totally ordered field containing every totally ordered field in existence) and adjoining the element $i$, which is analagous to how you construct $\mathbb{C}$ by adjoining $i$ to $\mathbb{R}$.

Extending further is possible, but you loose certain nice properties. Quaternions are no more commutative, octonions no more associative. In higher dimensions, you cannot expect to have any kind of "reasonably nice multiplication" and "norm", because higher dimensional spheres are not fully paralelizable. It means, you cannot choose a basis of the tangent space of $1$ in the sphere $S^{n-1}$ of unit vectors and extend it to $n-1$ independent nonzero vector fields by left multiplication. (except from $n=1,2,4,8$; this is known from algebraic topology).

On the other hand, there are some nice structures in dimensions $2^n$, such as Clifford algebras, or the Cayley-Dickson construcion.

I think what you really want to know is the fact that there is no (strict) extension of $\mathbb C$ which is a field. (In fact, I think you can even say that the only field extensions of $\mathbb R$ are $\mathbb R$ and $\mathbb C$.)

Edit: For the record, the answer above is wrong as it stands (see comment by Asaf Karagila below). It should have said that there are no finite (meaning finite-dimensional) field extensions of $\mathbb C$. The same modification applies to the parenthetical remark in the original answer.

• That's plain false. There are no algebraic extensions, though. – Asaf Karagila May 16 '14 at 19:42
• Yes, you're right of course. I guess I should modify my answer. I really had in mind finite field extensions but was trying to use simple language. – MPW May 16 '14 at 20:14