# Complex numbers - are they the 'ultimate', or are there any complex complex numbers

Complex numbers - are they the 'ultimate', or are there any "complex complex" numbers

When we try to calculate the root of a negative number, we come to the idea to introduce complex numbers. Is there any operation for which complex numbers wouldn't be enough, so there's a need to introduce complex complex numbers, complex ^ n numbers, etc?

cristo
Staff Emeritus

When we try to calculate the root of a negative number, we come to the idea to introduce complex numbers. Is there any operation for which complex numbers wouldn't be enough, so there's a need to introduce complex complex numbers, complex ^ n numbers, etc?

• John M. Carr
D H
Staff Emeritus

... or the octonions, or sedenions, and so on. Every power of 2 yields a different algebraic structure. Something is lost with each step. The complex numbers are not orderable, the quaternions are not commutative, the octonions are not associative, the sedonions are not alternative, and so on.

The complex numbers fill an algebraic hole in the reals. The real numbers form a complete field, but they are not algebraically closed (there is no real solution to x2+1=0). The complex numbers fills that gap. Unlike the reals, complex numbers are algebraically closed. Like the reals, the complex numbers form a field. Moreover, the complex numbers are algebraically closed. Those larger algebraic structures, unlike the complex numbers, do not fill a hole. They are instead invented structures motivated by the complex numbers.

• John M. Carr
Hurkyl
Staff Emeritus
Gold Member

When we try to calculate the root of a negative number, we come to the idea to introduce complex numbers. Is there any operation for which complex numbers wouldn't be enough, so there's a need to introduce complex complex numbers, complex ^ n numbers, etc?
There are lots of different number systems we use, geared for different purposes.

You have learned the complexes, which contain the reals, which contain the rationals, which contains the integers, which contain the natural numbers... but don't be mislead by this: the number systems aren't arranged in a neat hierarchy like this. Instead, they branch off in all sorts of directions.

Probably the most common number systems in use, other than the ones you already explicitly know, are:

. The cardinal numbers
. The ordinal numbers
. The extended real numbers
. The projective real numbers
. The projective complex numbers

I would assert things like
. Polynomial rings
. Vector spaces
. The "algebra" of all matrices
. The "algebra" of all Abelian groups
are essentially number systems too, although I would probably get more disagreement about that.

gel

If you start with natural numbers and try adding properties, you end up at the complex numbers.
- If you want to be able to subtract any number from any other, you have to add negative numbers to get the set of integers.
- If you want to divide, you need the rationals.
- If you want to take limits, you need the reals.
- To take square roots (or solve general polynomials), then you have to go to the complex numbers.

The complex numbers are a complete and algebraically closed field, so there's not really much reason to go further. It is possible to use quaternions and octonions. I don't know if you can derive them by trying to complete the complex numbers under some operation, although you do lose properties.
With quaternions you lose commutativity, ab != ba.
With octonions you also lose associativity, (ab)c != a(bc).

matt grime
Homework Helper

... or the octonions, or sedenions, and so on. Every power of 2 yields a different algebraic structure.
No it doesn't, not in any real sense. It stops there I believe - only a few powers of two yield decent algebraic structures (a result of differential geometry I seem to rememeber to do with the possible space of vector fields).

EDIT - I think that the 'real sense' means something like you lose enough structure for it not satisfy some property - I can't find the result I'm thinking of. There are of course algebras of arbitrary degree over C.

EDIT EDIT I have it - proved by Bott Periodicity, the only division algebras over R are C, quarternions and octonions.

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D H
Staff Emeritus

No it doesn't, not in any real sense. It stops there I believe - only a few powers of two yield decent algebraic structures (a result of differential geometry I seem to rememeber to do with the possible space of vector fields).
The Cayley–Dickson does go on forever, but as you noted, the utility of the constructed algebras pretty much comes to an end because each of the first few steps takes away some very useful property. There isn't much algebraic structure left beyond the octonions. The sedenions aren't even alternate but are still power associative. Beyond that, about all that is left is the ability to construct a conjugate and a norm.

gel

The sedenions aren't even alternate but are still power associative. Beyond that, about all that is left is the ability to construct a conjugate and a norm.
Wikipedia has an entry, http://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction" [Broken]. According to them, the algebras are all power associative, and not just up to the sedenions.

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