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.
[Math] Extending the set of complex numbers
complex numbersreal numbers
Best Answer
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).
See http://en.wikipedia.org/wiki/Hypercomplex_number