Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I was thinking about complex numbers and how they came about and someting interesting occured to me:

the formation of complex numbers occurs because there exists a function (namely $f(x)=x^2$) that maps reals to a smaller subset of reals and therefore the by setting the equation $x^2 = p$ in $\{x\in\Bbb R:x\not\in f(\Bbb R)\}$. We are forced to create a new mathemtical value, $i$ to handle the equation.

Couldn't this extension occur elsewhere? For example, say there is some function $f_{\Bbb C}$ that maps the complex numbers to a smaller subset of the complex numbers then the expression

$f_{\Bbb C}(x) = j$ in $\{z\in\Bbb C:z\not\in f_{\Bbb C}(\Bbb C)\}$

will require the creation of a new mathematical constant.

Is this possible? Has it ever happened? This differs greatly from quaternions and other tools which are considered extensions of the complex plane. The most basic idea that comes to mind is the absolute value function.

The equation: $|x| = c$ where $c$ is not a positive real number or $0$ but is a member of the complex plane.

You would get numbers $p$, such that $|p| = -1$, and $t$ such that $|t| = i$. From there the expression $p\cdot t$ would be such that $|p\cdot t| = -i$. Then an entire algebra from these can be constructed.

But the only reason I'm not satisfied with this answer is because Complex numbers were naturally developed from the basic hyperoperators (addition -- multiplication -- powers (produces complex numbers) --> tetration ---> pentation etc...) Is there some hyperoperator down the line that will produce the next complex number?

share|cite|improve this question
*edit to the last line, produce the next complex extension – frogeyedpeas Jan 15 '13 at 23:28

First of all we have the fundamental theorem of algebra which says that the the roots of polynomial functions with complex coefficients all lie in the field of complex numbers.

You are intimating that integers are required to solve

$x+a=b$ with $a>b$ are natural numbers.

Fractions are required to solve

$ax=b$ where $a,\,b\in\mathbb{N}$ with $a\neq 0$.

Real numbers are required to solve

$x^2=a$ with $a\in\mathbb{N}$


My answer to your question is don't disregard the quaternions so easily: which could be considered to be required to solve an equation of the form:


where $a\neq 0$.

share|cite|improve this answer
when considering the field (or is it just a set?) R: x^2 = -1, is not an invalid statement, it just has no well known solution whereas when considering the field C, the equation xy - yx = 7 ---> 0 = 7 which is an invalid statement in C. – frogeyedpeas Jan 16 '13 at 2:04
I want to be able to do a sort of extension of C without assuming anything about my extension. This happened naturally from R to C. There was nothing invalid in the statement x^2 = -1, it was perfectly well defined, obeyed all rules etc..., but has no solution in R. I wonder if there exists such an equation in C that is perfectly well defined, violated no basic rules yet has no solution nor limit of a solution in C. – frogeyedpeas Jan 16 '13 at 2:06
Once you put the structure of an ordered field on the real numbers $x^2=-1$ does not obey the proposition for real numbers that $x^2\geq 0$. Similarly $xy-yx=`7'$ does not obey the commutativity of the field complex numbers. – Jp McCarthy Jan 16 '13 at 2:40
sorry to necromance the point but could one make the following argument: – frogeyedpeas Feb 4 '13 at 21:43
x^2 = -1 when simplified over the real numbers is still x^2 = -1 whereas xy - yx = 7 when simplified over the complex numbers is 0 = 7, an invalid statement over the complex numbers? – frogeyedpeas Feb 4 '13 at 21:43

There are some things that come to mind when you ask this question. One example would be the Riemann Sphere - basically, you have the complex plane $\mathbb{C}$, and a very natural function like $f(z) = \frac{1}{z}$, and you have two basic problems:

  1. $f(z)$ is never $0$.
  2. $f(0)$ is undefined.

So you add the point $\infty$ to your plane and define the new plane's topology in a way that lets $f(z)$ be defined everywhere.

However, when you do that, you lose the beautiful algebraic properties of $\mathbb{C}$, as this is not even a field anymore. In fact, because $\mathbb{C}$ is an algebraically closed field, it has no finite extensions, meaning that if you want to extend $\mathbb{C}$ and still get a field, you will have to add infinitely many independent elements - and that wouldn't be "pretty", in much the same way that $\mathbb{R}$ is not a "pretty" extension of $\mathbb{Q}$ - it adds uncountable many constants, most of which don't have a nice representation.

Other functions have even more complicated and weirder Riemann surfaces that enable them to make more sense than on the standard complex plane, but all of these are of geometric nature and lack the beauty of the algebraic extension of $\mathbb{R}$ to $\mathbb{C}$

share|cite|improve this answer
When you say algebraically closed do you mean closed to all elementary operations or do you mean fully closed as in tetration (repeated exponents), pentation (repeated tetrations), etc... are all closed under C. – frogeyedpeas Jan 16 '13 at 2:09
@frogeyedpeas By algebraically closed we mean that every root of every polynomial in $\mathbb{C}$ is in $\mathbb{C}$. As for tetration, etc. - I'm not sure how this is defined for non-natural numbers, or what you mean by them being closed under $\mathbb{C}$. – Alfonso Fernandez Jan 16 '13 at 2:39

Picard's theorem states that any entire function on $ \mathbb{C}$ that omits more than 1 points is actually constant. i.e. either the range of the function is 1 point (function is constant), the range omits precisely one point (for example: $ x \rightarrow e^x$ is never $ 0$), or has as range all complex numbers (examples: all polynomials, $ \sin$ and $ \cos$, etc.)

so in your construction, we'd either end up with (generalisations) of quaternions, like Clifford algebras or higher order analogues, or we'd have to start with a non-holomorphic function (like you did with the absolute value function). Anyway, if you demand that it is a field, and contains the reals in some way, you'd end up with the reals, the complex numbers, the quaternions, or the octonions, due to Hurwitz's theorem, meaning that we either end up with something where we can't divide, or something already known.

share|cite|improve this answer
what exactly do you mean by constant? Is that to say that any entire function in C has all of C contained in its range? – frogeyedpeas Jan 16 '13 at 2:07
@frogeyedpeas: I added a sentence explaining that. Also the 2 points was mistaken, it should be 1 point. – MathJJ Jan 17 '13 at 23:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.