Using subtract operation with natural numbers would yield integers.
Similarly, using division operation with integers would yield rational numbers.
Then, applying division operation again with rational numbers would yield real numbers.
Continuing, applying exponent operation with real numbers would yield complex numbers.

But from here, what operation over complex numbers would yield hypercomplex numbers? Or if none, is this an artificial extension of complex numbers?

  • 1
    $\begingroup$ (1) division only gives you rational numbers. not reals. (2) in what way does "applying exponent" get you from reals to complex? not clear what you mean here. $\endgroup$
    – Simon
    Jan 13, 2022 at 8:37
  • 1
    $\begingroup$ If by hypercomplex numbers you mean quaternions, then roughly the answer is yes, because one of the main interpretations of quaternions by Hamilton and others who worked with them in the 1800s was as a kind of division of two-dimensional vectors, which in turn correspond (in certain ways) to complex numbers. See the various (freely available) items I cite in my answer to What set of criteria led Hamilton to discover the quaternions? $\endgroup$ Jan 13, 2022 at 8:43
  • $\begingroup$ @Simon (2) by getting to complex, I meant getting out of real. For example (-1)^1/2 where domain is real, it gives non-real i as result which indeed is a complex. $\endgroup$ Jan 13, 2022 at 8:44
  • $\begingroup$ @Simon (1) edited the question now. Verify it :) $\endgroup$ Jan 13, 2022 at 8:58
  • $\begingroup$ Have a look at this answer of mine here $\endgroup$
    – Jean Marie
    Jan 13, 2022 at 9:02

2 Answers 2


from real to complex

You state that "taking the square root of $-1$" gets you out of the reals and into the complex numbers. That is kinda true, but also misleading. Without complex numbers, the "square root" functions is simply only defined for non-negative real numbers, so there is simply no way to take the square root of negative one. The same way as division $a/b$ is only defined for $b\neq 0$.

It is cleaner to state the process like this: We are looking for an extension of the real numbers such that:

  • The extended set of numbers still forms a "field" (i.e. laws like commuativitiy and associativity still hold)
  • There is a number "$i$" such that $i\cdot i=-1$.
  • The extended field should be as small as possible given these previous conditions.

It turns out, that this has a unique answer, which are the complex numbers.

from complex to quaternions

Here again, you are asking for an extension. So you have to state what new objects you want to have, and what properties you want to preserve. Here is one version of that:

  • we want numbers $i,j,k$ such that \begin{align} (ai+bj+ck)^2 = \text{some constant} \cdot (a^2 + b^2 + c^2) \end{align} for all real numbers $a,b,c$. You could either view this as an artificial challenge, but actually, problems like these do turn up in particle physics for example (related to "spin").

Now the problem is, that it is mathematically impossible to have these numbers $i,j,k$ and also preserve all properties of a field. But if you are willing to give up commutativity (while preserving associativity and the law of distribution), the unique smallest extension is the quaternions.

  • $\begingroup$ Quaternions are absolutely not a unique extension. And you can extend complex numbers without sacrificing commutativity in many ways. $\endgroup$
    – Anixx
    May 21, 2022 at 19:14
  • $\begingroup$ @Anixx Well, it depends what kind of structure you want exactly. As I stated it is impossible to have the numbers $i,j,k$ with the stated property in a field, so we have to give up some property of a field. If we only give up commutativity, we essentially have a "divison algebra". And then the Frobenius theorem states that there are exactly three different finite-dimensional division algebras over $\mathbb{R}$, namely (1) the real numbers itself, (2) complex numbers, and (3) quaternions. In this sense, quaternions are unique. But of course there was a choice in which property we gave up. $\endgroup$
    – Simon
    May 22, 2022 at 12:28

Solving equation $|z|=i$ is impossible in complex numbers, but has two solutions in tessarines, which is an extension of complex numbers. The solutions are $j$ and $-j$, where $j$ is hyperbolic (split-complex) unity. This depends on the generalization of modulus function to tessarines though.

The canonical definition is $|a+bi+cj+dij|=\sqrt{a^2+b^2+d^2-c^2}$.


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