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Let $f: \mathbb R \longrightarrow [0,\infty[$ be some positive-valued function and consider the equation $$f(x) = -1.\tag{1}$$ Looks weird, but for $f(x)=x^2$ one has $x^2=-1$ which leads to the set of complex numbers by defining $\mathbb i$ to be the solution. Now consider other simple algebraic functions like $f(x)=\sqrt{x}$ or $f(x)=|x|$, such that (1) cannot be solved by real or complex $x$. By defining $\mathbb j$ such that $$f(\mathbb j)=-1$$ we get a new set $\mathbb J$ of numbers $a + \mathbb j b, \ a,b \in \mathbb C$.

Complex numbers led to a new era of mathematics. I like thinking outside of the box.

Let us collect some ideas about this approach. Is it known? Is it nonsense, inconsistent, non-valuable? If not, what properties do these numbers have?

Edit:

When $f$ is a polynomial (1) leads to $\mathbb C$.

Let $f$ be arbitrary and $\mathbb J =\{a + \mathbb j b \ |\ a,b \in \mathbb C\}$. To make it a field we need $x\cdot y \in \mathbb J$ for all $x,y \in \mathbb J$. This requires either

  1. $\mathbb j^2 =0$,

  2. $\mathbb j^2 \approx \mathbb j$ or

  3. $\mathbb j^2 \in \mathbb C$.

For the latter let wlog $|\mathbb j^2|_{\mathbb C}=1$ such that $\mathbb j^2 = \exp(\mathbb i \theta)$, for some $\theta \in [0,2\pi]$. By the series expansion we get $$\exp(\mathbb j z)= \cosh(z \exp(\mathbb i \theta/2)) + \mathbb j \sinh(z \exp(\mathbb i \theta/2)), \ z\in \mathbb C.$$

For the second let wlog $\mathbb j^2 =\mathbb j$. If $f$ has a series expansion around $\mathbb j$ then $$f(\mathbb j)= f(0)+ \mathbb j (f(1) -1),$$ which requires $f(0)=-1$ and $f(1)=1$.

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("Ring" means commutative ring with identity.)

The function $f$ doesn't matter. What does matter is that you demanded that $\mathbb{J}$ be generated by linear combinations of $(1,\mathbb{j})$ over its base ring.

Let $k$ be a ring, let $R$ be a ring equipped with an injective homomorphism $k\to R$, and let $\alpha$ be an element of $R$. Then if $k[\alpha]$ is finitely generated over $k$ with at most $n$ generators, there must be a relation $$\alpha^n=c_1\alpha^{n-1}+c_2\alpha^{n-2}+\cdots+c_n$$ where $c_1,\ldots c_n\in k$ (and vice versa, if there is such a relation then $k[\alpha]$ is generated over $k$ with at most $n$ generators). We say that $\alpha$ is integral over $k$. This notion goes at least as far back as Dedekind (1877).

If the base ring is $\mathbb{R}$ we must have some relation $$\mathbb{j}^2=c_1\mathbb{j}+c_2$$ for $c_1,c_2\in \mathbb{R}$. However, without loss of generality we can shift and rescale $\mathbb{j}$ by elements of $\mathbb{R}$ to find that there are only three inequivalent cases corresponding to negative, zero, and positive discriminant:

If the base ring is $\mathbb{\mathbb{C}}$ then there only two cases, corresponding to zero and nonzero discriminant:

  • $\mathrm{\varepsilon}^2=0$. These are "dual complex numbers", although the terminology is overloaded.
  • $\mathrm{j}^2=1$. These are the bicomplex numbers, which Cockle (1848) introduced as tessarines.

Other avenues of generalization with a similar flavor include the composition algebras and their Cayley–Dickson construction (Dickson (1919)), Clifford algebras (Sylvester (1882)), and Jordan algebras (Albert (1946), McCrimmon (1966)).

You'll note that most of the literature is from the 19th century. During the latter half of this period, there was a time when constructing and naming new number systems was the fashion, giving us exotic-sounding names: nonions, sedations, etc. In a sense, Hurwitz's theorem (1923) put an end to this industry, demonstrating that only four such systems (reals, complexes, quaternions, and octonions) really have the norm-preserving multiplication needed to get their calculus off the ground.

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  • $\begingroup$ Really good answer. If think $\mathbb j^2 = \mathbb j$ is another case, but not promising as shown in my edit. $\endgroup$ – fwgb Sep 21 at 16:02
  • $\begingroup$ That isn't a separate case. You get an algebra isomorphic to the split-complex numbers (with scalars from $\mathbb{R}$) or bicomplex numbers (scalars from $\mathbb{C}$), because $x$ is a solution of $x^2=x$ if and only if $y=1-2x$ is a solution of $y^2=1$. $\endgroup$ – K B Dave Sep 21 at 16:09
  • $\begingroup$ Okay, i see it! $\endgroup$ – fwgb Sep 21 at 16:14
  • $\begingroup$ ' norm-preserving multiplication ' is an important point. Already for split-complex numbers we dont have it, since $\|\cdot\|$ extended is not a norm anymore. $\endgroup$ – fwgb Sep 21 at 16:19
  • $\begingroup$ What about the case $\mathbb j^2 = \mathbb i$? $\endgroup$ – fwgb Sep 21 at 17:11

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