# Unconventional hypercomplex numbers

I was learning a lot about hypercomplex numbers lately. I've seen articles about complex numbers, double numbers, dual numbers, binarions, quaternions, octonions etc.

But one thing in common about all these is, first basis in all of these is 1. Why so? For eg, can't I have a number with basis $$\{ i, j, k \}, i^2 = j^2 = k^2 = ijk = -1$$? Also, why all hypercomplex numbers have a dimension of power of 2 over reals? Is 3, 5, or 6 dimensional algebra not possible?

Also, why only consider $$i^2$$? Can I define a number of the form $$a+ib, i^3=-1, i^2=0, i\neq-1, i\neq0$$. Is this a new number? or is it isomorphic to some existing system?

• Note that with every extension, you lose a nice property : The complex numbers have no order , the quaternions are not multiplicatively commutative and so on. In the case of the complex numbers however , we gain a nice property , a field being algebraically closed. Mar 28, 2022 at 11:00
• @Peter Since you were talking about algebraic closure, see this question of mine. Mar 28, 2022 at 11:04
• Even exponentiation is problematic for non-real numbers. The main purpose of tetration is producing extremely large numbers, and there are interesting properties for the infinite power tower $x\uparrow x\uparrow x\cdots$ for positive real numbers $x$. But what should $3\uparrow \uparrow \frac{1}{2}$ be ? Or even $i\uparrow \uparrow i$ ? I do not think that we can make this meaningful. Mar 28, 2022 at 11:07
• This question has three different questions inside of it. I bet at least one has been answered elsewhere on this site. Can you focus on just one per post? Mar 28, 2022 at 11:30
• @SouravKannanthaB I tried to write a single answer where I didn't retread concepts too much when addressing different parts. Let me know if anything is unclear or if you have other questions about it. Apr 1, 2022 at 11:46

What most of the algebras that you mentioned have in common is that they are all examples of unital algebras. A unital algebra is an algebra with a multiplicative identity element, and that multiplicative identity element is denoted with the symbol $$1,$$ though this is a bit of an abuse of notation. We really should denote it as $$\vec1.$$

I do not know what the double numbers are, and the binarions are just another name for the complex numbers. The dual numbers are an alternative to the complex numbers. Consider the the vector space $$\mathbb{R}^2$$ with basis $$\{e_0,e_1\}.$$ To form an algebra, we want a bilinear product $$\cdot,$$ and so if we can determine the products of the basis vectors, bilinearity is sufficient to determine all the products in the vector space. Bilinearity, to put it simply, is a more technical way of addressing distributivity. We also want the algebra to be unital, i.e, that is, for $$e_0\cdot{x}=x\cdot{e_0}=x.$$ Hence we know $$e_0\cdot{e_0}=e_0$$ and $$e_0\cdot{e_1}=e_1\cdot{e_0}=e_1.$$ The remaining product to determine is $$e_1\cdot{e_1}.$$ We know that $$e_1\cdot{e_1}=e_1^2=ae_0+be_1,$$ so $$e_1^2-be_1-ae_0=0.$$ We can complete the square, so that $$e_1^2-be_1+\frac{b^2}4e_0=\left(\frac{b^2}4+a\right)e_0=\left(e_1-\frac{b}2e_0\right)^2.$$ Since $$\frac{b^2}4+a$$ are themselves real numbers, this limits what the possible products $$e_1^2$$ can be. The three different cases are $$\frac{b^2}4+a\lt0,$$ $$\frac{b^2}4+a=0,$$ and $$\frac{b^2}4+a\gt0.$$ These three cases determine the three possible unital algebras over $$\mathbb{R}^2$$ up to isomorphism. These cases give the complex numbers, the dual numbers, and the split-complex numbers, respectively.

Of course, nothing is limiting you to make your algebra unital. You can have other number systems, those systems are just not named and not particularly interesting as extensions of the real numbers. But you can study them and they are perfectly valid structures to work with.

Now, let us talk about quaternions and octonions. You should look into the Cayley-Dickson construction. The construction is a construction that systematically extends the real numbers to the complex numbers, and applying the construction again to the complex numbers produces the quaternions, and applying it to the quaternions produces the octonions, and so on. Thus, the reason these algebras have a dimension that is a power of $$2$$ is because of the way the Cayley-Dickson construction works.

But if you are willing to work with algebras not built from the Cayley-Dickson construction, then there is absolutely nothing wrong with using algebras of other dimensions. On YouTube, there is a very good video about the triplex numbers, a unital associative and commutative algebra over $$\mathbb{R}^3.$$ The basis is $$\{1,i,j\}$$ and one has $$i\cdot{j}=j\cdot{i}=1,$$ and $$i^2=j$$ and $$j^2=i.$$ This actually equivalent to making the basis $$\{1,i,i^2\}$$ and noting the rule that $$i^3=1,$$ thus uniquely determining the multiplication table.

Now, to answer your last question: why only consider $$i^2$$? Well, not all algebras focus on only $$i^2,$$ as shown above. However, if you do uniquely determine $$i^2=r$$ where $$r\in\mathbb{R},$$ then you cannot separately control what $$i^3$$ is, because then $$i^3=ri,$$ necessarily. You cannot have it any other way. In general, if $$i^n=r$$ with $$r\in\mathbb{R}$$ for some $$n\in\mathbb{N},$$ then $$i^{m+n}$$ is automatically determined for all $$m\in\mathbb{Z},$$ so you have no freedom to choose those individually. In the example you gave, you suggested, $$i^2=0,$$ which makes your algebra equivalent to the dual numbers. Thus, it is already impossible that $$i^3=-1,$$ because $$i^3=i\cdot{i^2}=i0=0.$$ You can have some number system where $$i^3=-1$$ and $$i^2=0,$$ but that number system will not be an algebra over anything, and the multiplication will not distribute over addition. This opens up a whole can of worms beyond the scope of your question.

If you want more on the subject, you should read on the subject of Clifford algebras, the theory that unifies all of these concepts together.

EDIT 1: I would also like to point out that if you want to get really creative with number systems, you are not limited to algebras over $$\mathbb{R}.$$ You can form algebras over any field. Why not try making $$2$$-dimensional algebras over the field with two elements $$\mathbb{Z}_2$$? What about algebras over $$\mathbb{Q}$$?

• Woww, this is really a great answer for the question. Thanks for such an answer. I will see about unital algebra. I've seen a bit about Clifford algebras but have to see in depth that also. Mar 28, 2022 at 19:45
• @SouravKannanthaB By definition, no. By definition, an algebra is a vector space $A$ over a field equipped with a bilinear product $A^2\to{A}.$ If the product is not bilinear, then the algebraic structure in question is, by definition, not an algebra. And if the product is not bilinear, then you need to provide a precise definition by what is even meant by the word "product" in that case. Mar 30, 2022 at 14:17
• @SouravKannanthaB The cross product on $\mathbb{R}^3$ is bilinear. I have no idea what a non-bilinear product looks like for a vector space. Mar 30, 2022 at 15:31
• @SouravKannanthaB When I think of the word "product", I think "distributes over addition", which is going to end up bilinear in all reasonable cases. H-space operations are kind of like a "product" that needn't be bilinear, but it turns out that basically the relevant ones come from bilinear products anyway. So the question remains, "what would you mean by product of vectors if not something that distributes over addition?" Apr 1, 2022 at 1:12
• @MarkS. That's good point. I was just thinking product as a second operation '*' after addition. Apr 1, 2022 at 8:02

# Overview

Given the examples of number systems mentioned in the OP, and the use of the word "basis", I'm only going to discuss number systems that are vector spaces with scalar multiplication by real numbers, where multiplication satisfies the distributive law on both sides and scalar multiplication commutes with vector multiplication (so we don't have to worry about $$i(3j)$$ potentially being different from $$3(ij)$$).

Confusingly, such a number system is called an "algebra" (over the reals). For certain parts of the discussion below, I may add in extra assumptions like being finite-dimensional or being "unital" (having a multiplicative identity).

# Q1: $$1$$ in the basis?

## Attempt at non-real basis

For eg, can't I have a number with basis $$\{ i, j, k \}, i^2 = j^2 = k^2 = ijk = -1$$?

Suppose you make these numbers based on that equation. Since you wrote $$ijk$$ rather than $$(ij)k$$, I'll assume multiplication is associative. Unfortunately, the proposed basis $$\{i,j,k\}$$ can't be a basis; the reason is that if you do a bunch of algebra, you can't express everything that you've said exists (e.g. $$-1$$) in the proposed basis.

### $$i,j,k$$ are like quaternions

Multiplying $$ijk=-1$$ by $$i$$ on the left yields $$jk=i$$. Multiplying it by $$k$$ on the right yields $$ij=k$$, so that (multiplying by $$k$$ on the left) $$kij=-1$$. Doing analogous manipulations further yield $$jki=-1$$ and $$ki=j$$.

For the out-of-order products, note that $$-(k(ij))(ji)=-k(i(jj)i)$$, but the left side simplifies to $$-k$$ and the right side simplifies to $$ji$$. Similarly, $$ik=-j$$ and $$kj=-i$$. All together, this means that $$i,j,k$$ multiply together like quaternions.

### $$\{i,j,k\}$$ is not a basis

We have $$-i^2=1$$, so $$1$$ is certainly one of these numbers. But if the basis is supposed to be $$\{i,j,k\}$$, then we need $$1=ai+bj+ck$$ for some real numbers $$a,b,c$$. Multiplying by $$i$$ on the right yields $$i=-a-bk+cj$$. $$i$$ is supposed to square to $$-1$$, but $$(-a-bk+cj)^2=a^2-b^2-c^2+2abk-2acj$$. This means that $$-1-a^2+b^2+c^2=2abk-2acj$$. In other words, either $$a=0$$ (so we had $$1=bj+ck$$ to begin with), or $$b=c=0$$ (so we had $$1=ai$$ to begin with, which would mean $$(1/a)^2=-1$$, which is impossible), or at least one of the coefficients on the right is nonzero.

In all possible cases, we have an equation of the form $$A=Bj+Ck$$ for real numbers $$A,B,C$$ with $$B,C$$ not both zero. Then, multiplying by $$k$$ on the right and left, we have $$Ak=Bi-C$$ and $$Ak=-Bi-C$$, so that $$B=0$$. But doing the same thing on the right yields $$C=0$$, contradicting our assumption that $$B$$ and $$C$$ aren't both zero.

## Why 1 in basis?

But one thing in common about all these is, first basis in all of these is 1. Why so?

In general, if you're going to define an algebra by equations, you're often* going to have an equation with a nonzero real number in it. Once you do, that implies that the real numbers sit inside your algebra, and so it's a convenient choice (it will simplify a bunch of algebra) to take $$1$$ as a basis element.

You're certainly welcome to write the quaternions in a form like $$ai+bj+ck+dijk$$ or $$ai+bj+cij+d((ij)^2+(ji)^2)$$, etc. But that's going to make the algebra a lot uglier than just having $$1$$ as a basis element.

### Exceptions?

Now, I said "often", but you could, in theory, define something like "the droll numbers" with basis "{$$\varepsilon$$}" and the defining equation $$\varepsilon^2=0$$. Then every number is of the form $$r\varepsilon$$, every product of droll numbers is $$0$$ (which makes it a "zero algebra" - in particular, there's no multiplicative identity), and every sum of droll numbers is like a sum of real numbers except that there are $$\varepsilon$$s attached.

But an algebra without a multiplicative identity is not convenient; and if an algebra has/you add in a multiplicative identity to make it unital, you may as well call the identity "$$1$$" and make it a basis element.

# Q2: Other Powers

## Attempt in OP

Can I define a number of the form $$a+ib, i^3=-1, i^2=0, i\neq-1, i\neq0$$. Is this a new number? or is it isomorphic to some existing system?

The distributive law forces products with $$0$$ to be $$0$$, so $$i^3=i(i^2)=i*0=0\ne-1$$. This contradiction shows that we must be careful when picking equations to define an algebra.

## Other Approaches

Also, why only consider $$i^2$$?

This is largely just a matter of how these things are grouped together so you're only seeing the $$i^2$$ ones. That said, in the special case of a single basis generator, there are mathematical reasons why $$i^2$$ would deserve special attention.

### Why do we only see $$i^2$$?

The short/main answer is that, for historical reasons, a certain chunk of the algebras focused on $$i^2$$ are the ones called "hypercomplex numbers" (in part in analogy with $$i^2=-1$$ in the complexes), and lots of other algebras exist; they're just not part of that "hypercomplex" umbrella. Those other ones might be referred to with other names/be part of other collections, or have no special name at all.

### Why are other powers less accessible?

One class of ways to make algebras, which certainly allows powers other than squares, is to use a single polynomial equation to define the algebra, like $$i^3=1$$ (so that $$\{1,i,i^2\}$$ is a basis). For convenience, we can move things to one side so that the algebra is determined by a single polynomial, like $$x^3-1$$. This sort of construction (with whatever polynomial) forms a quotient ring of the polynomial ring $$\mathbb R[x]$$.

If you pick a constant polynomial, then you're saying $$0=0$$ (and you just get the infinite-dimensional space of all polynomials) or you're saying $$a=0$$ for some nonzero real, which is a contradiction. If you pick a linear polynomial like $$ax+b$$, then $$x=-b/a$$ and you've actually constructed the reals. And unfortunately, if you pick a polynomial of degree higher than $$2$$, then you always get "zero divisors" like $$(i+1)(i^2-i)=0$$ in my $$i^3=1$$ example, because polynomials of degree higher than $$2$$ can be factored into lower degree polynomials with real coefficients. See, e.g., this answer to Show that if $$p(x)$$ is reducible in $$F[x]$$, then $$F[x]/(p(x))$$ is not an integral domain. for a proof.

So the only interesting case that doesn't have annoying zero divisors is the case of a quadratic polynomial. As Michael Penn shows in the YouTube video The complex number family, some algebra shows that all of the ones you get with a quadratic polynomial are the complex numbers, the dual numbers, and the split-complex/"double" numbers.

# Q3: Numbers of dimensions

Also, why all hypercomplex numbers have a dimension of power of 2 over reals? Is 3, 5, or 6 dimensional algebra not possible?

There are several different factors at play here. Like with the previous question, part of the answer is related to convenience, historical accident, and other things related to human actions. But another part of the answer is related to some very deep mathematics.

## Historical Reasons

I am not a historian by any means, but I think the main historical point is that the quaternions and bicomplexes/tessarines and octonions were conceived before most other algebras (e.g. matrix algebras) were thought of as algebras in their own right. e.g. Cayley who popularized octonions only then worked on figuring out some basic facts/notation for matrix calculations.

And even after that initial work, Dickson helped develop what is known as "the Cayley-Dickson construction". That construction and its generalization only double the dimension of certain algebras, so if you're building up from the reals or something two-dimensional, you're only going to get algebras whose dimension is a power of two.

Another overlapping class of constructions of many algebras is that of the Clifford algebras whose dimensions are also always a power of 2. Clifford algebras are always associative and unital and include the quaternions as a special case. In fact, all of the Clifford algebras over the reals have been classified, and they're all matrices or pairs of matrices of real, complex, or quaternion numbers. (They follow a nice period-8 sort of pattern known as Bott periodicity which has many formulations.)

As mentioned earlier, there are certainly algebras of dimensions other than a power of 2, but they're not going to be building on the legacy of Cayley et al., and wouldn't be called "hypercomplex numbers", so you might not hear about them in the same contexts as you'd read about the quaternions.

## Mathematical Reasons

However, there remains an unanswered question: Why didn't people like Hamilton, Graves, and Cayley also develop algebras of other dimensions around the same time as the quaternions and octonions? Actually, Hamilton famously tried and failed to find a 3-dimensional algebra that satisfied him, before going on to develop quaternions (and, later, biquaternions).

It turns out that there is a confluence of theorems which limit various nice properties to dimension being a power of two. And many of them limit the algebra further to being complexes, the quaternions, and sometimes the octonions. It would take us too far afield to give a summary of all of the relevant definitions and weakest-possible assumptions, so I'll just mention some key theorems.

### Nice settings

If we make significant assumptions about how nice an algebra is, then the possibilities for it are quite constrained. The Frobenius theorem (for real division algebras) says that a finite-dimensional unital associative algebra where every element has a multiplicative inverse must be the reals, the complexes, or the quaternions. Hurwitz's Theorem for composition algebras says that a finite-dimensional algebra with a sort of real-valued norm obeying $$\Vert xy\Vert=\Vert x\Vert\Vert y\Vert$$ and $$x\ne0\Rightarrow\Vert x\Vert>0$$ must be one of those three or the octonions. And the Hopf invariant one theorem says that the only spheres that are "H-spaces" (i.e. that have a sort of continuous almost-unital multiplication) are the spheres of dimensions $$0,1,3,7$$, and a multiplication is given by regular multiplication with the norm-1 reals, complexes, quaternions, and octonions, respectively.

### More general settings

If we relax conditions a bit, then there are still a number of theorems forcing the dimension to be a power of two. Studying this gets complicated because a lot of different characterizations of things like division algebra start to separate depending on whether you assume your algebra is unital, alternative, finite-dimensional, etc.

Hopf showed that any finite-dimension (real) division algebra has dimension a power of two, and then Kervaire and Milnor showed that the dimension must be 1,2,4, or 8. The MathematicsSE question Unital nonalternative real division algebras of dimension 8, its answer, and the paper How to obtain division algebras from a generalized Cayley-Dickson doubling process discuss some examples of dimension 2 or 8 that aren't the more well-known algebras like the octonions.

And while everything above has been over the reals, a generalization of Hurwitz's Theorem says that unital "composition algebras" over fields other than the real numbers still have to have dimension 1, 2, 4, or 8.

• You really had many things to say when you said to split the question into three separate. Btw, I've heard the words algebra over reals and algebra over complex. But never heard something like algebra over quaternions. Do they exist? Apr 1, 2022 at 18:13
• Even among powers of two, most of the 'nice settings' you talked about only applies where all basis apart from first square to -1. Is there a specialty about them which makes them nicer than their siblings like dual numbers, double numbers, tesserines, coquaternions etc? Apr 1, 2022 at 18:30
• Thanks for a very elaborate answer for my question. This is a great answer. Can you give an example for an algebra whose base is not defined in the powers of 2. Apr 1, 2022 at 19:02
• @SouravKannanthaB "over quaternions?" I suppose you could define such a thing, but it's a pain to define an algebra over something that's not commutative. "nicer than siblings?" If $j^2=1$ then $(1+j)(1-j)=0$ even though the factors aren't zero, which causes problems. Similarly for $\varepsilon^2=0$ even though $\varepsilon\ne0$. "give an example..." Edit: I misunderstood the question in my first response, give me a minute... Apr 2, 2022 at 1:40
• @SouravKannanthaB I think "an example for an algebra whose base is not defined in the powers of 2" would be the "triplex numbers" that Angel mentioned in their answer and that I mentioned as an example of a polynomial quotient ring, as defined by "$i^3=1$", and which are featured in this YouTube video that Angel referred to. Apr 2, 2022 at 1:49

Intuition. Basically, if you add two complex dimensions to reals, say $$i$$ and $$j$$, you automatically get a fourth dimension $$ij$$ because this number cannot be expressed using only the three dimensions. The system you get then is called bicomplex numbers and 4-dimensional.

On the other hand, if you add two split-complex dimensions to reals, say $$j$$ and $$k$$, you do not get a fourth dimension automatically because $$jk=j+k-1$$ can be expressed in the already existing 3 dimensions. Thus, you get a 3D algebra.

3D numbers. Example 1

Take $$\mathbb{R}^3$$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication.

Now assign $$(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$$.

A number would be written in the form $$a+bj+ck$$. Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $$(j-1)(k-1)=0$$.

Here is a Mathematica code to experiment with:

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} -> x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &;  Using this code one can see that $$j^2=k^2=1$$ $$jk=j+k-1$$ $$\log (j+k+1)=\frac{1}{2} j \log (3)+\frac{1}{2} k \log (3)$$ $$j^j=j^k=j$$ $$k^k=k^j=k$$ $$\sqrt{j+k}=\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}}$$ $$0^{j+k}=1-\frac{j}{2}-\frac{k}{2}$$ The division formula would be: $$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{j}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1-b_1+c_1}{a_2-b_2+c_2}\right)+\frac{k}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1+b_1-c_1}{a_2+b_2-c_2}\right)+\frac{a_1+b_1-c_1}{2 \left(a_2+b_2-c_2\right)}+\frac{a_1-b_1+c_1}{2 \left(a_2-b_2+c_2\right)}$$ If we add a complex unity $$i$$, we will get a 6-dimensional number system. Particularly, we will see that $$i^{j+k}=1-j-k$$ and $$\log (j k)=i\pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$$ 3D numbers. Example 2 This is a realization of triplex numbers, described in this video. Here, $$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right)$$ $$k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)$$ Unprotect[Power]; Power[0, 0] = 1; Protect[Power];$Pre = (# /. ({j -> {1, E^(2 I \[Pi]/3)},
k -> {1, E^(-2 I \[Pi]/3)}}) /. {x_, y_} ->
FullSimplify[(x/3 + Im[y]/Sqrt - Re[y]/3) j + (x/3 -
Im[y]/Sqrt - Re[y]/3) k +
1/3 (x + y + Conjugate[y])] // FullSimplify // Expand) &;


Particularly, we will see that

$$j^2=k$$, $$k^2=j$$, $$jk=1$$

$$j^k=-\frac{1}{3} 2 e^{\frac{\pi }{\sqrt{3}}} j+\frac{j}{3}+\frac{1}{3} e^{\frac{\pi }{\sqrt{3}}} k+\frac{k}{3}+\frac{e^{\frac{\pi }{\sqrt{3}}}}{3}+\frac{1}{3}$$

$$0^{j + k + 1}=-\frac{j}{3}-\frac{k}{3}+\frac{2}{3}$$

$$\log j = \frac{2 \pi j}{3 \sqrt{3}}-\frac{2 \pi k}{3 \sqrt{3}}$$

$$\log(j+k)=\frac{\pi j}{\sqrt{3}}+\frac{1}{3} j \log (2)-\frac{\pi k}{\sqrt{3}}+\frac{1}{3} k \log (2)+\frac{\log (2)}{3}$$

The division formula is

$$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b j_2+c j_1\right)-a_2 \left(b_2 \left(a j_2+b j_1+c\right)+c_2 \left(a j_1+b+c j_2\right)\right)+c_2^2 \left(a j_2+b j_1+c\right)-b_2 c_2 \left(a+b j_2+c j_1\right)+b_2^2 \left(a j_1+b+c j_2\right)}{a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2}$$

If we add complex unity, we will see that

$$i^{j+k}=-\frac{j}{3}-\frac{j}{\sqrt{3}}-\frac{k}{3}+\frac{k}{\sqrt{3}}-\frac{1}{3}$$