Is there a third dimension of numbers like real numbers, imaginary numbers, [blank] numbers?

15I guess this is related: http://en.wikipedia.org/wiki/Quaternion – Aryabhata Apr 10 '11 at 16:30

1Wow, that is annoying. Are threedimensional numbers even possible/necessary? – awesomeguy Apr 10 '11 at 16:42

2Quaternions are fourdimensional, but they are used in computer graphics to deal with 3D situations. – André Nicolas Apr 10 '11 at 16:54

7A good hint that there is no three dimensional numbers is the fact that $(a_1^3 + b_1^3 + c_1^3)(a_1^3 + b_2^3 + c_2^3)$ doesn't always equal the sum of three cube numbers (the axiom that product of two three cubes is another three cube). The quaternions barely satisfy the condition for four dimensional except for commutativity. – Mark Apr 10 '11 at 20:27

8EDIT for above comment: I accidentally wrote cubes for some reason. I meant the product of three squares $(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)$ doesn't always equal the sum of three square numbers. It's well known in any elementary number theory book that Euler proved its true for product of four square hinting that four dimensional numbers exist. – Mark Apr 10 '11 at 22:36

See also the eighth dimensional [Octonions](http://en.wikipedia.org/wiki/Octonion) – Bennett Gardiner Nov 11 '13 at 12:58

@AndréNicolas, hypercomplex numbers are fourdimensional themselves. But, they can expand into more than just 4 dimensions. – Obinna Nwakwue Jul 12 '16 at 20:59

Linking this to closely related threads [1](https://math.stackexchange.com/q/3135371/11619) and [2](https://math.stackexchange.com/q/1784166/11619). The reason why 3D does not work is very simple to describe, and does not need any deep theory. – Jyrki Lahtonen Mar 15 '19 at 21:36
8 Answers
Alas, there are no algebraically coherent "triplexes". The next step in the construction as has been said already are "quaternions" with 4 dimensions.
Many young aspiring mathematicians have tried to find them since Hamilton in the 19th century. This impossibility links geometric dimensionality, fundamental properties of polynomial equations, algebraic systems and many other aspects of mathematics. It is really worth studying.
A quite recent book by modern mathematicians which details all this for advanced college undergraduates is Numbers by Ebbinghaus, Hermes, Hirzebruch, Koecher, Mainzer, Neukirch, Prestel, Remmert, and Ewing.
However, the set of quaternions with zero real part is an interesting system of dimension 3 with very interesting properties, linked to the composition of rotations in space.
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What do you mean by "algebraically coherent"? Is this algebraically coherent or not? https://math.stackexchange.com/a/4436389/2513 – Anixx May 17 '22 at 13:18

@Annix : I used this phrase "algebraically coherent" precisely because I wanted to convey a global idea, not the list of all properties or schemas that such a system should verify or conserve. The idea described in your stackexchange answer is new to me and deserves careful study before I can comment on it. Thanks a lot for drawing my attention. – ogerard May 18 '22 at 18:15

There is a video regarding example 2 in the post: https://www.youtube.com/watch?v=dvI7dGXFgm8 So we have 2 distinct commutative associative 3dimensional algebras over reals. Here is some algebraic comparison with a table: https://math.stackexchange.com/a/4453434/2513 – Anixx May 18 '22 at 18:20
You may also find of interest some more general results besides the mentioned Frobenius Theorem. Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative extension ring of $\mathbb R$ without nilpotents ($x^n = 0\,\Rightarrow\, x = 0$) is isomorphic as a ring to a direct sum of copies of $\rm\,\mathbb R\,$ and $\rm\,\mathbb C.\,$ Wedderburn and Artin proved a generalization that every finitedimensional associative algebra without nilpotent elements over a field $\rm\,F\,$ is a finite direct sum of fields.
Such structure theoretic results greatly simplify classifying such rings when they arise in the wild. For example, I applied a special case of these results last week to prove that a finite ring is a field if its units $\cup\ \{0\}$ comprise a field of characteristic $\ne 2.\,$ For another example, a sci.math reader once proposed an extension of the real numbers with multiple "signs". This turns out to be a very simple case of the above results. Below is my 2009.6.16 sci.math post on these "PolySign" numbers.
The results in Eitzen's paper Understanding PolySign Numbers the Standard Way, characterizing Tim Golden's socalled PolySign numbers as ring direct sums of $\mathbb R$ and $\mathbb C$, have been known for over a century and a half. Namely that $\rm\,P_n =\, \mathbb R[x]/(1+x+x^2+\,\cdots\, + x^{n1})\ $ is isomorphic to a certain ring direct sum of $\,\mathbb R$ and $\,\mathbb C,\,$ is just a special case of more general results due to Weierstrass and Dedekind in the 1860s. These classic results are so wellknown that you will find them mentioned even in many elementary textbooks on number systems and their generalizations. For example, in Numbers by Ebbinghaus et.al. p.120:
Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative ring extension of R with unit element but without nilpotent elements, is isomorphic to a ring direct sum of copies of R and C.
Ditto for historical expositions, e.g. Bourbaki's Elements of the History of Mathematics, p. 119:
By 1861, Weierstrass, making precise a remark of Gauss, had, in his lectures, characterized commutative algebras without nilpotent elements over R or C as direct products of fields (isomorphic to R or C); Dedekind had on his side reached the same conclusions around 1870, in connection with his "hypercomplex" conception of the theory of commutative fields, their proofs were published in 188485 [1,2]. [...] These methods rely above all on the consideration of the characteristic polynomial of an element of the algebra relative to its regular representation (a polynomial already met in the work of Weierstrass and Dedekind quoted earlier) and on the decomposition of the polynomial into irreducible factors.
Nowadays these fundamental results are merely special cases of more general structure theories for algebras that are part of any first course on algebras (but not always met in a first course on abstract algebra). A web search turns up more on the subsequent history, e.g. excerpted from
Y. M. Ryabukhin, Algebras without nilpotent elements, I,
Algebra i Logika, Vol. 8, No. 2, pp. 181214, MarchApril, 1969
https://doi.org/10.1007/BF02219831
Algebras without nilpotent elements have been studied long ago. So, Weierstrass characterized in his lectures in 1861 finitedimensional associativecommutative algebras without nilpotent elements over the field of real or complex numbers as finite direct sums of fields. To be exact, some nonessential restrictions have there been imposed. In 1870 Dedekind removed those nonessential restrictions. The following theorem of WeierstrassDedekind is now considered as a classical one: every finitedimensional associativecommutative algebra without nilpotent elements over a field F is a finite direct sum of fields. The results of Weierstrass and Dedekind (for the case when F is the field of complex or real numbers) have been published in [1,2]. The results of works of Molien, Cartan, Wedderburn and Artin [36] imply that Dedekind's theorem holds for any field F. Moreover, the following theorem of WedderburnArtin holds: every finitedimensional associative algebra without nilpotent elements over a field F is a finite direct sum of fields." [...]
 K. Weierstrass, "Zur Theorie der aus n Haupteinheiten gebildeten complexen Grossen," Gott. Nachr. (1884).
 R. Dedekind, "Zur Theorie der aus n Haupteinheiten gebildeten complexen Grossen," Gott. Nachr. (1885).
 F. Molien, "Ueber Systeme hoherer complexer Zahlen," Math. Ann., XLI, 83156 (1893).
 E. Cartan, "Les groupes bilineaires et les systemes de nombres complexes," Ann. Fac. Sci., Toulouse (1898).
 J. Wedderburn, "On hypercomplex numbers," Proc. London Math. Soc. (2), VI, 349352 (1908).
 E. Artin, "Zur Theorie der hyperkomplexen Zahlen," Abh. Math. Sere. Univ. Hamburg, 5, 251260 (1927).
and excerpted from its sequel
Y.M. Ryabukhin, Algebras without nilpotent elements, II,
Algebra i Logika, Vol. 8, No. 2, pp. 215240, MarchApril, 1969
https://doi.org/10.1007/BF02219832
In [1] we proved structural theorems on the decomposition of algebras without nilpotent elements into direct sums of division algebras; certain chain conditions were imposed on these algebras.
Yet it is possible to prove structural theorems also without imposing any chain conditions. In this case the direct sums are replaced by subdirect sums and instead of division algebras we shall consider algebras without zero divisors.
The first structural theorem of this kind is apparently the classical theorem of Krull [2]:
Any associativecommutative ring without nilpotent elements can be represented by a subdirect sum of rings without zero divisors.
Krull's theorem was subsequently extended to the case of any associative ring. This was done by various authors and in various directions. In [3], Thierrin came very close to a final generalization of Krull's theorem to the associative, but not commutative case. The final result was obtained in [4]:
Any associative ring without nilpotent elements can be represented by a subdirect sum of rings without zero divisors.
At the Ninth AllUnion Conference on General Algebra (held at Gomel'), I. V. L'vov reported an even stronger result:
Any alternative ring without nilpotent elements can be represented by a subdirect sum of rings without zero divisors.
It could be assumed that the theorem on decomposition into a subdirect sum of algebras without zero divisors holds for any ring. Yet this assumption is erroneous (see [1]), since there exists a finitedimensional simple, special Jordan algebra without nilpotent elements that has zero divisors and cannot therefore be decomposed into a subdirect sum of algebras (or rings) without zero divisors.
There naturally arises the following question: what conditions must a ring without nilpotent elements satisfy to permit its representation by a subdirect sum of rings without zero divisors?
In this paper we answer this question:
An algebra R over an associativecommutative ring F with unity can be represented by a subdirect sum of rings without zero divisors, iff it is a conditionally associative algebra without nilpotent elements.
Let us recall that an algebra R is said to be conditionally associative, iff we have in R the conditional identity x(yz) = 0 iff (xy)z = 0.
We say a (not necessarily associative) algebra R does not have nilpotent elements, iff in R we have the conditional identity x^2 = 0 iff x = 0.
From this theorem we easily obtain the abovementioned results of [24], as well as the result of L'vov (it suffices to take as the ring F the ring Z of integers). [...]
 Yu. M. Ryabukhin, "Algebras without nilpotent elements,I," this issue, pp. 215240.
 W. Krull, "Subdirect representations of sums of integral domains," Math. Z., 52, 810823 (1950).
 J. Thierrin, "Completely simple ideals of a ring," Acad. Belg. Bull. C1. Sci., 5 N 43, 124132 (1957}.
 V. A. Andrunakievich and Yu. M. Ryabukhin, "Rings without nilpotent elements in completely simple ideals," DAN SSSR, 180, No. 1, 9 (1968).
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Bill, something seems to be wrong with the title of Dedekind's paper. Two blanks are missing. When you add these, the title becomes identical to Weierstrass' which also seems weird to me. – Rasmus Apr 10 '11 at 20:15

1@Ras No, it's correct. The papers do in fact have the same titles. – Bill Dubuque Apr 10 '11 at 21:04


10

Generally speaking, I upvote any answer that is multiple pages long. +1. – Aidan F. Pierce Jun 02 '14 at 13:54
Every finitedimensional division algebra over $\mathbb{R}$ is one of $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. This is what is called the Frobenius Theorem. You may refer to here for details.
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11Every finite dimensional associative division algebra over $\mathbb R$ is one of the above. The octonions $\mathbb O$ is also a division algebra, but it is not associative. – Calle Apr 10 '11 at 17:29

1@sis440: You mean Frobenius. More broadly, you might also like to lookup theorems by Dickson, Artin and Wederburn which gives a broader picture of the classification of algebra. – ogerard Apr 10 '11 at 17:35

3Oh, I forgot that division algebra need not be associaitive... You are right, $\mathbb{O}$ has to be added if associativity is omitted. – Sangchul Lee Apr 10 '11 at 17:35
You might look up quaternions.
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I was hovering over the +1 and it said "This answer is useful". I agree(d), and clicked! ..and quoted the popup $$$$ :) – The Chaz 2.0 Jul 13 '11 at 03:39

3Will try hovering over one of your answers, and do what the gnome in the machine asks me to do. – André Nicolas Jul 13 '11 at 04:31
In addition to complex numbers and quaternions, you might want to look up Clifford Algebras which encapsulate both and extend to arbitrary dimension. Complex and quaternions are subalgebras of the Clifford Algebras over $\mathbb{R}^2$ and $\mathbb{R}^3$ respectively.
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It is a nice consequence of the hairy ball theorem that certain 3dimensional numberanalogues cannot exist.
Suppose we can define nice addition and multiplication on $V\cong \Bbb R^3$. Under some natural conditions on these operations (see below), one finds that there is an element $i\in V$ (named accordingly to the complex unit $i\in\Bbb C$) so that multiplication with $i$ in $V$ behaves like a $90^\circ$ rotation. That is, $x$ and $ix$ are orthogonal to each other for all $x\in V$.
But then $\mathbb S^2 \to \mathbb S^2, x\mapsto ix$ defines a unit vector field on the 2sphere, which is known to not exist by the hairy ball theorem. The same argument works for all odddimensional cases, but is not strong enough to exclude most evendimensional cases (as is necessary by the other answers).
Some details
Some natural requirements on such a number system are the following (there are others, but many turn out to be equivalent or stronger):
The addition is componentwise, that is, $(x,y,z)+(x',y',z')=(x+x',y+y',z+z')$.
The number system extends the real numbers in a natural way, e.g. each real numbers $x\in\Bbb R$ is represented by $(x,0,0)\in V$ or the like (as $x\mapsto x+0i$ embedds $\Bbb R$ in $\Bbb C$).
There is an "absolute value" $\cdot$ that assigns to any $(x,y,z)\in\Bbb R^3$ a real number, its absolute value. Clearly this exists in $\Bbb R$, and also in $\Bbb C$ via $x+yi=(x^2+y^2)^{1/2}$. A natural property of the absolut value is that it interacts well with the multiplication of the system, that is $uv=u \cdotv$, and that it behaves as expected on the embedded $\Bbb R$, that is $x$ is the usual absolute value for $x\in\Bbb R$ when considering $x$ as an element of $V$.
One thing that already follows from this is that $\cdot$ is a norm on the real 3dimensional vector space $V$, and one can define an inner product
$$\langle u,v\rangle := \frac12(x+y^2x^2y^2)$$
using the socalled polarization identities. In particular, there must be an element $i\in V$ with $\langle i,1\rangle=0$. Then follows
$$\langle ix,x\rangle\> = \frac12(ix+x^2ix^2x^2) = \frac{x}2(i+1^2i^21^2) = x\cdot\langle i,1\rangle = 0.$$
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There can be 3dimensional commutative hypercomplex number systems. I will give two examples.
3D numbers. Example 1
Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with elementwise 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 $(j1)(k1)=0$.
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
$j^2=k^2=1$
$jk=j+k1$
$\log (j+k+1)=\frac{1}{2} j \log (3)+\frac{1}{2} k \log (3)$
$j^j=j^k=j$
$k^k=k^j=k$
$\sqrt{j+k}=\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}}$
$0^{j+k}=1\frac{j}{2}\frac{k}{2}$
The division formula would be:
$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{j}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}\frac{a_1b_1+c_1}{a_2b_2+c_2}\right)+\frac{k}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}\frac{a_1+b_1c_1}{a_2+b_2c_2}\right)+\frac{a_1+b_1c_1}{2 \left(a_2+b_2c_2\right)}+\frac{a_1b_1+c_1}{2 \left(a_2b_2+c_2\right)}$
If we add a complex unity $i$, we will get a 6dimensional number system.
Particularly, we will see that
$i^{j+k}=1jk$
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[3]  Re[y]/3) j + (x/3 
Im[y]/Sqrt[3]  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
$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b j_2+c j_1\right)a_2 \left(b_2 \left(a j_2+b j_1+c\right)+c_2 \left(a j_1+b+c j_2\right)\right)+c_2^2 \left(a j_2+b j_1+c\right)b_2 c_2 \left(a+b j_2+c j_1\right)+b_2^2 \left(a j_1+b+c j_2\right)}{a_2^3+b_2^3+c_2^33 a_2 b_2 c_2}$
If we add complex unity, we will see that
$i^{j+k}=\frac{j}{3}\frac{j}{\sqrt{3}}\frac{k}{3}+\frac{k}{\sqrt{3}}\frac{1}{3}$
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Short Answer: No,
A lot of responses bring up the fact the next closed algebraic set are quaternions. But these aren't perfect since there is no problem to which Quaternions naturally arise as a solution:
For example we know $i$ naturally is formed as the solution to the previously unsolvable $\sqrt{1}$ that being said all polynomials have roots in the complex plane so we are guaranteed no other such new algebraic unit (like) $i$ will be naturally formed.
Now I personally don't know of a theorem that says this CANNOT happen again. For example it could be that:
$$x^x = i$$ Or (if we define ${}^xx$ to mean tetration)
$${}^xx = i$$
Etc... following the pattern, may not have a solution in C. In that case we are now free to generate a new elementary unit $j$ however this elementary unit will be quite strange since it is associated with a higher operator so expressions such as:
$$j, j^2, j^3 ...$$
Are all unique and simplified, leading us to a now infinite dimensional system of numbers
So 3D is not possible, but I believe infinite dimensional is still possible.
Unless you like the unnatural creations that are the quaternions etc... (I only dislike them because of their lack of natural formation)
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