In the construction of types of numbers, we have the following sequence:

$$\mathbb{R} \subset \mathbb{C} \subset \mathbb{H} \subset \mathbb{O} \subset \mathbb{S}$$


$$2^0 \mathrm{-ions} \subset 2^1 \mathrm{-ions} \subset 2^2 \mathrm{-ions} \subset 2^3 \mathrm{-ions} \subset 2^4 \mathrm{-ions} $$


"Reals" $\subset$ "Complex" $\subset$ "Quaternions" $\subset$ "Octonions" $\subset$ "Sedenions"

With the following "properties":

  • From $\mathbb{R}$ to $\mathbb{C}$ you gain "algebraic-closure"-ness (but you throw away ordering).
  • From $\mathbb{C}$ to $\mathbb{H}$ we throw away commutativity.
  • From $\mathbb{H}$ to $\mathbb{O}$ we throw away associativity.
  • From $\mathbb{O}$ to $\mathbb{S}$ we throw away multiplicative normedness.

The question is, what lies on the right side of $\mathbb{S}$, and what do you lose when you go from $\mathbb{S}$ to one of these objects ?

Dominic Michaelis
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Willem Noorduin
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    One useful generalization starting from $\mathbb{H}$ and extending to all powers of $2$ is Clifford algebras: http://en.wikipedia.org/wiki/Clifford_algebra . You can also keep applying the Cayley-Dickson construction past $\mathbb{S}$ (http://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction) although I don't know if this is useful. – Qiaochu Yuan Nov 28 '11 at 17:13
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    Dragons. Beyond sedenions there are dragons. Beware. – Mariano Suárez-Álvarez Nov 28 '11 at 17:15
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    If you want to keep with your "throw away" theme, then from $\mathbb{R}$ to $\mathbb{C}$ you lose the ordering – Jason DeVito Nov 28 '11 at 19:23
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    (somewhat) related: http://mathoverflow.net/questions/19929/19975#19975 – Grigory M Nov 28 '11 at 23:02
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    @Jason: Thanks, I have add this to the first action from $\mathbb{R}$ to $\mathbb{C}$) – Willem Noorduin Nov 29 '11 at 09:26
  • AFAICS, http://arxiv.org/abs/1010.2156 claims to answer OPs last question – Grigory M Nov 29 '11 at 14:10
  • At some point I'd wager you're just left with a plain old vector space, and that multiplication doesn't do anything useful. – Joe Z. Feb 13 '14 at 19:40
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    Each of those systems is constructed using conjugates from the previous system. Each member of R is its own conjugate. That's why C is commutative. In C, not all numbers are their own conjugate and that's why H is not commutative. However, H is still associative because C is commutative. O is not associative because H is not commutative. However, I believe another special property holds for O because H is associative. I believe that S even loses that property because O is not associative. – Timothy Apr 08 '20 at 01:57
  • @Timothy “However, I believe another special property holds for O because H is associative. I believe that S even loses that property because O is not associative.” You are correct. That property is called “[alternativity](https://math.stackexchange.com/questions/3557626/do-any-significant-changes-happen-in-hypercomplex-numbers-beyond-the-eight-dimen)”. – Radial Arm Saw Aug 19 '20 at 22:33
  • For naming you can visit [this answer](https://english.stackexchange.com/a/234621) – V.P. Dec 26 '20 at 09:00

4 Answers4


What you are talking about is precisely the Cayley-Dickson construction.

Remark: I am left wondering what is gained by going past Octonions. The the first 4 are very special as they are the unique 4 normed divison algebras over $\mathbb{R}$. Perhaps someone with more knowledge can point out the possible uses of the Sedenions and their higher counterparts.

Eric Naslund
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  • Wikipedia states, "The Cayley–Dickson construction can be carried on ad infinitum, at each step producing a power-associative algebra whose dimension is double that of algebra of the preceding step." – ttt Nov 28 '11 at 17:27
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    As for applications of sedenions, I am quoting Wikipedia: "Moreno (1998) showed that the space of norm $1$ zero-divisors of the sedenions is homeomorphic to the compact form of the exceptional Lie group $G_2$." – darij grinberg Nov 28 '11 at 17:44
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    @darij: that can't be right. Clifford algebras are associative and the octonions aren't. – Qiaochu Yuan Nov 28 '11 at 19:30
  • Oh right, I mixed up things there. Comment deleted. – darij grinberg Nov 28 '11 at 23:12
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    Just for info, after Degen's 8-square, there is the _non-bilinear_ [Pfister's 16-square identity](http://en.wikipedia.org/wiki/Pfister%27s_sixteen-square_identity). – Tito Piezas III Aug 14 '13 at 04:21
  • As of 2020, someone tried using sedenions in neural networks https://ieeexplore.ieee.org/document/9160921 – Phoenix Jan 16 '22 at 21:54

If with "beyond" you mean something like "after" the answer is yes: the Cayley-Dickson's construction has no end and you can always extend your number system to a higher dimension one (always with $2^n$ dimensions with $n\in \Bbb N$), after Sedenions $\Bbb S$ come Trigintaduonions $\Bbb T$ which someone say are useful in electrical and computer engineering (I can't really figure out how) but they don't loose any other properties more than Sedenions do, this set and the others that came after it are not very interesting (with an only mathematical point of view) cause of that, they just have some subgroups that look like the Sedenions.

If with "beyond" you mean something like "other than" the answer is still yes: there are a lot of hypercomplex numbers with 16 dimensions system like the extension of tessarines, duals and hyperbolic number and so on.

Renato Faraone
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    You can always find properties that are lost at each stage - it's just a matter of how hard you look. For example the trigintaduonions loose the property of being sedenions... – skyking Jun 30 '16 at 13:05
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    I think it useful in general for each number of spatial dimensions we need to work with. In electrical physics it seem we can treat properties of electrical particle that seem related and can be rotated around each other as 5 spatial dimensions. Therefor in String theory with 10-11 dimensions might use $2^{11}$-nion – Thaina Mar 07 '17 at 06:03
  • @skyking with Tessarines you do not lose much: any $2^n$-dimensional tessarines are commutative and associative. For instance, 16-dimensional ones. – Anixx Mar 10 '21 at 17:21
  • @Anixx Not loosing much is NOT the same as loosing nothing. Furthermore it was talk about spaces constructed with CD-construct, for example already quaternion are non-commutative and octonions are not associative, sedenions are not alternative. As mentioned it's only a matter of how hard you look for lost properties. – skyking Mar 10 '21 at 20:52
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    I don't think tessarines lose something after dimension 4. – Anixx Mar 10 '21 at 20:54

I believe John Baez has answered your question in a series of short articles/blogposts which you could begin for example here.

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You can build tessarines of any dimension of the for $2^n$. They still will be commutative ans associative. This is used in digital signal processing.

Cayley-Dickson's construction, used in quaternions, on the other hand, loses most of its useful properties with the growth of the number of dimensions.

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