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Questions tagged [sedenions]

The sedenions are a 16 dimensional nonassociative algebra over the reals.

2
votes
0answers
26 views

Java Library that supports Quaternions Octonions, Sedenions?

I would like to experiment with multi dimensional complex numbers such as quaternions octonions, sedenions. I know Apache Commons Maths supports Quaternions, and I've found (although cannot download) ...
7
votes
2answers
233 views

Why do division algebras always have a number of dimensions which is a power of $2$?

Why do number systems always have a number of dimensions which is a power of $2$? Real numbers: $2^0 = 1$ dimension. Complex numbers: $2^1 = 2$ dimensions. Quaternions: $2^2 = 4$ dimensions. ...
6
votes
0answers
156 views

Are complex split-octonions isomorphic to a more easily-defined algebra?

I write fiction and nonfiction, both which are mathy. My fiction is not usually supermathy but I'm working on a fictional story that has some math in it, and I prefer accuracy to mathbabble. I'm ...
27
votes
2answers
2k views

What specific algebraic properties are broken at each Cayley-Dickson stage beyond octonions?

I'm starting to come around to an understanding of hypercomplex numbers, and I'm particularly fascinated by the fact that certain algebraic properties are broken as we move through each of the $2^n$ ...
4
votes
3answers
505 views

Math beyond Quaternions

Quaternions remove the commutative property and octonions eliminate the associative property can we go any higher and eliminate more properties?
21
votes
2answers
3k views

Why are properties lost in the Cayley–Dickson construction?

Motivating question: What lies beyond the Sedenions? I'm aware that one can construct a hierarchy of number systems via the Cayley–Dickson process: $$\mathbb{R} \subset \mathbb{C} \subset \mathbb{H}...
46
votes
3answers
7k views

What lies beyond the Sedenions

In the construction of types of numbers, we have the following sequence: $$\mathbb{R} \subset \mathbb{C} \subset \mathbb{H} \subset \mathbb{O} \subset \mathbb{S}$$ or: $$2^0 \mathrm{-ions} \subset ...