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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:

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

or:

"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 ?

  • 12
    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.2011-11-28
  • 75
    Dragons. Beyond sedenions there are dragons. Beware.2011-11-28
  • 12
    If you want to keep with your "throw away" theme, then from $\mathbb{R}$ to $\mathbb{C}$ you lose the ordering2011-11-28
  • 1
    (somewhat) related: http://mathoverflow.net/questions/19929/19975#199752011-11-28
  • 3
    @Jason: Thanks, I have add this to the first action from $\mathbb{R}$ to $\mathbb{C}$)2011-11-29
  • 0
    AFAICS, http://arxiv.org/abs/1010.2156 claims to answer OPs last question2011-11-29
  • 0
    At some point I'd wager you're just left with a plain old vector space, and that multiplication doesn't do anything useful.2014-02-13

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