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>it's just that multiplication and division lose most of their useful properties. Specifically they lose the property of not having zero divisors. There exist
by Maybestring 8y ago
>it's just that multiplication and division lose most of their useful properties.
Specifically they lose the property of not having zero divisors.
There exists sedonions a,b != 0 such that ab = 0
- OscarCunningham 8y agoAnother (related) property that fails is that inverses stop being useful for cancellation. Inverses still exist, for every p there's a q with pq = qp = 1, but if you've got an equation ap = b you can't cancel to get a = bq, because we don't have associativity. The left hand side (ap)q doesn't equal a(pq), so you can't reduce it to a. Of course associativity doesn't hold in the octonions either, but it holds just enough for cancellation to work.
- deleted 8y ago[deleted]
- anaphylactic 8y agoYeah, octonion multiplication is alternative.
- qubex 8y agoThat happens in general for matrices too.
- anaphylactic 8y agoYes, and this is a bit obvious, but reals, complex numbers, split complex numbers, quaternions, octonions, sedenions, can all be represented as matrices of the appropriate form.
- gjm11 8y agoThat's at most sort-of-true. It's not possible to represent octonions by matrices of numbers in such a way that multiplication of matrices corresponds to multiplication of octonions, because matrix multiplication is associative and octonion multiplication isn't.
- Iwan-Zotow 8y agocan't be true, unless there is some special matrix multiplication rule - afaik, standard matrix multiplication is associative
- SubiculumCode 8y agoAn n-dimensional matrix of octonions
- gowld 8y agoAre there 32-onions that lack power associativity? https://en.wikipedia.org/wiki/Power_associativity https://en.wikipedia.org/wiki/Power_associativity https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_constru...
- bonzini 8y agoNo, power associativity is never lost, after sedenions the properties remain the same.
- OscarCunningham 8y agoThey don't remain exactly the same. The answers to this (https://math.stackexchange.com/questions/641809/what-specific-algebraic-properties-are-broken-at-each-cayley-dickson-stage-beyon https://math.stackexchange.com/questions/641809/what-specifi...) question provide some interesting starting-points.