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Real numbers satisfy a(bc) = (ab)c as well as ab = ba. They are also comparable.

Generalising to complex numbers, everything stays the same, except the numbers lose their comparibility.

Generalising to quaternions, the ab = ba no longer holds

Generalising to octonions, a(bc) = (ab)c no longer holds.

My question is, can you keep generalising indefinitely, and is there ever a point where the resulting number system has lost so many useful relations/identities that it just becomes a useless mush? (As a speculative example, perhaps it may become so generalised that something as simple as "a + b" becomes meaningless or uncomputable)

rschwieb
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TheIronKnuckle
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    This type of question has been asked many times on this site. See for instance this, this, this or this. – J.-E. Pin Sep 23 '20 at 07:31
  • And if it weren't a duplicate, it would be opinion-based (because 'reasonable' is subjective.) – rschwieb Sep 23 '20 at 12:43
  • You definitely can have a hypercomplex number system which is associative and commutative of any dimension you want. But you will get zero divisors (which is not that much bad though). The simplest example is $\mathbb{R}^n$, each number being represented by sets of $n$ reals like $(a,b,c,d,e)$. with element-wise addition and multiplication. In such system $(a,a,a,a,a)$ would be equal to real number $a$. And you will have zero divisors (any set with a zero but not all zeros). – Anixx Nov 14 '22 at 14:01

2 Answers2

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I am no expert on this topic, but this hierarchy is generated by the Cayley-Dickson construction. When you apply it to the octonions to make the sedenions, you lose alternativity, which is a weaker form of associativity: it requires $x(xy) = (xx)y$ and $y(xx) = (yx)x$. You also gain zero-divisors. At this stage, you can keep going (as many times as you want, actually), but there's almost no nice properties left.

J.G.
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It'sNotALie.
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Just to expand on @It'sNotALie's discussion of the Cayley–Dickson_construction, it can be used to double the dimension of an any algebraic structure satisfying certain conditions; these are sometimes called algebras with involutions. They have commutative and associative addition with identity $0$, multiplication that's power-associative (i.e. $x$ commutes with $xx$) and left- and right-distributive over addition with identity $1$, and an involution ${}^\ast$ with $(x+y)^\ast=x^\ast+y^\ast,\,(xy)^\ast=y^\ast x^\ast$. We can prove the following:

  • The new algebra satisfies the same axioms;
  • The new one contains elements satisfying $x^\ast\ne x$, i.e. is not self-conjugate;
  • The new one commutes iff the old one was self-conjugate, associates iff the old one commmuted, and alternates iff the old one associated.

With this, we can explain why properties are shedded as they are by each of $\Bbb C$, $\Bbb H$ (quaternions), $\Bbb O$ (octonions), $\Bbb S$ (sedenions, which also introduces zero divisors, i.e. solutions to $xy=0,\,x\ne0,\,y\ne0$; this costs us a norm). Beyond $\Bbb S$, we preserve all properties so far discussed that survive to $\Bbb S$.

There are other options where we don't just use square roots of $-1$ to double the dimension over $\Bbb R$.

J.G.
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