10 questions linked to/from What lies beyond the Sedenions
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### Complex Number, Quaternions and Octonions [duplicate]

There are complex $\mathbb C$, quaternions $\mathbb H$ and octonions $\mathbb O$. Is there any higher dimensional generalization of them, such in the $\mathbb R^{16}$? Or why do we just study three ...
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### Do you lose any more equational identities when you go past sedenions? [duplicate]

Every Cayley-Dickson algebra can be viewed as a $(+,-,*,0,1)$ algebra. The reals and the complexes share the same equational identities. The quaternions have a subset of the equational identities, ...
8k views

### What is lost when we move from reals to complex numbers? [duplicate]

As I know when you move to "bigger" number systems (such as from complex to quaternions) you lose some properties (e.g. moving from complex to quaternions requires loss of commutativity), but does it ...
8k views

### Why are addition and multiplication commutative, but not exponentiation?

We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not. My question is why. As background there are plenty of mathematical schemes that ...
3k views

### What is the motivation for quaternions?

I know imaginary numbers solve $x^2 +1=0$, but what is the motivation for quaternions?
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### 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}...
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### 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 $2^n$ ...
216 views

### Is $\mathbb{R}^n$ a field?

Is $\mathbb{R}^n$ a field for all $n$? I suppose for n=1 and 2 the result is clear. What about higher values of $n$.
As we know, $i$ = $\sqrt{-1}$, a simple complex unit. In complex space of two dimensions, you graph an axis of $a+bi$ where $i$ is your second dimension axis. Now, you also know, in three and four ...
So in $\mathbb{R}$ and $\mathbb{C}$ you have both associative and commutative property, but as you extend to $\mathbb{H}$ you lose the commutative property, and $\mathbb{O}$ loses the associativity. ...