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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.
  • Octonions: $2^3 = 8$ dimensions.
  • Sedenions: $2^4 = 16$ dimensions.
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    $\begingroup$ "The sedenions have a multiplicative identity element e0 and multiplicative inverses but they are not a division algebra because they have zero divisors. " en.wikipedia.org/wiki/Sedenion $\endgroup$
    – Thomas Andrews
    Apr 14, 2016 at 0:49
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    $\begingroup$ The original statement of this fact is from Hopf, Heinz. "Ein topologischer Beitrag zur reellen Algebra." Commentarii mathematici Helvetici 13 (1940/41): 219-239. See Satz IV: gdz.sub.uni-goettingen.de/dms/load/img/… or digizeitschriften.de/dms/img/… $\endgroup$
    – Chris Culter
    Apr 14, 2016 at 0:56
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    $\begingroup$ recommend bookstore.ams.org/gsm-67 $\endgroup$
    – Will Jagy
    Apr 14, 2016 at 1:02
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    $\begingroup$ The sedenions are not a division algebra. Also, it may help to distinguish division algebras from composition algebras (normed division algebras). Division algebras (over $\Bbb R$) are more plentiful (although still restricted to the same dimensions 1,2,4,8) and considerably harder to prove things about than composition algebras. $\endgroup$
    – anon
    Apr 14, 2016 at 1:14
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    $\begingroup$ The short answer is that a division algebra structure (even without a norm) induces a trivialization of the sphere $S^{n-1}$, and considering the multiplication as a map $\mathbb{RP}^{n-1} \to \mathbb{RP}^{n-1} \to \mathbb{RP}^{n-1}$ proves that $n$ must be a power of $2$. With even more machinery, one can show that $n$ must be $1, 2, 4$, or $8$. Unfortunately, I'm not aware of a more elementary proof of that fact (at least, in full generality). $\endgroup$
    – anomaly
    Apr 14, 2016 at 1:25

2 Answers 2

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They don't. Here is a 9-dimensional associative non-commutative division algebra (over $\Bbb{Q}$): $$ D=\left\{\left(\begin{array}{ccc} x_1&\sigma(x_2)&\sigma^2(x_3)\\ 2x_3&\sigma(x_1)&\sigma^2(x_2)\\ 2x_2&2\sigma(x_3)&\sigma^2(x_1) \end{array}\right)\bigg\vert\ x_1,x_2,x_3\in E\right\}, $$ where $E=\Bbb{Q}(\cos2\pi/7)$ and $\sigma$ is the automorphism defined by $\sigma(\cos2\pi/7)=\cos4\pi/7$.

Only over the reals are we so constrained. Topology makes a huge difference. Or, more precisely, the fact that odd degree polynomials with real coefficients always have a real zero.

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The particular family of algebras you are talking about has dimension over $\Bbb R$ a power of $2$ by construction: the Cayley-Dickson construction to be precise.

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