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.
  • 1
    $\begingroup$ "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 $\endgroup$ Apr 14, 2016 at 0:49
  • 1
    $\begingroup$ 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/… $\endgroup$ Apr 14, 2016 at 0:56
  • 1
    $\begingroup$ recommend bookstore.ams.org/gsm-67 $\endgroup$
    – Will Jagy
    Apr 14, 2016 at 1:02
  • 2
    $\begingroup$ 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
  • 4
    $\begingroup$ 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


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.