Proof of Sedenion's nonalternativity I have to prove that the Sedenion is a non-alternative algebra so I have to show that the associator e not alternative. So far I have made many attempts, more often than not I find out that many paris of basis elements associates (mostly because they come from reals or complex). Considering the standard basis $\{e_0,...,e_{15}\}$ I've found that $(e_2,e_{13},e_5)\neq0$ and it seemed to help me prove the result but I had no success. Any hint is appreciated
 A: After a laborious but trivial calculation I figured that:
$x=-e_3-e_{12}$ and $y=-e_5-e_{10}$ are such that $(x,y,x)\neq0$.
A: Here's how we can figure out what to test.
The sedenions $\mathbb{S}$ are alternative (every pair of elements generates an associative algebra) iff


*

*$x(xy)=(xx)y$

*$x(yx)=(xy)x$

*$(yx)x=y(xx)$


But $\mathbb{S}$ is alternative if and only if the associator $[x,y,z]=(xy)z-x(yz)$ is alternating (vanishes when arguments are repeated), so any  two identities imply the other (since two transpositions generate $S_3$), and moreover the first and third identity are equivalent for $\mathbb{S}$ because conjugation is an anti-automorphism. Therefore, we only need to check the first identity.
Write


*

*$x=a+bs$

*$y=c+ds$


where $\mathbb{S}=\mathbb{O}\oplus\mathbb{O}s$. If $a,b,c,d$ lie in a quaternionic subalgebra of $\mathbb{O}$, then with $s$ they generate an octonionic subalgebra of $\mathbb{S}$ which will be alternative, so we must analyze a different case. In particular, if we write $\mathbb{O}=\{1,e_1,\cdots,e_7\}$ with $e_1e_2=e_4$ and $e_ie_j=e_k\,\Rightarrow\, e_{i+1}e_{j+1}=e_{k+1}$ (indices mod $7$) then $x=e_1+e_2s$ and $y=e_3$ should work.
