In a paper of mine you can read here paper #73 at http://rangevoting.org/WarrenSmithPages/homepage/works.html I explain the "right" way to keep the sequence reals (1D), complex numbers (2D), quaternions (4D), octonions (8D)... going forever. [The "Cayley Dickson algebras" are not the right way, although are much better known than the 2^n-ons due to the way history went.] Anyway, my "2^n"ons stop being commutative at 4, stop being associative at 8, stop being distributive at 16, lose generic ability to divide and get unique answer at 32, etc But they keep many nice properties forever, such as having an identity element 1 and always being distributive on one side, let's say the right: a*(b+c)=ab+ac. Furthermore, they are distributive on the left provided either a, or both b and c, are "niners" meaning their last 2^n-9 coordinates all are 0. THEOREM (Inspired by RC Schroeppel): If a,b,c,... are each 2^n-ons which all happen to be niners, and there are min(2^n, 9) or more letters in all, then sum(all even perms of the letters) (((a*b)*c)*d)*... = same sum using odd perms. PROOF: By right-distributive and niner-left-distributive laws it suffices to prove it in the case where each letter is one of the min(2^n, 9) basis elements. (Reducing to finite set of cases.) (i) If at least one letter is 1, then since 1 associates, commutes, and distributes with everything we can move the 1 wherever we want, in particular swapping it with the letter cyclically to its left which is an odd perm, without changing anything. (ii) If at least two letters are same basis element, then we can swap those 2 letters (an odd perm) without changing anything. By pigeonhole principle at least one of cases (i) or (ii) must apply. Either way the identity is valid. QED WARNING: I have not checked this, I only proved it :). Unlike in my paper, where practically everything was checked by computer. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)