08/05/2018, 02:10 AM
(07/31/2018, 01:08 PM)Xorter Wrote: I have been writing a paper about Cayley-Dickson algebra and its higher 2^n-ion spaces. By xor extensions (from binary xoring to ternary, tetranary, pentanary,... etc.) we get 3^n-ions (Trions), 4^n-ions (Tetrions), ... . For this operation it is needed to check the existance of triads (or tetrads, pentads, hexads,... and so on to (b+1)-ads) of the (b^n)-ions. For instance the Bions (2^n-ions) exist, because there are plenty of triads, like (1;2;3). Trions are not exist, because there are no tetrads anywhere. Tetrions (4^n-ions) also exist, because there are pentads like (10;27;51). And (2^m)^n-ions exist, too.
How do I calculate it? I have extended complex bitwise (binary) operator xor to tritwise, tetratwise, pentatwise,... n-twise (n-ary) operator xor[n]. The n-ad is a set of numbers (x;y;z), iff: (x xor[n] y = z) and (z xor[n] x = y) and (y xor[n] z = x).
Let us suppose that a xor[n] b = (a+b) n-twise mod n.
Thanks for answering my question. I have been busy, but now I have time to ask another question. I still don't understand why (1;2;3) is a triad or why (10;27;51) is a pentad. Let me explain my thinking so you can correct me:
(1;2;3) = (1;2;10) in base 3. 1 xor[3] 2 = 0. 0 does not equal 10.
(10;27;51) = (20;102;201) in base 5. 20 xor[5] 102 = 122. 122 does not equal 201.
Thanks,
11Keith22.