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Zero divisor Example:

To find the zerodivisor, Z,  of  S  :=  e2  -  2 e4  -  e7 +  2 e9  -  e11  +  e13   .
 

First, choose the following octonion multiplication rule for the basal elements, ij .
                 ij ik  =  - djk + S(jk) ejkl im   ,

where the summation is over j and k for a fixed m, and  the structure constant, ejkm , is determined by the following cycles,

(1,2,3) , (1,4,5) , (1,7,6) , (2,4,6) , (2,5,7) , (3,4,7) , (3,6,5)  ,

and is given by the usual permutation rule for a 3-indexed antisymmetric tensor with values 0 or ± 1, i.e.,
 

  ijkm    =  {   1 ,  if (j,k,m) is an even permutation
 -1,   if (j,k,m) is an odd permutation 
  0 ,  otherwise.		 } of the octonion cycle.

Writing S in terms of octonions A and B,  i.e., S := ( A ; B ) , where
    A =  i 2 - 2 i 4 - i 7  ,     B =  2 i 1  - i 3 + i 5
we see that S satisfies the zerodivisor vector-closure conditions as well as the following conditions discussed in our publication
    A . B = 0,                n(A) = n(B),          w :=  ( n (S) )2 / n (A,B) = 1 .

Choosing x 1 , x 5 , x 6 , x 7 , as the free parameters, we obtain
    X = 18x 1i 1 + 6 ( 14x 1 - 4x 5 - 4x 6  + 3 x 7 ) i 2 + 18 ( 2x 1 +  x 5 ) i 3  +
                6 ( 7x 1  - 2x 5  -  2x 6  ) i 4  + 18x 5 i 5   + 18x 6  i 6  + 18x 7 i 7  ,
and
    Y =  ( - 6x1 - 6x5 + 3x6 ) i 1 + ( 2x1 - 16x5 - 7x6 - 12x7 ) i 2  + ( 38x1 - 16x5 + 2x6 + 6x7 ) i 3
              +( -34x1+2x5 + 11x6 - 12x7 ) i 4 + ( 50x 1 - 4x 5 - 4x 6 + 6x 7 ) i 5
              + ( -20x 1 - 20x 5  - 2x 6 + 12x 7 )  i 6   + ( 70x 1 - 20x 5 - 299x 6 + 12x7 ) i 7
 

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