On Triple Ratios: 3 of 6

          Reciprocal Additions

         

          Define ‘reciprocal addition’, a.k.a. ‘reduction’ as:

          x [+] y   =  1 / ( (1/x) + (1/y) )   =   xy  /  (y+x)

          Reduction is addition conjugated by reciprocal.

          It is commutative, associative, with identity infinity and attractor zero. Multiplication distributes over it:

          x*(y[+]z)  = (x*y)[+](x*z)

          Reciprocal turn addition and reduction into each other:

          1 / (x+y)       =  1/x  [+]  1/y

          1 / (x[+]y)    =    1/x  +  1/y

          There are three reductions on the triple ratios:

          x [+]₁ y   =  1 / ( (1/x) +₁ (1/y) )   =   xy  /  (y+₁x)

x [+]₂ y   =  1 / ( (1/x) +₂ (1/y) )   =   xy  /  (y+₂x)

x [+]₃ y   =  1 / ( (1/x) +₃ (1/y) )   =   xy  /  (y+₃x)

          Therefore “Complementarity”:

          (x [+]₁ y) * (x +₁ y)       =       x*y

          (x [+]₂ y) * (x +₂ y)       =       x*y

          (x [+]₃ y) * (x +₃ y)       =       x*y

          (x [+]₁ y)    =      

( (x₁y₂+x₂y₁)(x₁y₃+x₃y₁)  ;   x₂y₂(x₁y₃+x₃y₁)   ;   x₃y₃(x₁y₂+x₂y₁)  )

          (x [+]₂ y)    =      

(  x₁y₁(x₂y₃+x₃y₂)  ;  (x₁y₂+x₂y₁)(x₂y₃+x₃y₂)  ; x₃y₃(x₁y₂+x₂y₁) )

          (x [+]₃ y)    =      

(  x₁y₁(x₂y₃+x₃y₂)   ;   x₂y₂(x₁y₃+x₃y₁)     ;  (x₁y₃+x₃y₁)(x₂y₃+x₃y₂) )

          We can interpret reciprocal addition in terms of dual arithmetic:

          x [+] ey      =       xey / (x + ey)  =  (yx/(y²-x²))(y – ex)

          x [+]₁ ey     =       xey / (x +₁ ey)  =  (yx/(y²-₁x²))(y –₁ ex)

          x [+]₂ ey     =       xey / (x +₂ ey)  =  (yx/(y²-₂x²))(y –₂ ex)

          x [+]₃ ey     =       xey / (x +₃ ey)  =  (yx/(y²-₃x²))(y –₃ ex)