## Wednesday, June 28, 2017

### On Triple Ratios: 3 of 6

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)