Monday, June 26, 2017

On Triple Ratios: 1 of 6



          On Triple Ratios



          Triple Ratios and their Arithmetics


          A triple ratio is (a;b;c), with this equality rule:
                   (a;b;c) = (A;B;C)          if and only if
                   aB=Ab  and  bC=Bc  and cA=Ca.
          So if A, B and C are not zero, then
                   a/A  =  b/B  =  c/C

          This is triple equal proportion. It applies, for instance, to the sides of similar triangles; and also the Sine Law:
          (a;b;c)   =  (sin(α), sin (β), sin(γ))

The definition of equality implies the Cancellation Law:
(ka;kb;kc)  =  (a;b;c)

This in turn implies, if a, b, and c are all nonzero:
(a;b;c)  =  (1 ; b/a ; c/a)  =  (a/b ; 1 ; c/b)  =  (a/c ; b/c ; 1)

We define multiplication, unity and inverses this way:

(a;b;c)*(x;y;z)      =       (ax; by; cz)
(1;1;1)                 =       1
1/(a;b;c)               =       (bc; ca; ab)
So if a, b and c are all nonzero:
1/(a;b;c)               =       (1/a; 1/b; 1/c)


Define these trios of zeros, infinities and negatives:
01  =  (1;0;0)                
02  =  (0;1;0)                
03  =  (0;0;1)      
1  =  (0;1;1)               
2  =  (1;0;1)      
3  =  (1;1;0)      
-11  =  (-1;1;1)     =  (1;-1;-1)
-12  =  (1;-1;1)     =  (-1;1;-1)
-13  =  (1;1;-1)     =  (-1;-1;1)

(0;0;0) is the indefinite triple ratio. It equals all ratios, and it is the only one that does so.

For any n = 1, 2, or 3, and if abca, then
(0n)2   =   0n
0a0b = 0b0c =  0c0a =  (0;0;0)
1/0n  =  (0;0;0)
(∞n)2   =   n
ab = 0c 
bc = 0a 
ca = 0b 
abc  =  (0;0;0)
1/∞n  =  0n
(-1n)2     =    1
-1a*-1b  =  -1c   
-1b*-1c  =  -1a
-1c*-1a  =  -1b 
-1a*-1b*-1c   =   1
1/-1n     =    -1n

For any real number R, define these triple ratios:
R1     =       (1; R; R)
R2     =       (R; 1; R)
R3     =       (R; R; 1)

Then:
R1 R2 R3    =       1
R1 R2          =       1/R3 
R2 R3          =       1/R1 
R3 R1          =       1/R2 

Define these three additions:

(a;b;c) +1 (x;y;z)  =       (ax; bx+ya ; cx+za)
(a;b;c) +2 (x;y;z)  =       (ay+xb; by; cy+zb)
(a;b;c) +3 (x;y;z)  =       (az+xc; bz+yc ; cz)
“Two-Denominators Rule”

For each +n, the nth term is the denominator, and the other two terms are independent numerators. These rules follow:

(a;b;c) +1 (a;y;z)  =       (a; b+y ; c+z)
(a;b;c) +2 (x;b;z)            =       (a+x; b; c+z)
(a;b;c) +3 (x;y;c)  =       (a+x; b+y ; c)
“Common Denominators Rule”

(1;b;c) +1 (1;y;z)           =       (1; b+y ; c+z)
(a;1;c) +2 (x;1;z)           =       (a+x; 1; c+z)
(a;b;1) +3 (x;y;1)           =       (a+x; b+y ; 1)
“Unit Denominators Rule”


From +n and -1n, define –n:
x –1 y          =       x  +1 (-11)y
x –2 y          =       x  +2 (-12)y
x –3 y          =       x  +3 (-13)y

          Each of the three additions, subtractions, units and zeros form a ring with * and reciprocal:

          +n is commutative, associative, has identity 0n and negative (-1n)x
          Distribution works: a*(b+nc)  = (a*b)+n(a*c)
          Multiplication is commutative, associative, has identity 1.

          However reciprocal is problematic with the zeros and the infinities. The reciprocal of an infinity is a zero, the reciprocal of a zero is the indefinite ratio, and an infinity times its zero is indefinite.

          In the 3 arithmetic, any triple ratio (a;b;c) is either an infinity (a;b;0) or it equals (a/c; b/c; 1). In the unit-denominator ratios, all operators work independently on the first two terms, and leave the last term equal to one:
          (a;b;1)*(A;B;1)   =  (aA; bB; 1)
          1 / (a;b;1)             =  (1/a; 1/b; 1)
          (a;b;1)+3(A;B;1)   =  (a+A; b+B; 1)
          (a;b;1)-3(A;B;1)   =  (a-A; b-B; 1)

          So unit-third-term ratios under the third arithmetic are isomorphic to pairs of numbers operating in parallel; the dual, or hyperbolic numbers. Similarly with unit-first-term ratios under the first arithmetic, and unit-second-term ratios under the second arithmetic.


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