Friday, November 1, 2013

On Reduction, 10 of 11



     Logistic
    


     Zero is a problem for reduction, just as infinity is a problem for addition. By the Rabbit Rule;

           1/0 + 1/0 = (0*1+1*0)/(0*0)  = 0/0,

     the indefinite ratio; and by the Stool Rule;

           0 <+> 0  =  0/1  <+>  0/1  =  (0*0)/(0*1+1*0)  = 0/0.

     This makes sense if infinity and zero are unsigned, for then

       1/0 + 1/0  =  1/0 – 1/0    =  0/0 , by common denominators;
       0/1 <+> 0/1 =  0/1 <-> 0/1  =  0/0 , by common numerators.

     If we give signs to infinity and zero, then we can reduce some of the indefiniteness:

           +∞  +  +∞    =    +∞
           -∞  +  -∞    =    -∞
           +∞  +  -∞    =    0/0
           +0  <+>  +0  =    +0
           -0  <+>  -0  =    -0
           +0  <+>  -0  =    0/0

     That implies these tables:

           +  |  +0 +∞     <+>| +0   +∞      x | 1/x  
           --------------  ---------------  ----------
           +0 |   0       +0 |  0    0     +0 | 
           +∞ |          +∞ |  0        +∞ |  0

     This is isomorphic to Boolean logic, under this matching:

           0    --------------- true

              --------------- false

          +    --------------- and

          <+>  --------------- or

          1/x  --------------- not

     Let F be the generic positive finite quotient. (So F = n/m, with n and m both positive integers.) Then we get these tables:


          +   |  0   F             <+>|  0  F      x  | 1/x
          ------------------    -------------     ---------
          0   |  0   F              0  |  0  0  0     0  | 
          F   |  F   F              F  |  0  F  F     F  |  F
             |                     |  0  F        |  0


     This duplicates 3-valued Kleenean logic, under this matching:


          0    ------------  true

          F    ------------  intermediate

              ------------  false

          +    ------------  and

          <+>  ------------  or

          1/x  ------------  not


     Addition, reduction and reciprocal are a fragment of arithmetic; when applied to {0, finite, ∞}, they are isomorphic to Kleenean logic; when applied to the extended positives, they resemble a fuzzy logic.
     Under this matching, the DeMorgan laws:

           not(x and y)   =  (not x) or (not y)

           not(x or y)    =  (not x) and (not y)

     have these arithmetical counterparts:

           1/(x + y)      =  (1/x) <+> (1/y)

           1/(x <+> y)    =  (1/x) + (1/y)

     Therefore the name, "De Morgan distribution".



     Define the “Mediant” operator (x@y@z) this way:

           x @ y @ z  =    (x*y*z)/((x+y+z)(x<+>y<+>z))

     This implies:

           x*y*z      =    (x+y+z)(x@y@z)(x<+>y<+>z)

           x@y@z      =    (xy+yz+zx)/(x+y+z)

           x@y@z      =    (xy<+>yz<+>zx)/(x<+>y<+>z)

           w*(x@y@z)  =    (w*x)@(w*y)@(w*z)

           1/(x@y@z)  =    (1/x)@(1/y)@(1/z)

           x@0@y      =    x <+> y

           x@∞@y      =    x + y

     The last three laws show that the mediant corresponds to the majority operator in logic, M(x,y,z); since

           Not(M(x,y,z))   =    M(not x, not y, not z)

           M(x,T,y)        =    x or y

           M(x,F,y)        =    x and y


     Now consider the “equivalence” operator:

           x iff y  =  (x or not y) and (y or not x)

                    =  (x and y) or (not x and not y)

     It has this arithmetical counterpart:

               (x <+> 1/y) + (y <+> 1/x) 

           =   (x + y) <+> (1/x + 1/y) 

           =   (1<+>xy)/(x<+>y)

           =    (x+y)/(1+xy)

     - otherwise known as relativistic velocity addition! Denote it by  “x~y”, and call it “equivalence”, or “Einstein addition”. Its laws are:

          x ~ 0           =       x

          x ~ -x          =       0

          x~y             =      y~x

          (x~y)~z          =    x~(y~z)     =    x~y~z

           x~y~z           =   (x+y+z+xyz)/(xy+yz+zx+1)

           x~y~z        =   (x<+>y<+>z<+>xyz)/(xy<+>yz<+>zx<+>1)

           x ~           =      1/x

          x ~ -1/x       =      

           ~           =       0

          1/(x~y)         =   (1/x)~y   =   x~(1/y)  =  x~y~

           1/(x~y)     = (x<+>y)+(1/x<+>1/y) = (x+1/y)<+>(y+1/x) 

           1/(x~y)         =   (x<+>y)/(1<+>xy) = (1+xy)/(x+y)

          (1/x)~(1/y)     =      x~y

          -(x~y)          =    (-x)~(-y)

          -(x~y~z)         =    (-x)~(-y)~(-z)

          1/(x~y~z)        =    (1/x)~(1/y)~(1/z)

          x ~ 1           =       1

          x ~ -1         =      -1

          1 ~ -1         =      0/0

           (A~x) = B       if and only if        x = (-A~B)

           tanh(x) ~ tanh(y)     =    tanh (x + y)

           x ~ y          =    tanh (tanh-1x + tanh-1y)

     So Einstein addition is addition conjugated by hyperbolic tangent:

(x+y)/(1+xy)  =  tanh( arctanh(x) + arctanh(y) )
The left-side definition works even for x and y not in the domain of arctanh; so equiv is more than just an addition. Einstein addition is more like a multiplication than an addition, for both have order-two elements:     
x~~ =  x   ;   x*-1*-1 =  x

Now consider these ‘diagonal turn’ functions:
b(x)  =  tan(arctan(x) - 45o)
=   (x - 1) / (x + 1)
q(x)  =  tan(arctan(x) + 45o)
=   (x + 1) / (-x + 1)
      If x is the slope of a line, then b(x) is the slope of that line turned clockwise 45 degrees; and q(x) is the slope of that line turned counterclockwise 45 degrees. b sends 0 to -1 to to 1 to 0; and q sends 0 to 1 to to -1 to 0; a period 4 cycle.
The diagonal turn operators b and q are also Einstein additions conjugated by complex multiplication:
b(x) =  i * ( (-1/i) ~ (x/i) )
q(x) =  i * (  (1/i) ~ (x/i) )

In general;
b(q(x)) = q(b(x)) = x
b(b(x)) = q(q(x)) = -1/x
b(b(b(x))) = q(x)   ;  q(q(q(x))) = b(x)
b(b(b(b(x)))) = q(q(q(q(x))))  =  x
b and q are inverses, both with period 4. They are also negative reciprocals:      q(x) * b(x) =  -1

We can derive these identities and conjugations:
b(x*y) = b(x)~b(y)
b(q(x)*q(y)) = x~y
b(x~y) = -b(x)*b(y)                
b(q(x)~q(y)) = -x*y                
q(-x*y) = q(x)~q(y)
q(-b(x)*b(y)) = x~y
q(x~y) = q(x)*q(y)
q(b(x)~b(y)) = x*y

b(-x)  = - q(x) =  1/b(x)
b(1/x) = 1/q(x) =  - b(x)
b(-q(x))  =  1/x
b(1/q(x)) =  - x

q(-x)  = - b(x) =  1/q(x)
q(1/x) = 1/b(x) =  - q(x)
            q(-b(x))  =  1/x
q(1/b(x)) =  - x             

Like exponentials, b and q turn minus into reciprocal; like logarithms, they turn reciprocal into minus. These 45-degree turns also turn multiplication into Einstein addition, and vice versa. So reciprocal is diagonal minus, and vice versa; and Einstein addition is diagonal multiplication, and vice versa.





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