## 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)

(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.