Tuesday, September 1, 2015

Wilsonian Quantifiers



          Wilsonian Quantifiers

          The science-fiction writer Robert Anton Wilson invented a word, sumbunol, meaning “some but not all”. He did so as a way to combat the temptation to over-generalize.
Sumbunol’s formal definition is:
          Sbn(x)(P(x))        =       Exists(x)(P(x))  and  Exists(y)(not P(y))
                                      =       Exists(x,y)( P(x) xor P(y) )
          The last equation says; “sumbunol things are P” is equivalent to “P differs on some two things”. So sumbunol = varying.
          The negation of sumbunol is nunnerol, meaning “none or all”, with this formal definition:
          Nrl(x)(P(x))        =       All(x)(P(x))  or  All(y)(not P(y))
                                      =       All(x,y)( P(x) iff P(y) )
          The last equation says; “nunnerol things are P” is equivalent to “P is the same for any two things”.  So nunnerol = constant.
          These equations hold:
          Negation:
          Sbn(x)(P(x))  =  Sbn(x)(not P(x))  =  not Nrl(x)(P(x))
          Nrl(x)(P(x))  =  Nrl(x)(not P(x))  =  not Sbn(x)(P(x))

          Partial Distribution:  
A and Sbn(x)(P(x))   =  Sbn(x)( A and P(x) )
          A or Nrl(x)(P(x))   =  Nrl(x)( A or P(x) )
          “And” distributes over sumbunol, and “or” distributes over nunnerol; but “and” does not distribute over nunnerol; nor does “or” distribute over sumbunol:
True  or  Sbn(P(x)) = True;   but Sbn( True or P(x) ) = False.
False and Nrl(P(x)) = False;  but  Nrl( False and P(x) ) = True.

          Sumbunol and nunnerol have these Equivalence Rules:

Nrl(x)( P(x) iff Q(x) )    and     Nrl(x)(Q(x) iff R(x))
Implies       Nrl(x)(P(x) iff R(x))   
If “P iff Q” and “Q iff R” are constant, then “P iff R” is constant.

Nrl(x)( P(x) iff Q(x) )    and     Nrl(x)(Q(x))
Implies       Nrl(x)(P(x))       
If “P iff Q” is constant, and Q is constant, then P is constant.

                   Sbn(x)(P(x))  and  Nrl(x)(Q(x))     
implies   Sbn(x)( P(x) iff Q(x) )
          If P varies and Q is constant, then “P iff Q” varies.  

Nrl(x)( P(x) iff Q(x) )
                   Implies       Nrl(x)(P(x))   iff    Nrl(x)(Q(x))    
          If “P iff Q” is constant, then P and Q are equally constant.

                   Sbn(x)(P(x))   xor   Sbn(x)(Q(x))  
implies       Sbn(x)( P(x) xor Q(x) )
If P varies or else Q varies, then “P or else Q” varies.

          If F(p,q) is any function on Boolean logic, then
                   Nrl(x)(P(x))   and   Nrl(x)(Q(x))
                   Implies       Nrl(x) ( F(P(x),Q(x)) )
          This is “Constancy”: constant inputs imply a constant output.

If F(p,q) is any function on Boolean logic, then
                   Sbn(x) ( F(P(x),Q(x)) )
                   Implies       Sbn(x)(P(x))   or   Sbn(x)(Q(x))
          This is “Variability”: varying output implies a varying input.

                    For all a and b,
                   Nrl(x)(P(x)) and P(a)            
Implies       P(b)
          This is “Proof By Constancy Plus Example”.

                   For all a and b,
                   P(a)    and    not P(b)             
implies       Sbn(x)(P(x))
          This is “Opposing Examples”.


                   Nrl(x)(P(x))   and   Exists(x)(P(x))
implies                 All(x)(P(x))
          This is “Constancy and Existence implies Universality”.

                   Exists(x)(P(x))   
implies                 Sbn(x)(P(x))   or   All(x)(P(x))
          This is “Existence implies Variation or Universality”.
         
          The reverse implications require that the universe of discourse of the quantifiers be not empty; i.e. that something exists:  Exist(x)(x=x)
          Given this assumption, then the above implications are equations.

          If  Exist(x)(x=x),  then
All(x)(P(x))         iff      Nrl(x)(P(x))   and   Exists(x)(P(x))
                   Exists(x)(P(x))    iff      Sbn(x)(P(x))   or   All(x)(P(x))
          If anything exists, then   universality = constancy and existence
                                            and    existence = variation or universality

         

Monday, August 31, 2015

Fermat versus Fermat



          Fermat Versus Fermat
          The Theorem versus the Primes

          Consider Fermat’s Last Theorem:
                   AN + BN   =   CN     has no positive integer solutions if N>2.
          This theorem has finally been proven by Andrew Wiles.
          But consider these equations:
                   13 + 23  =  1 + 8  =  9  = 4 (mod 5)
                   43  =  64  =  4 (mod 5)
          So     13 + 23  =  43  (mod 5)   ;  a modular exception to Fermat’s Last Theorem! Call such an exception a Fermat Triple.
There are many Fermat triples; for modulo 5, the cubing function is 1-to-1 and onto:
          X  mod 5             0       1       2       3       4
          X3 mod 5             0       1       3       2       4
          Therefore cubing is invertible, mod 5. In fact it is its own inverse:
                   (X3)3   =   X  mod 5
          Let (x1/3) be the cube root of x, mod 5; and define the ‘cube root of addition, mod 5’, by conjugating addition by the cube root, as follows:
x +1/3 y     =    (x3 + y3)1/3   (mod 5)
          x +1/3 y  =  z       if and only if      x3+y3  =  z3 .
          The cube root of addition is a Fermat-triple addition; it solves  a3+b3 = c3 for any a and b.  Thus mod 5 maximally violates Fermat’s Last Theorem for cubes.
+1/3    0   1   2   3   4

0        0   1   2   3   4               A Fermat-triple addition
1        1   3   4   2   0        
2        2   4   1   0   3       
3        3   2   0   4   1
4        4   0   3   1   2

          The cube root of addition is commutative, associative, has identity 0, inverse –x, and multiplication distributes over it:
          x*(y +1/3 z)          =  x*(y3+z3)1/3               =       (x3*(y3+z3))1/3   
= ((xy)3+(xz)3)1/3 =       (xy) +1/3 (xz)
So the cube root of addition fits the field axioms. Mod 5 has not just an exception to Fermat’s Last Theorem: it has an entire arithmetical system against it!
Better yet: modulo 5, any odd power equals identity or cube:
x  =  x5  = x9  = …
x3 = x7 = x11 = …
So any odd power is invertible, mod 5; therefore there are Nth roots of addition, mod 5, for any odd N:
x +1/N y     =    (xN + yN)1/N   (mod 5)
x +1/N y  =  z       if and only if      xN+yN  =  zN .
+1/N also fits the field axioms including distribution. So mod 5 has a field of exceptions to Fermat’s Last Theorem for any odd power.
There are Fermat-triples for any odd power not just mod 5, but for any modulus in which all odd powers are 1-to-1 and onto. Which moduli are these? The Fermat primes; 2(2^n)+1. Thus this article’s title!

Theorem: Given any Fermat prime F, the integers mod F have one-to-one odd-power functions.
 Proof:
The integers mod F (a.k.a. ZF) is a field since F is a prime; so its multiplicative group (minus zero) is isomorphic to the cyclic addition group in ZF-1 = Z2^(2^n). Odd powers xN in ZF correspond to odd multiples Nx in Z2^(2^n); if N is odd then it is relatively prime to 2^(2^n), therefore it has a multiplicative inverse: an n such that nN = 1 mod 2^(2^n). Na=Nb implies nNa=nNb implies a=b; so Nx is one-to-one. N(na)=a; so Nx is onto. For odd N, the function Nx is one-to-one onto in Z2^(2^n); therefore xN is one-to-one onto in ZF. QED.

The integers modulo any Fermat prime have one-to-one onto odd power functions; therefore they also have all odd roots of addition. Thus Fermat primes maximally violate Fermat’s last theorem!

The five known Fermat primes are:
3,   5,   17,   257,   65537
         
By Fermat’s Little Theorem, x3 = x mod 3; so xodd = x mod 3. All odd powers are the identity mod 3, so any addition mod 3 is a Fermat triple for all odd powers.
         
We have seen odd roots of addition mod 5, above.


          Now for mod 17. Consider these tables:
X       0   1   2    3    4    5    6   7   8    9  10  11  12  13  14  15  16
X3     0   1   8   10  13   6  12   3   2  15  14   5   11    4    7    9  16
X5     0   1  15   5    4   14   7  11  9    8    6   10    3   13  12   2  16

          From these we can derive Z17  root-addition tables.


+1/3  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16

0    0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
1    1  8 15 12 10 14  4 13  7 16  9  5  6 11  2  3  0
2    2 15 16  1 13 10  7 12  3  5 11  4  8  6  9  0 14
3    3 12  1  7  6 16 11  4  6  2 14  9 13 10  0  8 15
4    4 10 13  6 15  8  2 16  9 12  3  1 14  0  7 11  6
5    5 14 10 16  8  6  1 15  2 13  7 12  0  3  4  9 11
6    6  4  7 11  2  1 14  9 10  3 15  0  5 16  8 13 12
7    7 13 12  4 16 15  9  5 11  1  0  8 10 14  3  6  8
8    8  7  3  6  9  2 10 11 13  0 16  2  4  5 15 12  1
9    9 16  5  2 12 13  3  1  0  4  6  7 15  8 11 14 10
10   10 9 11 14  3  7 15  0 16  6 12  8  2  1 13  5  4
11   11 5  4  9  1 12  0  8  2  7  8  3 16 15  6 10 13
12   12 6  8 13 14  0  5 10  4 15  2 16 11  9  1  7  3
13   13 11 6 10  0  3 16 14  5  8  1 15  9  2 12  4  7
14   14 2  9  0  7  4  8  3 15 11 13  6  1  1 10 16  5
15   15 3  0  8 11  9 13  6 12 14  5 19  7  7 16  1  2
16   16 0 14 15  6 11 12  8  1 10  4 13  3  3  5  2  9



+1/5  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
0    0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
1    1 15 16 10  3  2  9 14 11  8  6  7  4  5 13 12  0
2    2 16 13 12 15 13  3  8  6 10  4  9  1  7 11  0  5
3    3 10 12 11  8 15 14 16  5 13  7  2  9  1  0  6  4
4    4  3 15  8  9  1  7  2 13 14 11  5  6  0 16 10 12
5    5  2 13 15  1  7  4  9 10  3 12  6  0 11  8 16 13
6    6  9  3 14  7  4  5  1 16  2 13  0 11 12 15  8 10
7    7 14  8 16  2  9  1  3 12 15  0  4  5  6 10 13 11
8    8 11  6  5 13 10 16 12  1  0  2 15 14  3  4  7  9
9    9  8 10 13 14  3  2 15  0 16  5  1  7  4 12 11  6
10   10 6  4  7 11 12 13  0  2  5 14 16  8 15  1  9  3
11   11 7  9  2  5  6  0  4 15  1 16 12 13 10  3 14  8
12   12 4  1  9  6  0 11  5 14  7  8 13 10 16  2  3 15
13   13 5  7  1  0 11 12  6  3  4 15 10 16  8  9  2 14
14   14 13 11 0 16  8 15 10  4 12  1  3  2  9  6  5  7
15   15 12 0  6 10 16  8 13  7 11  9 14  3  2  5  4  1
16   16 0  5  4 12 13 10 11  9  6  3  8 15 14  7  1  2

          I leave computing tables for +1/7, +1/9, +1/11, +1/13, and +1/15, mod 17, as an exercise for the ambitious student.

          What of even powers?
X       0   1   2    3    4    5    6   7   8    9  10  11  12  13  14  15  16
X2     0   1   4    9   16   8    2  15  13  13 15   2    8    16   9    4   1
X4     0   1  16  13   1   13   4   4   16  16   4    4    13   1   13  16  1
X8     0   1   1   16   1   16  16  16  1    1  16  16   16    1  16   1    1
X16    0   1   1   1    1    1     1   1    1    1   1     1    1     1    1   1    1

There are many sums among the squares table: 1+8=9, 1+15=16, 1+16=0, 4+9=13, 4+13=0, 4+15=2, 9+16=8, 9+8=0, 9+13=5. These correspond to the classic Pythagorean triples:
(m2 – n2)2 + (2mn)2  =  (m2 + n2)2
m=2, n=1  yields the triple 32+42=52 ;  i.e. 9+16=8
m=3, n=2  yields the triple 52+122=132 ;  i.e. 8+8=16
m=3, n=1  yields the triple 82+62=102 ;  i.e. 13+2=15
m=4, n=1  yields the triple 152+82=172 ;  i.e. 4+13=0
m=4, n=2  yields the triple 122+162=32 ;  i.e. 8+1=9
m=5, n=4  yields the triple 92+62=72 ;  i.e. 13+2=15
and so on.

ZF supports higher Pythagorean triples; for if N is odd then:
(m2 +1/N (-n2))2N  +  (21/N mn)2N  =  (m2 +1/N n2)2N
          Proof: Left side   = (m2 +1/N (-n2))2N  +  (21/N mn)2N 
                   = (m2N + (-n2N))2  +  (2 mNnN)2 
                   = (m2N - n2N)2  +  (2 mNnN)2            if N is odd
                   = m4N – 2m2Nn2N+ n4N +  4 m2Nn2N
                   = m4N + 2m2Nn2N+ n4N
=   (m2N + n2N)2
=  (m2 +1/N n2)2N =       right side



          21/3 = 8 and 21/5 = 15, mod 17; therefore:
(m2 +1/3 (-n2))6  +  (8 mn)6   =   (m2 +1/3 n2)6         mod 17
(m2 +1/5 (-n2))10  +  (15 mn)10   =  (m2 +1/5 n2)10     mod 17
          so for instance;
                   m=2, n=1  yields          66 + 166   =   106        
                                      and             1210 + 1310 = 310
                   m=3, n=1  yields          106 + 76   =   166       
                                      and             610 + 1110 = 810
                   m=3, n=2  yields          86  +  146   =   126       
                                      and             410 + 510 = 1410          -   all mod 17.

          There are sums amongst the 4th and 8th powers: 1+16=0, 13+4=0 for 4th powers; 1+16=0 for 8th powers; but all 16th powers are 1 or 0, so the only possible sums amongst the 16th powers are trivial: 0+B=B.
          The higher binary powers involve 2kth roots of -1:
          If r(2^k) = -1 mod F, then    (a)(2^k) + (ra)(2^k)  =  0(2^k)
 And more generally    (a*(rodd))(2^k) + (a*(reven))(2^k)  =  0(2^k)
So since 24 = -1 mod 17, and 38 = -1 mod 17:
          (a*(2odd))4 + (a*(2even))4  =  04
(a*(3odd))8 + (a*(3even))8  =  08
These ‘hypercomplex’ triples include
14+24 = 24+44 = 34+64 =44+84 = 54+104 =64+124 =74+144 = 04
18+38 = 28+68 = 38+98 =48+128 = 58+158 =68+18 = 08
          Since (-1)odd = -1, hypercomplex triples for power 2k are also triples for power odd*2k. For instance, 4th roots of -1 are also 12th roots; therefore
112+212 = 212+412 = 312+612 =412+812 = 512+1012 =612+1212 =712+1412 = 012

Conjecture: modulo a Fermat prime F, the Fermat triples are:
For any power N: the trivial case   
0N + BN  =  BN
For odd power N: odd roots of addition: 
AN  +  BN    =    ( A  +1/N  B )N
For power 2: the Pythagorean triples:
(m2 - n2)2  +  (2 mn)2  =  (m2 + n2)2
For power Odd*2: the higher Pythagorean triples:
(m2 +1/N (-n2))2N  +  (21/N mn)2N  =  (m2 +1/N n2)2N
For power Odd*2k: hypercomplex triples:
          If r(2^k) = -1 mod F, then
(a)(2^k) + (ra)(2^k)  =  0(2^k)
And more generally
(a*(rodd))(Odd*2^k) + (a*(reven))(Odd*2^k)  =  0(Odd*2^k)

This seems to apply to mod 3, 5 and 17, but I haven’t checked mod 257, let alone mod 65537!