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Topic:
 SCI = SIC 
Replies:
1
Last Post:
Jul 24, 1998 9:50 AM




 SCI = SIC 
Posted:
Jul 23, 1998 4:01 AM


( Do not associate this title with Science = Sick ;) ( Rather: an exercise in function composition ) ^^^^^^^^^^^^^^^^^^^^
Consider in arithmetic ring Z(+,.) the two basic symmetries n and 1/n
as functions, in fact automorphisms of order 2, of Z(+) resp. Z(.):
Complement C(n) = n, Inverse I(n) = 1/n (n<>0)
And denote the successor function as: S(n) = n+1.
Notice that IC = CI (commute), but S does not commute with I or C.
(so indeed SCI = SIC ;)
Then compose all three functions in all possible (four) ways:
n(ICS) = 1 1/n
n(SCI) = 1/(1+n)
n(CSI) = 1/(1n)
n(ISC) = (1+ 1/n)
Watch notation: function composition is from left to right.  ^^^^^^^^^^^^^^^^^^ Call such composition a "dfs" (dual folded successor) function.
Theorem (basic 3loop of arithmetic):  The third iteration of each dfs function is the identity function E (nE = n, for all n<>0,1,1).
Proof:  Denote the ith iteration of a function F by F_i; show: (dfs)_3 = E.
Then by complete inspection the theorem follows easily.
For instance n(SCI)_3: substitute 1/(1+n) for n three times:
n(SCI)_2 = n(SCI SCI) = 1/[1 1/(1+n)] = (1+n)/[1+n 1] = (1+n)/n
n(SCI)_3 = 1/n 1 = (1+n) 1 = n ... SCB(simple comme bonjour).
Similarly, verify the other three dfs functions to have period three.
==========
An interesting consequence of this "3loop" property of arithmetic is
that for residues mod p^k (prime p>2, k>1):
In units group G_k of all residues coprime to p [ which is a cyclic group with G_k = (p1).p^{k1} ] we have:
Each 'a' generates a 3loop of three inverse pairs, called "triplet":
a+1 = 1/b > b = 1/(a+1) b+1 = 1/c > c = 1/(b+1) c+1 = 1/a > a = 1/(c+1) with abc=1 (all mod p^k)
(provided division by zero is avoided, so e.g: a+1 <>0, etc.)
Basic triplet example is in G(.) mod 9 = 2* = {2, 4, 8, 7, 5, 1}
where 8= 1, 7= 2, 5= 3 (mod 9).
Iteration (start value '1'):
1(SCI)_*: 1/(1+1)=4, 1/(4+1)=7, 1/(7+1)=1, with abc=1.4.7=1 mod 9
===========
Note: maximal period=3 is the number of symmetries +1 (coincidence?)
For this function composition result applied to a proof of FLTcase1, see paper ref[1]in .dvi on my homepage; short intro ref[3] in .htm
Comments (from any open mind;) are welcome.
Ciao, Nico Benschop (email\X)  AHA: One is Always Halfway Anyway http://www.iae.nl/users/benschop  xxxxxxxxxxxxxxxx.xxxxxxxxxxxxxxxx



