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

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Nico Benschop

Posts: 377
Registered: 12/12/04
--- SCI = SIC ---
Posted: Jul 23, 1998 4:01 AM
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--------( 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/(1-n)

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 3-loop 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 i-th 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 "3-loop" 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| = (p-1).p^{k-1} ] we have:

Each 'a' generates a 3-loop 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













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