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• The solution of Goldbach s conjectur 1)The conventional statement: BELOW every even natural number 2n 2 ,there is at least one prime pair( P_i , P_ j ) such
Message 1 of 1 , Jan 3, 2005
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The solution of Goldbach's conjectur

1)The conventional statement:

BELOW every even natural number 2n >2 ,there is at least one prime
pair( P_i , P_ j ) such that P_i + P_ j=2n .

2)The natural statement:

AROUND every non prime natural number q>1 ,there is at least one
prime pair( P_ i , P_ j ) such that P_i + d = q and
P_ j - d = q ,d=1,2,3,......,q - 1.

In the following proof it is the prime pairs of the natural
statement that is concerned.

Any series X is better viewed as an ordered collection of MEMBERs
and GAP NUMBERs as in the example below:

O= 1,3,5,7,............. = 1,(2),3,(4),5,(6),7,......... . The
numbers between brackets are the gap numbers.

A series X is called slideable for its gap numbers if every gap
number g has MEMBER pairs ( X_i , X _ j )around it such that
X_i + d = g and X_ j - d =g , d is a natural number.

3) The slideability statement:

The Prime number series is slideable for its gap numbers.

PROOF:

1) The prime number series P is formed by sequenstial exclusions
(conversions to gap numbers) of members of
N=2,3,4,5,..............as defined below:

nx = 2n,3n,4n,5n,......................=multiples of n except n
itself

"|" means except or exclude or seive.

N | _1x = N excluding multiples of the first member in N , ("_1"=2)
except itself = 2,3,(4),5,(6),7,(8),9,(10),11,.........

N| _1x|_2x = N|_1x excluding multiples of the second member in
N|_1x ,( "_2" =3) except itself.

P = N |_1x|_2x|_3x|_4x|...........................

Pk= N |_1x|_2x|_3x|..........|_kx

It is obvious that the members of Pk < ( _ [K+1] ) ^ 2 are all
prime numbers while the rest are either primes or non prime odds.

3) The exclusive process defined above " THE SEIVE" do not change
the slideability for gap numbers present in N |_1x, as a result each
Pk is also slideable for its gap numbers , and of course as a result
P would be also slideable for its gap numbers , and therefore
Godlbach's conjecture would be prooved.

BUT how do we proove that the seive is not a distroyer of
slideability?????....

Answer: The first step is that we should confirm that every Pk is
slideabile for its gap numbers " Pk:\$g" by testing large number of
Pk series from P2 till Pm were m is a very large positive integer.

Note:Although each Pk is an infinite series, the slideability of it
for ALL it's gap numbers can be prooved ,because each Pk consist of
infinite repeat of the same"MEMBER-GAP" segment, therefore if
slideability is prooved for the gap numbers of the first segments
then it can be generalized to all gap numbers of the series Pk, for
example N |_1x|_2 has a repeated segment of m( _ )m( _ )( _ ) ( _ ),

( m is a member and ( _ ) is a gap number ) , then if we proove
slideability for the first member -gap segments then we can
generalize the result to all other segments because they are
repeatitions of the first segment.

The gap size is the number of gap numbers between each two members ,
so the repeated segment better symbolized as m1m3 or simply { 1,3 }.
Example the repeated member-gap segment in P3={ 3,1,3,1,3,5,1,5}

The number of m in a single member - gap segment in

Pk = ( _2 - 1) ( _3 - 1 ) ( _4 - 1 ) ( _5 - 1 ) .........( _k - 1 )

Example: the number of members in a single member - gap segment in

P4=( _2 - 1) ( _3 - 1 ) ( _4 - 1 ) =(3-1)(5-1)( 7-1)= 48

The member - gap segment is denoted as [m-g] , the number of m in
[m-g] in Pk is mPk

Pk, [m-g]i is the ith [m-g] in Pk like for example

P2, [m-g]1 = 5,(6),7,(8),(9),(10)
P2, [m-g]2 =11,(12),13,(14),(15),(16)

Pk,[m-g]1_1 is the first member of the first [m-g] in Pk= _(k+1)

so P2,[m-g]1_1 =5, while p2,[m-g]2 _1 =11 , etc...

The size of Pk,[m-g] which is the number of natural numbers in[m-g]
wheather they are gap numbers or members is d= _k!p

!p is prime factorial so 7!p = 2x3x5x7, examples for P4,d=7!p=210 ;
P3,d=2x3x5=30 ; P2,d=2x3=6; ....etc

THE PAIRING LINES: The most important concept in the solution of
Goldbach's conjecture is the pairing lines of Pk !!!!!

Each Pk has mPk pairing lines, along each pairing line a member is
repeated at distance of Pk,d=_k!p ,

As an example see the figure below :

______________________________________ Pairing line 1
| | | | | "PL1"
P2=123*5*7***11*13***17*19***23*25***29*31
| | | | |
----------------------------------- Pairing line 2
"PL2"

Now one should learn how to view Pk series longitudinally as a
series of many subseries each series is called Pk, PLi

of course Pk,PL1 _ 1 = Pk,[m-g]1_1 ; Pk,PL2 _ 1 = Pk,[m-g]1_2 ;
Pk,PL3 _ 1 = Pk,[m-g]1_3

Now viewing Pk series as repeated segments [m-g] is called
horizontal view of Pk , while viewing Pk as a mother series of many
subseries each is a pairing line is called a longitudinal view. Also
one can name them as serial and parallel views respectively.

There are also other important horizontal views of Pk like veiwing
it as having a segment below Pk,[m-g]1_1 and the [m-g] repeated
segments starting by Pk,[m-g]1_1. Also there is the other important
horizontal view of the Pk < (_(k+1))^2 , were all the members are
prime, and Pk> =(_(k+1))^2 , were some of the members are prime and
others are non prime odds.

Back to prooving that the seive do not distroy slideability of
N |_1x , such that all series Pk generated by the seive on N |_1x,
retain slideabilty for gap numbers , therefore prooving Goldbach's
conjecture .

THE PROOF:

1)Each exclusive step of the stepwize sieve denoted "|_kx" , affects
all pairing lines of Pk in a similar manner, and exclusions of
members belonging to a pairing line do not affect pairing of members
belonging to other lines. This is the most important concept in
solving the conjecture.Therefore prooving should be made along one
pairing line since what happens at the others is similar.

2) Since pairing of members at a pairing line in Pk happens at
multiples of d= _k!p and exclusions forming Pk+1 from Pk are
happening at multiples of e= _(k+1) and since d/e is not a natural
number (for obvious reasons), then there is a pairing - exclusion
discripancy. It is obvious that for conversion of members AT THE
SAME PAIRING LINE in Pk to happen by the exclusion e it will take d
times of e to happen , ie exclusions happens at multiples of de , so
this will spare e-1 of every e member pairs( AT THE SAME PAIRING
LINE) around each converted member OF THAT PAIRING LINE inPk, and
spare minimally e-2 of every e member pairs AT THE SAME PAIRING LINE
around each member OF THAT PAIRING LINE inPk that remains as a
member in Pk+1, and minimally e-2 of every e member pairs AT THE
SAME OR DIFFERENT PAIRING LINES in Pk around each remaining gap
number of Pk in Pk+1.

The above statement can be simply seen by sketching these pairs in
any pairing line of a specific Pk.

Visualizing examples:

P2,PL1=5,11,17,23,29,35,41,47,53,59,65,71,77,83,89,95,101,107,113,119
,125,131,137,143,149,155,161,...............
P2,PL2=7,13,19,25,31,37,43,49,55,61,67,73,79,85,91,97,103,109,115,121
,127,133,139,145,151,157,163,................

At P3 = 5,11,17,23,29,(35),41,47,53,59,(65),71,77,83,89,
(95),101,107,113,119,(125),131,137,143,149,(155),161,...............
At P3 =7,13,19,(25),31,37,43,49,(55),61,67,73,79,(85),91,97,103,109,
(115),121,127,133,139,(145),151,157,163,................

P2 is the prior series(prior to being converted by |_k+1x),P3 is the
resulting series , k=2 ,_2=3 , k+1=3 ;_3=5

For number 85 (the converted member of P2 to a gap number in P3.)
every 5 pairs loos one pair ,the pairs lost are symmetrical (ie both
of each pair members are converted to gap numbers): (55,115);(
25,145).

For number 79 ( the remaining member of P2 in P3)every 5 pairs loos
two pairs, the lost pairs are asymmetrical (ie only one of each pair
members is converted to gap number thus eliminating that pair):
(73,85)(55,103)(43,115)(25,133)(13,145)

For number 82(the remaining gap number of P2 in P3)every 5 pairs
loos two pairs due to asymmetrical member pair lose , the lost pairs
are :(79,85) ( 55,109)( 49,115)(25,139)( 19,145)

For number 84 ( the remaining gap number of P2 in P3) it uses two
pairing lines since in P2 it lies between 85 and 83, so the pairing
distance is 1+ 6i like (77,91)(71,97).......,of each 5 of these
pairs two will be converted asymmetrically:(83,85)(65,103)(53,115)
(35,133)(23,145)

if the remaining gap number is paired by members from different
lines like number 84 in the example above, then we say that its
pairing at distance z+_k!p were z is calculated as below

k-2
z =1,2,3,6,7,8,9,30,31,32,33,36,37,38,39,.,Sum (Pk_i)!p +1
i= 1

All the above is applied for n were 2n is large enough to allow such
conversions weather symmetrical or not , however in the case were n
is small and 2n is smaller than the exclusive jump, then only
unitary asymmetrical pair lose will occure per pairing line .THESE
UNITARY PAIR LOSE WILL INCREASE AS 2n increases,accounting for the
increased pair spare tendancy as 2n increases.

The message of this proof is that exclusions happens at wide
distances that give large room for spairing pairs around each member
or gap number in Pk at all prior pairing
distances"d=1,2,6,30,......", therefor not altering slideability of
Pk for its gap numbers thus prooving Goldbach's conjecture.

The horizontal proof:

This proof is more difficult to be see than the previous one.

Every 2n , has minimum possible prime pairs around n calculated as
below:

Min (P_i,P_ j)=[INT{INT[INT(n/2)-1]/3 + f(R[INT(n/2)-1]/3)}/_3 ]
[( _3) - 2] +

f(R{INT[INT(n/2)-1]/3 + f(R[INT(n/2)-1]/3)})/_3............/_k

( _k)^2 < 2n <( _(k+1))^2 ; f(y)=0 for y< 2 ; f(y)=y-2 for y>1

INT stands for integer like INT 6.9 = 6
R stands for the remainder as below:

R(x/y)=x- yINT(x/y)

The above formula is very agressive though it is the most minimum,
another less agressive forumla would be like below:

Min (P_i,P_ j)=INT((n/2)-1 /3)(3/5)(5/7)(9/11)...([( _k ) - 2])/ _k

( _k)^2 < 2n <( _(k+1))^2

I have tested both formulae using computor programs till P9, 2n=600,
both were working.Min(P_i ,P_j)>=1 , for 2n above 64 for the first
formula and a much smaller number for the second formula, and since
P1 till P4 are all proved as slideable manually,
then Min(P_i ,P_j)>=1 prooves Goldbach's conjecture.INDEED AS 2n
increases, THE DIFFERENCE BETWEEN THE NUMBER OF GOLDBACH'S PRIME
PAIRS AND Min(P_i ,P_j)FOR EACH 2n INCREASES,SO IS THE RATIO BETWEEN
THEM,reflecting a general increament in pair sparing as 2n increases.

THEREFORE GOLDBACH'S CONJECTURE IS PROOVED!!!!!!!!!...........

Zuhair
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