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15848Final Corrected version of Trying solving Goldbachs' conjecture.

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  • zaljohar
    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.


      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

      "|" 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

      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"
      | | | | |
      ----------------------------------- Pairing line 2

      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:


      At P3 = 5,11,17,23,29,(35),41,47,53,59,(65),71,77,83,89,
      At P3 =7,13,19,(25),31,37,43,49,(55),61,67,73,79,(85),91,97,103,109,

      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);(

      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):

      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)

      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

      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

      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
      THEM,reflecting a general increament in pair sparing as 2n increases.