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prime symmetry

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  • Mike Antczak
    Hi, this may be a trivial (or well known) result, but when I first observed it I was quite surprised. NOTE: [] indicates subscript. Let c = q[1]q[2]...q[n],
    Message 1 of 4 , Jul 9, 2003
      Hi, this may be a trivial (or well known) result, but
      when I first observed it I was quite surprised.

      NOTE: [] indicates subscript.

      Let c = q[1]q[2]...q[n], where q is an odd prime
      number then c is an odd composite.

      Let P = {p[1],p[2],...,p[m]}, where p is an odd prime
      and gcd(p,q)=1, p[m] < 2c < p[m+1]

      Let A and B be partition cells of P, such that {p:p in
      A, p < c} and {p':p' in B, c < p' < 2c}.

      Case 1.
      If p[1]^2 > 2c, p[1] in A, then 2c - p = p', where p
      in A, and p' in B.

      Ex.
      c = 15 = 3*5
      2c = 30
      P = {7,11,13,17,19,23,29}
      A = {7,11,13}
      B = {17,19,23,29}

      30 - 7 = 23
      30 - 11 = 19
      30 - 13 = 17

      Case 2.
      If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p
      (except when p = 1), where p in A, and p' in B.

      Ex.
      c = 21 = 3*7
      2c = 42
      P = {5,11,13,17,19,23,29,31,37,41}
      A = {5,11,13,17,19}
      B = {23,29,31,37,41}

      42 - 23 = 19
      42 - 29 = 13
      42 - 31 = 11
      42 - 37 = 5

      What does this suggest? This suggests that there is a
      fundamental symmetry within the primes.

      Proof:

      NOTE: this proof is only valid for case 1 and with
      slight modification for case 2.

      Using the previous definitions unless specified:

      Let q' be the mirror image of q with respect to c,
      meaning that q'-c=c-q or q'=c+c-q (i.e. the distance
      from q to c is equal to the distance from q' to c).
      Recall that c=q[1]q[2]...q[n].

      Let p' be the mirror image of p with respect to c,
      meaning that p'-c=c-p or p'=c+c-p (i.e. the distance
      from p to c is equal to the distance from p' to c).
      Recall that p is an element in A and p' is an element
      in B.

      Since q|c and q|(c-q), then it follows that q|(c+c-q)
      or q|q'.

      Since p does not divide c and p does not divide c-p,
      then it follows that p does not divide c+c-p or p does
      not divide p'.

      Therefore p'=c+c-p or p'=2c-p and a p with equal
      distance from c exists on either side of c as was
      defined in case 1. Hence the symmetry.

      Q.E.D.

      A similar proof could be shown for case 2 but a
      general case, where p[1]^ < c, would begin to be
      incoherent because the elements in A would create
      composite numbers that when are subtracted from 2c
      yeild an element in B.

      Ex.
      c=25=5*5
      2c=50
      P={3,7,11,13,17,19,23,29,31,37,41,43,47}
      A={3,7,11,13,17,19,23}
      B={29,31,37,41,43,47}

      But...
      3*3=9 and 50-9=41, which is an element of B, therefore
      50-41 does not produce a prime.


      When all goes well, it is obvious that p+p'=2c, which
      implies Goldbach's Conjecture. And if GC is false,
      then all of the p must produce composite numbers that
      equal 2c-p'. Is this possible?

      -Mike


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    • Jose Ramón Brox
      Hi Mike: I think that you haven t prove that p is prime, only that it is not divisible by p. Jose ... From: Mike Antczak To: primenumbers@yahoogroups.com
      Message 2 of 4 , Jul 9, 2003
        Hi Mike:

        I think that you haven't prove that p' is prime, only that it is not divisible by p.

        Jose
        ----- Original Message -----
        From: Mike Antczak
        To: primenumbers@yahoogroups.com
        Sent: Wednesday, July 09, 2003 5:21 PM
        Subject: [PrimeNumbers] prime symmetry


        Hi, this may be a trivial (or well known) result, but
        when I first observed it I was quite surprised.

        NOTE: [] indicates subscript.

        Let c = q[1]q[2]...q[n], where q is an odd prime
        number then c is an odd composite.

        Let P = {p[1],p[2],...,p[m]}, where p is an odd prime
        and gcd(p,q)=1, p[m] < 2c < p[m+1]

        Let A and B be partition cells of P, such that {p:p in
        A, p < c} and {p':p' in B, c < p' < 2c}.

        Case 1.
        If p[1]^2 > 2c, p[1] in A, then 2c - p = p', where p
        in A, and p' in B.

        Ex.
        c = 15 = 3*5
        2c = 30
        P = {7,11,13,17,19,23,29}
        A = {7,11,13}
        B = {17,19,23,29}

        30 - 7 = 23
        30 - 11 = 19
        30 - 13 = 17

        Case 2.
        If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p
        (except when p = 1), where p in A, and p' in B.

        Ex.
        c = 21 = 3*7
        2c = 42
        P = {5,11,13,17,19,23,29,31,37,41}
        A = {5,11,13,17,19}
        B = {23,29,31,37,41}

        42 - 23 = 19
        42 - 29 = 13
        42 - 31 = 11
        42 - 37 = 5

        What does this suggest? This suggests that there is a
        fundamental symmetry within the primes.

        Proof:

        NOTE: this proof is only valid for case 1 and with
        slight modification for case 2.

        Using the previous definitions unless specified:

        Let q' be the mirror image of q with respect to c,
        meaning that q'-c=c-q or q'=c+c-q (i.e. the distance
        from q to c is equal to the distance from q' to c).
        Recall that c=q[1]q[2]...q[n].

        Let p' be the mirror image of p with respect to c,
        meaning that p'-c=c-p or p'=c+c-p (i.e. the distance
        from p to c is equal to the distance from p' to c).
        Recall that p is an element in A and p' is an element
        in B.

        Since q|c and q|(c-q), then it follows that q|(c+c-q)
        or q|q'.

        Since p does not divide c and p does not divide c-p,
        then it follows that p does not divide c+c-p or p does
        not divide p'.

        Therefore p'=c+c-p or p'=2c-p and a p with equal
        distance from c exists on either side of c as was
        defined in case 1. Hence the symmetry.

        Q.E.D.

        A similar proof could be shown for case 2 but a
        general case, where p[1]^ < c, would begin to be
        incoherent because the elements in A would create
        composite numbers that when are subtracted from 2c
        yeild an element in B.

        Ex.
        c=25=5*5
        2c=50
        P={3,7,11,13,17,19,23,29,31,37,41,43,47}
        A={3,7,11,13,17,19,23}
        B={29,31,37,41,43,47}

        But...
        3*3=9 and 50-9=41, which is an element of B, therefore
        50-41 does not produce a prime.


        When all goes well, it is obvious that p+p'=2c, which
        implies Goldbach's Conjecture. And if GC is false,
        then all of the p must produce composite numbers that
        equal 2c-p'. Is this possible?

        -Mike


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      • Mike Antczak
        ... Right, first of all p cannot be even because p =c+c-p. Since c is odd, odd+odd is even-odd is odd. Also, the other possiblity is that q|p . Since
        Message 3 of 4 , Jul 9, 2003
          --- Jose_Ram�n_Brox <ambroxius@...> wrote:
          > Hi Mike:
          >
          > I think that you haven't prove that p' is prime,
          > only that it is not divisible by p.
          >
          > Jose
          > ----- Original Message -----
          > From: Mike Antczak
          > To: primenumbers@yahoogroups.com
          > Sent: Wednesday, July 09, 2003 5:21 PM
          > Subject: [PrimeNumbers] prime symmetry
          >
          >
          > Hi, this may be a trivial (or well known) result,
          > but
          > when I first observed it I was quite surprised.
          >
          > NOTE: [] indicates subscript.
          >
          > Let c = q[1]q[2]...q[n], where q is an odd prime
          > number then c is an odd composite.
          >
          > Let P = {p[1],p[2],...,p[m]}, where p is an odd
          > prime
          > and gcd(p,q)=1, p[m] < 2c < p[m+1]
          >
          > Let A and B be partition cells of P, such that
          > {p:p in
          > A, p < c} and {p':p' in B, c < p' < 2c}.
          >
          > Case 1.
          > If p[1]^2 > 2c, p[1] in A, then 2c - p = p', where
          > p
          > in A, and p' in B.
          >
          > Ex.
          > c = 15 = 3*5
          > 2c = 30
          > P = {7,11,13,17,19,23,29}
          > A = {7,11,13}
          > B = {17,19,23,29}
          >
          > 30 - 7 = 23
          > 30 - 11 = 19
          > 30 - 13 = 17
          >
          > Case 2.
          > If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p
          > (except when p = 1), where p in A, and p' in B.
          >
          > Ex.
          > c = 21 = 3*7
          > 2c = 42
          > P = {5,11,13,17,19,23,29,31,37,41}
          > A = {5,11,13,17,19}
          > B = {23,29,31,37,41}
          >
          > 42 - 23 = 19
          > 42 - 29 = 13
          > 42 - 31 = 11
          > 42 - 37 = 5
          >
          > What does this suggest? This suggests that there
          > is a
          > fundamental symmetry within the primes.
          >
          > Proof:
          >
          > NOTE: this proof is only valid for case 1 and with
          > slight modification for case 2.
          >
          > Using the previous definitions unless specified:
          >
          > Let q' be the mirror image of q with respect to c,
          > meaning that q'-c=c-q or q'=c+c-q (i.e. the
          > distance
          > from q to c is equal to the distance from q' to
          > c).
          > Recall that c=q[1]q[2]...q[n].
          >
          > Let p' be the mirror image of p with respect to c,
          > meaning that p'-c=c-p or p'=c+c-p (i.e. the
          > distance
          > from p to c is equal to the distance from p' to
          > c).
          > Recall that p is an element in A and p' is an
          > element
          > in B.
          >
          > Since q|c and q|(c-q), then it follows that
          > q|(c+c-q)
          > or q|q'.
          >
          > Since p does not divide c and p does not divide
          > c-p,
          > then it follows that p does not divide c+c-p or p
          > does
          > not divide p'.
          >
          > Therefore p'=c+c-p or p'=2c-p and a p with equal
          > distance from c exists on either side of c as was
          > defined in case 1. Hence the symmetry.
          >
          > Q.E.D.
          >
          > A similar proof could be shown for case 2 but a
          > general case, where p[1]^ < c, would begin to be
          > incoherent because the elements in A would create
          > composite numbers that when are subtracted from 2c
          > yeild an element in B.
          >
          > Ex.
          > c=25=5*5
          > 2c=50
          > P={3,7,11,13,17,19,23,29,31,37,41,43,47}
          > A={3,7,11,13,17,19,23}
          > B={29,31,37,41,43,47}
          >
          > But...
          > 3*3=9 and 50-9=41, which is an element of B,
          > therefore
          > 50-41 does not produce a prime.
          >
          >
          > When all goes well, it is obvious that p+p'=2c,
          > which
          > implies Goldbach's Conjecture. And if GC is
          > false,
          > then all of the p must produce composite numbers
          > that
          > equal 2c-p'. Is this possible?
          >
          > -Mike

          Right, first of all p' cannot be even because
          p'=c+c-p. Since c is odd, odd+odd is even-odd is odd.
          Also, the other possiblity is that q|p'. Since
          p'=c+c-p then it follows that q|(c+c-p). Although
          q|c, q cannot divide c-p because p is a prime such
          that gcd(p,q)=1 (i.e. p and q are relatively prime).
          Therefore, since p' is odd and not divisible by p or
          q, p' is prime.

          And remember, I'm only proving case 1.

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        • Mike Antczak
          ... To show you what I mean... The best possibility for case 2 is c=105=3*5*7, 2c=210. Note that 11^2=121 which 105
          Message 4 of 4 , Jul 9, 2003
            > --- Jose_Ram�n_Brox <ambroxius@...> wrote:
            > > Hi Mike:
            > >
            > > I think that you haven't prove that p' is prime,
            > > only that it is not divisible by p.
            > >
            > > Jose
            > > ----- Original Message -----
            > > From: Mike Antczak
            > > To: primenumbers@yahoogroups.com
            > > Sent: Wednesday, July 09, 2003 5:21 PM
            > > Subject: [PrimeNumbers] prime symmetry
            > >
            > >
            > > Hi, this may be a trivial (or well known)
            > result,
            > > but
            > > when I first observed it I was quite surprised.
            > >
            > > NOTE: [] indicates subscript.
            > >
            > > Let c = q[1]q[2]...q[n], where q is an odd prime
            > > number then c is an odd composite.
            > >
            > > Let P = {p[1],p[2],...,p[m]}, where p is an odd
            > > prime
            > > and gcd(p,q)=1, p[m] < 2c < p[m+1]
            > >
            > > Let A and B be partition cells of P, such that
            > > {p:p in
            > > A, p < c} and {p':p' in B, c < p' < 2c}.
            > >
            > > Case 1.
            > > If p[1]^2 > 2c, p[1] in A, then 2c - p = p',
            > where
            > > p
            > > in A, and p' in B.
            > >
            > > Ex.
            > > c = 15 = 3*5
            > > 2c = 30
            > > P = {7,11,13,17,19,23,29}
            > > A = {7,11,13}
            > > B = {17,19,23,29}
            > >
            > > 30 - 7 = 23
            > > 30 - 11 = 19
            > > 30 - 13 = 17
            > >
            > > Case 2.
            > > If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p
            > > (except when p = 1), where p in A, and p' in B.
            > >
            > > Ex.
            > > c = 21 = 3*7
            > > 2c = 42
            > > P = {5,11,13,17,19,23,29,31,37,41}
            > > A = {5,11,13,17,19}
            > > B = {23,29,31,37,41}
            > >
            > > 42 - 23 = 19
            > > 42 - 29 = 13
            > > 42 - 31 = 11
            > > 42 - 37 = 5


            To show you what I mean...

            The best possibility for case 2 is c=105=3*5*7,
            2c=210. Note that 11^2=121 which 105 < 121 < 210,
            this means that 2c-p'=p which corresponds to the
            definition of case 2.

            All the primes greater than 105 and less than 210
            subtracted from 210 will yield a prime less than 105.
            Check it out.


            > > What does this suggest? This suggests that
            > there
            > > is a
            > > fundamental symmetry within the primes.
            > >
            > > Proof:
            > >
            > > NOTE: this proof is only valid for case 1 and
            > with
            > > slight modification for case 2.
            > >
            > > Using the previous definitions unless specified:
            > >
            > > Let q' be the mirror image of q with respect to
            > c,
            > > meaning that q'-c=c-q or q'=c+c-q (i.e. the
            > > distance
            > > from q to c is equal to the distance from q' to
            > > c).
            > > Recall that c=q[1]q[2]...q[n].
            > >
            > > Let p' be the mirror image of p with respect to
            > c,
            > > meaning that p'-c=c-p or p'=c+c-p (i.e. the
            > > distance
            > > from p to c is equal to the distance from p' to
            > > c).
            > > Recall that p is an element in A and p' is an
            > > element
            > > in B.
            > >
            > > Since q|c and q|(c-q), then it follows that
            > > q|(c+c-q)
            > > or q|q'.
            > >
            > > Since p does not divide c and p does not divide
            > > c-p,
            > > then it follows that p does not divide c+c-p or
            > p
            > > does
            > > not divide p'.
            > >
            > > Therefore p'=c+c-p or p'=2c-p and a p with equal
            > > distance from c exists on either side of c as
            > was
            > > defined in case 1. Hence the symmetry.
            > >
            > > Q.E.D.
            > >
            > > A similar proof could be shown for case 2 but a
            > > general case, where p[1]^ < c, would begin to be
            > > incoherent because the elements in A would
            > create
            > > composite numbers that when are subtracted from
            > 2c
            > > yeild an element in B.
            > >
            > > Ex.
            > > c=25=5*5
            > > 2c=50
            > > P={3,7,11,13,17,19,23,29,31,37,41,43,47}
            > > A={3,7,11,13,17,19,23}
            > > B={29,31,37,41,43,47}
            > >
            > > But...
            > > 3*3=9 and 50-9=41, which is an element of B,
            > > therefore
            > > 50-41 does not produce a prime.
            > >
            > >
            > > When all goes well, it is obvious that p+p'=2c,
            > > which
            > > implies Goldbach's Conjecture. And if GC is
            > > false,
            > > then all of the p must produce composite numbers
            > > that
            > > equal 2c-p'. Is this possible?
            > >
            > > -Mike
            >
            > Right, first of all p' cannot be even because
            > p'=c+c-p. Since c is odd, odd+odd is even-odd is
            > odd.
            > Also, the other possiblity is that q|p'. Since
            > p'=c+c-p then it follows that q|(c+c-p). Although
            > q|c, q cannot divide c-p because p is a prime such
            > that gcd(p,q)=1 (i.e. p and q are relatively prime).
            >
            > Therefore, since p' is odd and not divisible by p or
            > q, p' is prime.
            >
            > And remember, I'm only proving case 1.


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