Loading ...
Sorry, an error occurred while loading the content.

Problems with Tom's paper on the Goldbach Conjecture

Expand Messages
  • richard_in_reading
    On a quick read I spotted that on page 9 n and n+1 are not twin primes. You meant p_n and p_(n+1) Just before your statement of Theorem 1 on page 3 you say
    Message 1 of 14 , Sep 3, 2009
    • 0 Attachment
      On a quick read I spotted that on page 9 n and n+1 are not twin primes. You meant p_n and p_(n+1)

      Just before your statement of Theorem 1 on page 3 you say
      "For any set J of primes, this will imply maxima and minima in regard to the number of integers k for which Div(J, k) = {k}, given the set of r for which ..."
      What's the relevance of r? Presumably the clause "given the set of r for which" is meant to constrain something but it's not clear what, why or how.
      Also according to your definition on the bottom of page 2 Div(J,k) is just the members of J dividing k. Given that J are primes, Div(J,k)={k} means that k is a prime in J and your statement becomes rather trivial.
      Even if I consider that you got your definition of Div wrong I still can't make it make non-trivial sense.

      The last paragraph on page 9 is important but I can't understand it. Could you explain what you are asserting and preferably some of the connecting logic more clearly.

      At the top of page 12 shortly before you say QED to the Goldbach conjecture you mention two disjoint subsets of Q, K and K'. Don't you need to constrain the union of the subsets to be Q otherwise your following statement regarding an inequality of a product is likely to be false. Also it's not clear how the two statements you make result in Theorem 2 being proved.

      What this "corollary" to theorem 1? A corollarly is a noteworthy result that follows easily from a previous theorem. It's not clear what your corollary to theorem 1 means, how it's relevant or how it follows.

      The functions f(n) and a(n,r) are they seriously just introduced to help you prove that the product of (p-2) is less than the product of (p-1) for the elements p of K ?

      It seems that your document details a rather longwinded proof of some very elementary results and then rushes over the interesting bits. I would hope to see building blocks for the proof of the Goldbach Conjecture to be set out explicitly as theorems and then for it to be made very clear how these blocks fit together to prove the main result.

      I understand how difficult it is to write a good paper and how relatively easy it is to find mistakes. Keep plugging away at it!

      Richard Heylen
    • marku606
      ... Thank you Richard. I m glad you were able to penetrate it enough to see specific problems. I was able to understand Tom s colloquial version, and
      Message 2 of 14 , Sep 3, 2009
      • 0 Attachment
        --- In primenumbers@yahoogroups.com, "richard_in_reading" <richard_in_reading@...> wrote:
        >[snip]
        > It seems that your document details a rather longwinded proof of some very elementary results and then rushes over the interesting bits. I would hope to see building blocks for the proof of the Goldbach Conjecture to be set out explicitly as theorems and then for it to be made very clear how these blocks fit together to prove the main result.
        >


        Thank you Richard. I'm glad you were able to penetrate it enough to see specific problems. I was able to understand Tom's 'colloquial' version, and realized that the core of his proof went from a very elementary result to a QED in two sentences flat. Tom told me that the missing link of proof was found in the full version, but I could not penetrate the set theory nomenclature. My eyes glazed over, even with - note - coffee *and* chocolate by my side.

        But the folding idea is an appealing picture, lending itself to a nice rephrasing Goldbach's conjecture: Every integer greater than one is the mean of two primes.


        But here's a reality check for attempts at solving GB's conjecture.
        We want to show that 2n is the sum of two primes. We draw a line from 0 to 2n, with n in the center:

        0.......n.......2n

        GB conjecture would have that for every n>1 there is a prime equidistant on either side of n.

        On the way to prove such, of course it would have to be proven that there are indeed primes from n to 2n. Such a little thing. :)


        Mark
      • djbroadhurst
        ... Indeed :-) I sometimes wonder why purported provers of the G*ldb*ch conjecture don t criticize Erdos for using central binomial coefficients in his proof
        Message 3 of 14 , Sep 3, 2009
        • 0 Attachment
          --- In primenumbers@yahoogroups.com, "marku606"
          <mark.underwood@...> wrote:

          > On the way to prove such, of course it would have to be proven
          > that there are indeed primes from n to 2n.
          > Such a little thing. :)

          Indeed :-)

          I sometimes wonder why purported provers of the G*ldb*ch conjecture
          don't criticize Erdos for using central binomial coefficients in
          his proof of Bertrand's postulate, nicely caught here
          http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/bertrand.pdf
          by Robin Chapman. The close reasoning in this proof of a result
          immensely weaker than the G*ldb*ch conjecture ought to offer
          some sort of warning, one might have thought? Of course
          it is not beyond the bounds of reason that Chebyshev and Erdos
          fooled themselves and the rest us into thinking that even
          Betrand's postulate is this hard to prove.

          David
        • Chris Caldwell
          ... Nice point mark! ... Yes, and that is one of the simplest proofs. It would be kind for these provers to offer us a half-page proof of the Bertrand result
          Message 4 of 14 , Sep 3, 2009
          • 0 Attachment
            >> On the way to prove such, of course it would have to be proven
            >> that there are indeed primes from n to 2n.
            >> Such a little thing. :)

            Nice point mark!

            > I sometimes wonder why purported provers of the G*ldb*ch conjecture
            > don't criticize Erdos for using central binomial coefficients in

            Yes, and that is one of the simplest proofs. It would be kind for these
            provers to offer us a half-page proof of the Bertrand result which is so

            much simpler than Goldbach. What a fine way to catch the attention of
            mathematicians that would be... CC
          Your message has been successfully submitted and would be delivered to recipients shortly.