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RE reciprocal consecutive primes

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  • Jose Ramón Brox
    ... From: Milton Brown Actually, it doesn t seem that hard: 1/p =p+2 r =q+2 =p+4 It must be shown that 1/p
    Message 1 of 3 , Nov 3, 2005
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      ----- Original Message -----
      From: "Milton Brown" <miltbrown@...>


      Actually, it doesn't seem that hard:

      1/p < 1/q + 1/r

      q>=p+2
      r>=q+2>=p+4

      It must be shown that 1/p < 1/(p+2) + 1/(p+4)

      ----------------------------------------

      FALSE. If r>=p+4, then 1/r <= 1/(p+4). Considering the case where 1/r = 1/(p+4) and 1/q =
      1/(p+2) is a best case study, not a worst case one.

      Jose Brox
    • Jose Ramón Brox
      ... From: Jan van Oort p^2 mn which is true iff ( m a ) ( condition i ) ...
      Message 2 of 3 , Nov 3, 2005
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        ----- Original Message -----
        From: "Jan van Oort" <glorifier@...>

        p^2 > mn

        which is true iff

        ( m < p AND n < p ) OR ( m = p + a AND n = p - b AND b > a ) ( condition i )

        ----------------------------------------------------
        Jose:

        That's not true.

        Consider b = a-1, then we have

        m*n = (p+a)(p-b) = (p+a)(p-a+1) = p^2 +p-a(a+1)

        If a(a+1) > p, then m*n < p^2, with b < a, m=p+a, n = p-b. You can get similar results
        with different b < a.

        ----------------------------------------------
        Jan:

        now condition ( i ) only holds iff Bertrand's postulate holds
        ------------------------------------------------------------
        Jose:

        Can you explain why? I can't see it at the moment.


        Jose Brox
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