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RE reciprocal consecutive primes

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  • Jose Ramón Brox
    ... From: theo2357 Let p,q,r be three consecutive primes. How can be proven that 1/p
    Message 1 of 3 , Nov 2, 2005
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      ----- Original Message -----
      From: "theo2357" <Theo.3.1415@...>


      Let p,q,r be three consecutive primes. How can be proven that
      1/p < 1/q + 1/r? It's NOT trivial!

      ========================================

      Bertrand's Postulate states that if n>2 then there's a prime strictly between n/2 and n.
      Therefore:

      q > r/2

      1/q < 2/r

      1/r + 1/q > 1/p <--> 1/r + 1/q - 1/p > 0 <-->
      <--> 2(1/r + 1/q - 1/p) > 0 <--> 2/r + 2/q - 2/p > 0

      Using that 2/r > 1/q:

      2/r + 2/q - 2/p > 3/q - 2/p

      3/q - 2/p = (3p - 2q ) / (pq) > 0 <--> 3p - 2q > 0 <--> q/p < 3/2

      That's asymptotically true cause p(n) ~ n / log(n) -->
      --> p(n+1) / p(n) ~ (n+1)/n * log(n) / log(n+1) = 1 and therefore you can find a natural
      n0 so that if n > n0, then |p(n+1)/p(n) - 1| < 1/2 --> p(n+1)/p(n) < 3/2 for every n > n0.

      That is, for n "bigger enough", 1/r + 1/q > 1/p always holds.

      But it seems to hold for every (p,q,r), not only for the "bigger" ones. q/p < 3/2 is a
      simpler necessary condition that seems to hold if p>7. I don't know now how to prove it
      without using analysis (to prove the trueness for every 3-uple of consecutive primes).
      Maybe there's a simple and clever argument, I recommend to search for better boundings to
      the distribution of primes than Bertrand's Postulate (I'm sorry, I'm sooo lazy...).

      Nice observation, btw.

      Regards. Jose Brox.
    • 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 2 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 3 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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