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`Re: [PrimeNumbers] Re: Andrica

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  • John W. Nicholson
    I am wondering if this is positive or saying that it is not strong enough. I also think I may be unclear so I restated it again below. ... From: djbroadhurst
    Message 1 of 3 , Aug 3, 2002
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      I am wondering if this is positive or saying that it is not strong enough. I
      also think I may be unclear so I restated it again below.

      ----- Original Message -----
      From: "djbroadhurst" <d.broadhurst@...>
      To: <primenumbers@yahoogroups.com>
      Sent: Saturday, August 03, 2002 4:05 AM
      Subject: [PrimeNumbers] Re: Andrica


      > Let p and q be successive primes.
      > Andrica conjectures that
      > q - p < sqrt(q) + sqrt(p)
      > whereas Tschebysheff proved merely that
      > q - p < p
      > which is far weaker.

      rewriten as q < 2p
      include p itself p < q < 2p

      take sqrt, sqrt (p) < sqrt (q) < sqrt(2)*sqrt(p) --- length of side with
      square q
      take out the factor sqrt (p), 1< sqrt (q) / sqrt(p) < sqrt(2)

      subtract the area p from q by
      take the sqrt, sqrt (p)
      take out the factor sqrt (p), 1 --- length of side with square p
      so
      1-1 = 0 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1
      1 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1 < 1

      using the difference of two squares
      q - p = (sqrt(q) + sqrt(p)) (sqrt(q) - sqrt(p))
      (q - p) / (sqrt(q) + sqrt(p)) = (sqrt(q) - sqrt(p))
      and
      (sqrt (q) - sqrt(p)) < (sqrt(2) sqrt(p)) - (sqrt(p))
      (sqrt (q) - sqrt(p))/sqrt(p) < (sqrt(2) - 1)(sqrt(p))/sqrt(p)
      S = sqrt (q) / sqrt(p) -1 < sqrt(2) - 1 < 1
      so the first number with p = 2 is < 1 and the limit S as p -> oo = 0.
      QED

      I'm just a college student in physics so it would help if I could get some
      feedback.

      Thanks David.

      John
    • Jon Perry
      What David is saying is that you have merely re-worked Tschebysheff s proof. Without having put in anything extra into the system, all you have done is find
      Message 2 of 3 , Aug 4, 2002
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        What David is saying is that you have merely re-worked Tschebysheff's proof.

        Without having put in anything extra into the system, all you have done is
        find new expressions for existing formula.

        As a Mathematician, your proof is lacking in several areas:

        #rewriten as q < 2p
        #include p itself p < q < 2p

        This could be written as: there exists a prime q, p<q<2p

        #take sqrt, sqrt (p) < sqrt (q) < sqrt(2)*sqrt(p) --- length of side with
        #square q
        #take out the factor sqrt (p), 1< sqrt (q) / sqrt(p) < sqrt(2)

        OK.

        #subtract the area p from q by
        #take the sqrt, sqrt (p)
        #take out the factor sqrt (p), 1 --- length of side with square p
        #so
        #1-1 = 0 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1
        #1 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1 < 1

        Line 2 makes no sense, nor does line 3
        Line 5 is OK, but Line6 contains obvious errors.

        At this point what we have is sqrt(q)/sqrt(p) - 1 < 1, which you managed to
        get to in your first proof, but seem to have failed to in this one.

        From here, we can get sqrt(q)-sqrt(p)<sqrt(p), but this is far from
        Andrica's conjecture.

        #using the difference of two squares
        #q - p = (sqrt(q) + sqrt(p)) (sqrt(q) - sqrt(p))
        #(q - p) / (sqrt(q) + sqrt(p)) = (sqrt(q) - sqrt(p))
        #and
        #(sqrt (q) - sqrt(p)) < (sqrt(2) sqrt(p)) - (sqrt(p))
        #(sqrt (q) - sqrt(p))/sqrt(p) < (sqrt(2) - 1)(sqrt(p))/sqrt(p)
        #S = sqrt (q) / sqrt(p) -1 < sqrt(2) - 1 < 1
        #so the first number with p = 2 is < 1 and the limit S as p -> oo = 0.
        #QED

        With such a dodgy start, most Mathematicians would be rolling about on the
        floor by now in tears, but persistence is to be encouraged.

        Line 1 is good, but you can begin to see why this conjecture is considered
        hard.
        Line 2 is good.

        After here, you need to explain where your subtitutions are coming from.
        Also needed is a statement declaring how the algebra will lead to a proof of
        the conjecture. This assists any reader in following your work, and takes
        the pressure off the reader in trying to follow the work.

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com
      • Phil Carmody
        ... Jon, I don t know if you re trying to be Me Too to David s Big Dog , http://www.winternet.com/~mikelr/flame3.html ? However, you re not really in a
        Message 3 of 3 , Aug 4, 2002
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          --- Jon Perry <perry@...> wrote:
          > With such a dodgy start, most Mathematicians would be rolling about
          > on the
          > floor by now in tears, but persistence is to be encouraged.

          Jon,
          I don't know if you're trying to be "Me Too" to David's "Big Dog",
          http://www.winternet.com/~mikelr/flame3.html ?
          However, you're not really in a position to make snide comments about
          proofs, particularly in such a patronising way, and especially
          considering some analyses of your own proof style (
          http://groups.yahoo.com/group/primenumbers/message/7546 )

          Phil
          (No bonus points for spotting the irony herein!)


          =====
          --
          The good Christian should beware of mathematicians, and all those who make
          empty prophecies. The danger already exists that the mathematicians have
          made a covenant with the devil to darken the spirit and to confine man in
          the bonds of Hell. -- Common mistranslation of St. Augustine (354-430)

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