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RE: [PrimeNumbers] Re: Andrica

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  • 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 1 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 2 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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