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Re: [PrimeNumbers] Re: Proof of D. Andrica conjecture. ver II

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  • John W. Nicholson
    One small but big change See it near bottom. ... is ... Change the following Let us replace (a-1) with a real number k. So there is an real 0
    Message 1 of 3 , Aug 5, 2002
      One small but big change See it near bottom.

      >
      > Thanks to Dick and David Broadhurst,
      >
      > You have me doing a change to even a better way of showing this:
      >
      > Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2 where a, b,
      > and c are a real numbers.
      > Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we assume
      is
      > < c + a
      > Take the square root to find c
      > c = a + b
      > s^2 = 2ab + b^2 < a + b + a = 2a + b
      > Or 2ab + b^2 < 2a + b
      > So 0 < b < 1
      > Replace b with y = 1/b so 1 < y however
      > 2a + y must be < a^2 by Bertrand/Tschebycheff:
      > So y < a^2 - 2a
      > By adding 1 on the right side we complete the square without going >= a^2
      > So y <= (a - 1)^2
      > This is the largest y value; the smallest is y => (a - (a - 1))^2

      Change the following
      Let us replace (a-1) with a real number k.
      So there is an real 0 < k < a such that
      b = 1/(a - k)^2, c = a + b, and s^2 = 2ab + b^2 < 2a + b

      QED

      Do I have it this time?
    • richard042
      ... where a, b, ... assume is
      Message 2 of 3 , Aug 6, 2002
        --- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>
        wrote:

        > > Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2
        where a, b,
        > > and c are a real numbers.
        > > Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we
        assume is < c + a

        This is essentially an assumption that c-a < 1
        which is an assumption that Andrica's conjecture is true.
        Is this supposed to be an "assume for now" type statement as
        if it will be proven correct or somehow justified later in the proof?

        > > Take the square root to find c
        > > c = a + b
        > > s^2 = 2ab + b^2 < a + b + a = 2a + b
        > > Or 2ab + b^2 < 2a + b
        > > So 0 < b < 1
        > > Replace b with y = 1/b so 1 < y however
        > > 2a + y must be < a^2 by Bertrand/Tschebycheff:

        This step above is a trouble spot,
        how do you arrive at it?
        By B/T the gap=2ab+b^2 < a^2=P(n)
        since b=1/y, 2a/y + 1/y^2 < a^2
        thus 2a+y < y*a^2-1/y+y

        -[snip]- the rest.

        The only way to prove Andrica's Conjecture
        via Bertrand/Tschebycheff is to prove that
        it is always true that
        sqrt(2*p)-sqrt(p)<1
        sqrt(2)-1<1/sqrt(p)
        which clearly, is not true for all p>5.

        You need stronger mojo baby.

        -Dick
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