Re: [PrimeNumbers] Re: Proof of D. Andrica conjecture. ver II
- 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
> < c + aChange the following
> 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
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
Do I have it this time?
- --- In primenumbers@y..., "John W. Nicholson" <johnw.nicholson@a...>
> > Let P(n) = a^2 and P(n+1) = c^2 = (a + b)^2 = a^2 + 2ab + b^2where a, b,
> > and c are a real numbers.assume is < c + a
> > Let the gap s^2 = c^2 - a^2 = (c - a)(c + a) = 2ab + b^2 which we
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 cThis step above is a trouble spot,
> > 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:
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
which clearly, is not true for all p>5.
You need stronger mojo baby.