[ExI] Goldbach Conjecture

[Giulio Prisco (2nd email)]

I think the Goldbach conjecture is probably false, with probability 1
(that means, certainly false). Here is why:

Apparently there is nothing in the laws of arithmetics that forces an
even number to be the sum of two prime numbers. The conjecture is true
for all even numbers on which it has been tested, but these are an
infinitesimal part of the total (any finite number is infinitesimal
wrt infinite). Hence, if there is no proof, the probability of he
Goldbach conjecture being true is zero.

2009/11/28 Will Steinberg <asyluman at gmail.com>:
> I think I may really have got it now.
>
> Since we can represent every situation like this by a doublesieve, we need
> only prove all doublesieves will have holes in them.  Here is how I do
> this.  A sieve of length n--let's say 16--has a set of spaces A that are at
> 2k, 3k, 5k...mk where k>1 and mk<n.  It also has a set of spaces in the
> opposite manner, which can be given by nmod2+2k, nmod3+3k...nmodm+mk where
> once again k>1 and mk<n.  All we need to prove is one necessary hole between
> those spaces.  It is given thus:  Choose a prime in the set that is under
> any 2k (let's choose 5) that has the property n mod p =1. We cannot create 5
> from our 2k, 3k,spaces, because the only one to do so would be 5k, and k
> must be greater than 2.  So we must find a nmodp+pk that is equal to 5, k of
> course being greater than 2 again.  But alas!  We find that the only way to
> have nmodp+pk=5 be if p was less than 5.  so we have nmod2+2k which yields 4
> and nmod3+3k which yields a minimum of seven!  The point is here:  Construct
> the sieve for any length.  Choose a low prime.  As long as none of the set B
> spaces are equal to it, we have a hole.  The nmodp+pk must be equal to our P
> for the whole to be covered.   But there is always a P inexpressible by
> nmodp+pk--simply choose a low p with a modulus of 1 for n.  then, if nmodp
> is not 1, nmodp+pk is greater than P.  But if nmodp IS one, nmodp+pk can
> never equal our P.
>
>
> On Fri, Nov 27, 2009 at 9:30 PM, Will Steinberg <asyluman at gmail.com> wrote:
>>
>> I think I may have ended up merely restating something, but I think
>> approaching the problem in a new paradigm is important--namely, instead of
>> having each even number expressed as the sum of two primes, have each number
>> n have 2 distinct primes an equal distance from it (because if Px + Py = 2n,
>> then n - m = Px and n + m = Py.  So I wrote out the numbers from a to a
>> number 2n-1 (say 9) then wrote the numbers 9 to 1 below them, so each column
>> sums to 2n (10).
>>
>> e.g. 1 2 3 4 5 6 7 8 9
>>       9 8 7 6 5 4 3 2 1
>>
>>   Then I ran a sieve of eratosthenes each way.  The "holes" after a
>> forwards and backwards run-through reveal the sets of primes which are
>> equidistant from n and thus sum to 2n, or every even number.
>>
>> Now, if we know there is one prime left unscathed by our doublesieve, it
>> must have a mirror partner (also prime because of our sieving).  And we KNOW
>> that there is at least one prime p between n and 2n given Bertrand's
>> postulate.  Since this is a mirror, we can consider only one half of the
>> line, from 1 to n.  Now, we have already marked off the composites given by
>> the sieve.  But now lots are recast and even some primes must be lost.
>> These are determined by taking whatever composite numbers are between n and
>> 2n and subtracting them from 2n.  Now we have a set that looks like this:
>> {composites between 1 and n, 2n-composites between n and 2n}.
>>
>> Visually, what we will have is a symmetrical pattern: two sieves laid on
>> top of each other oppositeways.  Each hole will have its conjugate and so
>> the same for each filled in space.  In essence, if we can prove that that
>> pattern will always contain holes, then there will always be 2 primes able
>> to add up to 2n.
>>
>> I think the answer may lie with some sort of modulus thing.  If you look
>> at the center n, adding up prime numbers (to produce composites) will often
>> lay a stretch of number over n.  For example. the stretch of 9-12 lies over
>> 10, with 1 on the right and 2 on the left.  Consequently, all subsequent
>> additions of 3 will give 10+2+3k.  This means that the spaces with numbers
>> that equal 10-2-3k will be filled in on the right half, as well as 10-4-7k,
>> et cetera.  I think the actual answer lies near and I would love for someone
>> to give me insight (or tell me this has all already been done)
>>
>>
>>
>> On Fri, Nov 27, 2009 at 6:59 PM, spike <spike66 at att.net> wrote:
>>>
>>> On Behalf Of Will Steinberg
>>>        Subject: [ExI] Goldbach Conjecture
>>>
>>>
>>>         I think I found something really good about the Goldbach
>>> Conjecture; does anyone have a background in number theory or know
>>> somebody
>>> who does?  It is important
>>>
>>>        -Will
>>>
>>>
>>> Hi Will!  I am a Goldbach conjecture fan.  I know there are other
>>> Goldbachers here, Lee Corbin is one of our math superbrains, and several
>>> others too.  What did you find?
>>>
>>> spike

-- 
Giulio Prisco
http://cosmeng.org/index.php/Giulio_Prisco
aka Eschatoon Magic
http://cosmeng.org/index.php/Eschatoon