[ExI] Goldbach Conjecture

[Will Steinberg]

I think I've got it now.  P != 2nmodp +pk, right?  We can re-express 2nmodp
as 2n-p*[2n/p], where brackets denote the floor function.  So we have P !=
2n - p*[2n/p] +pk.  Dividing by p yields the following:

P/p != 2n/p - [2n/p] +k

2n/p-[2n/p] equals the fractional part of 2n/p.  And since k is any integer,
we can set it equal to the integer part of P/p.  Now we have this:

frac(P/p) != frac(2n/p).

Now: since P and p are prime, the fractional part will be irreducible.  And
since p is in the denominator of 2n/p, it too must have an irreducible
fractional part.  But since 2 is in the numerator, these parts must be
different!  The only case in which they would not be is if n=p, in which
case 2n would be a double of a prime, satisfying the GC.  I think this is
really it.

On Sat, Nov 28, 2009 at 11:55 AM, Will Steinberg <asyluman at gmail.com> wrote:

> If it can be proved that every two-way sieve of eratosthenes has at least
> one hole, the conjecture can be proven.  What this means is that (since oles
> are at 2k, 3k, 5k, nmod2+2k, nmod5+5k, etc.)  There has got to be some kind
> of proof saying that for any given number n, there is a prime in n than
> cannot be expressed by nmodp+pk.  What has to be looked at is the modulo
> values that will be given for ns. I think we can often choose 3, because the
> only case when 2 can be covered is if we have nmod2=1 and pk=2.  (or if the
> number is divisible by 3).  Otherwise we can simply continue moving up our
> primes.  In the case of 2n=22, we see holes at 3,5, and 11.  What is needed
> to continue is a way to prove there will always be a p that doesn't equal
> nmodp +pk
>
> 2009/11/28 spike <spike66 at att.net>
>
>>
>>
>> > -----Original Message-----
>> > From: extropy-chat-bounces at lists.extropy.org
>> > [mailto:extropy-chat-bounces at lists.extropy.org<extropy-chat-bounces at lists.extropy.org>]
>> On Behalf Of
>> > Giulio Prisco (2nd email)
>> > Sent: Friday, November 27, 2009 11:12 PM
>> > To: ExI chat list
>> > Subject: Re: [ExI] Goldbach Conjecture
>> >
>> > 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.
>>
>> I disagree sir, however I confess my line of reasoning is not as well
>> developed as the one you offer.
>>
>> I took the even numbers and calculated the number of ways each even number
>> (shown on the X axis) could be expressed as a the sum of two primes.  The
>> number of different ways is on the Y.  For Goldbach to have been wrong,
>> there is some super-anomaly way out there somewhere which departs from the
>> data trends shown.
>>
>> Yes I do know that this line of reasoning is not to be substituted for
>> actual mathematical logic, do forgive please.
>>
>> I plotted them to a few million on matlab, found there are striking
>> patterns in the data, such as the eye-catching streaks.
>>
>> spike
>>
>>
>>
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>>
>>
>
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