Here's a little bit of LaTeX delineating some stuff. Why the end works is that the overflow from P/p (to get the fractional part) will always be an even number (prime-prime) and the fractional part from 2n/p will be 2n-p which will always be odd, unless p is 2, in which case the first part is odd and the second even. In any case, the fparts are not equal and so there is a column in the matrix for ANY 2n satisfying Goldbach. QED.<br>
<br><div class="gmail_quote">On Sat, Nov 28, 2009 at 6:39 PM, Will Steinberg <span dir="ltr"><<a href="mailto:asyluman@gmail.com">asyluman@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
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:<br><br>P/p != 2n/p - [2n/p] +k<br>
<br>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:<br><br>frac(P/p) != frac(2n/p). <br><br>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.<div>
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<br><div class="gmail_quote">On Sat, Nov 28, 2009 at 11:55 AM, Will Steinberg <span dir="ltr"><<a href="mailto:asyluman@gmail.com" target="_blank">asyluman@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
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<br>
<br><div class="gmail_quote"><div>2009/11/28 spike <span dir="ltr"><<a href="mailto:spike66@att.net" target="_blank">spike66@att.net</a>></span><br></div><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
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<p></p><div><br><br><div><div></div><div>> -----Original Message-----<br>> From:
<a href="mailto:extropy-chat-bounces@lists.extropy.org" target="_blank">extropy-chat-bounces@lists.extropy.org</a><br>> [<a href="mailto:extropy-chat-bounces@lists.extropy.org" target="_blank">mailto:extropy-chat-bounces@lists.extropy.org</a>]
On Behalf Of<br>> Giulio Prisco (2nd email)<br>> Sent: Friday, November
27, 2009 11:12 PM<br>> To: ExI chat list<br>> Subject: Re: [ExI] Goldbach
Conjecture<br>><br>> I think the Goldbach conjecture is probably false,
with<br>> probability 1 (that means, certainly false). Here is
why:<br>><br>> Apparently there is nothing in the laws of arithmetics
that<br>> forces an even number to be the sum of two prime numbers.
The<br>> conjecture is true for all even numbers on which it has been<br>>
tested, but these are an infinitesimal part of the total (any<br>> finite
number is infinitesimal wrt infinite). Hence, if there<br>> is no proof, the
probability of he Goldbach conjecture being<br>> true is zero.<br><br></div></div></div><div><div></div><div>I
disagree sir, however I confess my line of reasoning is not as well developed as
the one you offer.
<p>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.</p>
<p>Yes I do know that this line of reasoning is not to be substituted for actual
mathematical logic, do forgive please.</p>
<p>I plotted them to a few million on matlab, found there are striking patterns
in the data, such as the eye-catching streaks.</p>
<p>spike<br><br><br><img src="cid:416201016@28112009-2F63" height="440" width="605"></p></div></div></font></div>
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