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Topic: Goldbach conjecture
Replies: 43   Last Post: Sep 26, 2000 8:55 AM

 Messages: [ Previous | Next ]
 Gerry Myerson Posts: 192 Registered: 12/8/04
Re: Goldbach conjecture
Posted: Nov 3, 1997 9:59 PM

In article <EJ3G2n.4Ep@syd.csa.com.au>, warwickp@syd.csa.com.au wrote:

> Euler saw that n^2 + n + 41 was prime for n = 0, 1, 2, ..., 39 and
> conjectured that it always produced prime values:

I can assure you that Euler did no such thing.

> ...you've missed the point of my post: although there's "so damn many"
> primes they end up being very sparsely distributed. Verifying a conjecture
> for a large number of examples might make the conjecture "obvious" to the
> lay person, but it shouldn't for the serious mathematician.

Then we're even, as you've missed the point of my post. It is not just that
the conjecture is verified for a large number of examples: it is that a
very much stronger conjecture is verified for as far out as any lay person
or serious mathematician can go. I stick to my position that there is no
serious doubt about, and no reason to doubt, the truth of the Goldbach
conjecture.

The primes are not sparsely distributed, not for the purposes of Goldbach;
they're just off from being all the numbers by a factor of log. Here's yet
another way to think of things: there are about 1200 primes below 10000.
If you chose 1200 odd numbers below 10000 at random (without replacement),
--- let's call them, Usenet-primes --- what is the probability that you
would be unable to express 10000 as a sum of two Usenet-primes?

Well, there are 5000 odd numbers to choose from, so the total number of ways
of choosing is (5000 choose 1200). Now, if you pair the odd numbers off,
like so: (1, 9999); (3, 9997); (5, 9995); etc., then in order to avoid two
Usenet-primes adding to 10000, you have to pick at most one number from
each pair. So, first pick 1200 pairs --- (2500 choose 1200) ways to do this
--- then pick one from each pair --- 2-to-the-1200 ways to do that. So, the
probability of falsifying the Usenet/Goldbach conjecture is

(2-to-the-1200) times (2500 choose 1200) over (5000 choose 1200).

That's about 1 in (4 times 10-to-the-83). That's what I meant when I said,
a few posts ago, that it would take a massive conspiracy among the primes
to falsify Goldbach. And it gets worse, as you go to higher numbers.

There is no similar argument in favor of Li(x) vs Pi(x), or x^2 + x + 41.

Gerry Myerson (gerry@mpce.mq.edu.au)

Date Subject Author
10/29/97 Gerry Myerson
10/30/97 David Ullrich
10/30/97 Gerry Myerson
11/2/97 Warwick Pulley
11/2/97 Ron Bloom
11/3/97 feldmann@bsi.fr
11/4/97 David Ullrich
11/10/97 Bob Runkel
11/11/97 Richard Carr
11/19/97 Richard Carr
11/3/97 Gerry Myerson
11/3/97 Warwick Pulley
11/3/97 Gerry Myerson
11/8/97 Andre Engels
11/9/97 Warwick Pulley
11/10/97 Meinte Boersma
11/13/97 Chris Thompson
9/16/00 Daniel McLaury
9/17/00 Fred Galvin
9/17/00 Jan Kristian Haugland
9/17/00 denis-feldmann
9/17/00 Erick Wong
9/26/00 John Rickard
11/6/97 Richard White (CS)
11/6/97 James Graham-Eagle
11/6/97 Legion
11/6/97 Gerry Myerson
11/7/97 Legion
11/7/97 Gerry Myerson
11/4/97 David Petry
11/3/97 Chris J. Bennardo
11/3/97 Gerry Myerson
11/8/97 Andre Engels
11/5/97 Robert Hill
10/30/97 goldbach
11/2/97 Orjan Johansen
11/2/97 John Rickard
8/25/98 Elijah Bishop
11/2/97 Orjan Johansen
10/30/97 Brian Hutchings
11/2/97 Gerry Myerson
11/6/97 Brian Hutchings
4/28/99 Papus