Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: Vindication of Goldbach's conjecture
Replies: 5   Last Post: Jul 25, 2012 9:51 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Gus Gassmann

Posts: 28
Registered: 9/11/09
Re: Vindication of Goldbach's conjecture
Posted: Jul 25, 2012 9:51 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Jul 25, 10:17 am, mluttgens <lutt...@gmail.com> wrote:
> On 24 juil, 17:01, Gus Gassmann <horand.gassm...@gmail.com> wrote:
>
>
>
>
>
>
>

> > On Jul 24, 10:20 am, mluttgens <lutt...@gmail.com> wrote:
>
> > > On 22 juil, 04:14, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>
> > > > mluttgens <lutt...@gmail.com> writes:
> > > > > On 21 juil, 15:32, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> > > > >> lutt...@gmail.com writes:
>
> > > > >> <snip>
>
> > > > >> > Both terms of 6 = 3 + 3 are primes.
> > > > >> > I considered the case where at least one of the terms is not prime.
> > > > >> > Your example is irrelevant!

>
> > > > >> Your claim is essentially the same as GC.  I thought you'd miss-worded
> > > > >> it which is why I thought there was a counter example.  Correctly worded
> > > > >> (as I think it is) proving it is equivalent to proving GC.

>
> > > > >> <snip>
>
> > > > >> >> ... counter examples may
> > > > >> >> be very hard to find, but that does not constitute a sound argument: you
> > > > >> >> can't prove X by noting that X follows from Y and challenging people to
> > > > >> >> disprove Y (but you know that, yes?).

>
> > > > >> > No, you did not.
>
> > > > >> What does that mean?  Is it a comment on my remark about your "proof by
> > > > >> you can't contradict me" method?

>
> > > > > Not at all.I was referring to some quibbling of you...
>
> > > > Ah, then better to put it next to the quibble.  In my opinion, the
> > > > comment about you proof structure ("look, you always get two primes if
> > > > you add and subtract some even number -- show me a counter example") was
> > > > much more than a quibble.

>
> > > > >> (By the way, can you get you newsreader to stop turning plain 7-bit
> > > > >> characters into HTML entities?)

> > > > <snip>
> > > > > Sorry, the new Goggle interface was responsible. For that reason,
> > > > > I have just went back to the older interface.

>
> > > > Thanks.  Much better.
>
> > > > <snip>
>
> > > > > Proof of the validity of Goldbach's conjecture
> > > > > _______________________________________

>
> > > > > According to the conjecture, every even integer greater than 4 can be
> > > > > expressed as the sum of two primes.

>
> > > > > Let?s consider the infinite series of uneven integers.
> > > > > Such series contains an infinite number of products p = ab, where a
> > > > > and b are primes.
> > > > > To each product p corresponds a single sum s = a + b, s being of
> > > > > course an even integer.
> > > > > This approach leads to all possible sums of two primes.

>
> > > > There's no point to this pre-amble.  It adds nothing to the discussion
> > > > and just looks like padding.

>
> > > > > By the way, some even integers can be the sum of two uneven integers,
> > > > > at least one of them not being a prime.

>
> > > > All even integers other than zero can be written as the sum of two odd
> > > > integers, at least one of them not being prime: 2k = 1 + (2k-1).  It
> > > > comes over as a bit odd to say "some" when you are stating an obvious
> > > > property of all numbers != 0.

>
> > > > > This leads to the bold assumption, that one or more even numbers
> > > > > greater than 4 could not necessarily be expressed as the sum of 2
> > > > > primes.

>
> > > > I'd start the argument here...  You don't need (or use) any of the
> > > > above.

>
> > > > > A sum s of two primes a and b greater than 3 can always be written as
> > > > > s = (a + n) + (b - n) or s = (a ? n) + (b + n), where n is an even
> > > > > integer.
> > > > > The obtained terms (a +/- n) and/or (b -/+ n) can be prime numbers,
> > > > > but being ordinary uneven numbers does not imply that an even integer
> > > > > cannot be a sum of two primes.
> > > > > Let?s notice that such method, which consists of adding or
> > > > > subtracting  the successive elements of  the series of even numbers n,
> > > > > can be applied for arbitrarily large sums s.
> > > > > It leads to all possible pairs of numbers: two primes, a prime and a
> > > > > uneven number, that is  not  a prime, or two uneven numbers, which are
> > > > > not prime.

>
> > > > ...and you are assured of getting two primes for all s, only if GC is
> > > > true.

>
> > > > > On the other hand, a sum s? of two uneven integers, where at least one
> > > > > of its terms is not prime, can be transformed into a sum s of primes
> > > > > by adding some even integer n to one of its terms and subtracting the
> > > > > same n from  its other term.

>
> > > > This statement needs a proof.  If GC is true is it's obviously true; if
> > > > GC is false, it's false.

>
> > > > > To determine n, it suffices to apply the above method to the sum s =
> > > > > s?. Then, one straightforwardly gets the value of n leading to the
> > > > > uneven terms of sum s?.

>
> > > > The above is not a method of getting two primes -- it's a method of
> > > > getting all pairs of odd numbers that sum to s.  One of these will
> > > > always be a pair of primes only if GC is true.

>
> > > > > Example:
>
> > > > > s? = 13 + 15 = 28 (s? is not the sum of two primes).
>
> > > > > From s = s? = 28, one gets
> > > > > s = 5+23 and 11+17, and also
> > > > > s = (5+8) + (23-8) = 13+15 = s?
> > > > > s = (11+2) + (17-2) = 13+15 = s?
> > > > > QED!

>
> > > > > The assumption that one or more even numbers greater than 4 could not
> > > > > be expressed as the sum of 2 primes is thus refuted.

>
> > > > > This leads to the conclusion that any even integer can indeed be
> > > > > expressed as the sum of two primes.

>
> > > > > Marcel Luttgens
>
> > > > > July 22, 2012
>
> > > > No need to date your posts.  Usenet records the date of posting in the
> > > > header.

>
> > > > --
> > > > Ben.

>
> > > Thank you! You are of course right.
>
> > > But my aim was to show that a sum s? = a + b of two uneven numbers, at
> > > least one of them not being a prime, could easily be transformed into
> > > a sum of two primes, simply by adding and subtracting some even number
> > > from its terms:

>
> > > The chosen example was:
>
> > > s? = 13 + 15 =  (13-8) + (15+8) = 5+23
> > >                     =  (13-2) + (15+2) = 11+17

>
> > > It has been claimed that such transformation could sometimes not be
> > > possible.
> > > I am wondering about which terms a and b should be chosen to justify
> > > that claim.

>
> > That is, of course, a good question. IF Goldbach's conjecture is
> > false, then of course there is a counterexample to your claim.
> > However, nobody knows at this point one way or the other. However,
> > asking for a counterexample and not receiving one is by no means
> > equivalent to having found a proof! In essence your approach is this:

>
> > Theorem: Goldbach's conjecture is true.
>
> > Proof: If it were false, there would be a counterexample. Nobody has
> > found one. So the theorem is proven.

>
> > That is not mathematics, and I hope you can see why it isn't.
>
> > > Till now, I didProof: If it were false, there would be a counterexample. Nobody has
>
> found one. So the theorem is proven.  not find a clue in the
> litterature, but you have
>

> > > perhaps a reference?
>
> > > Marcel Luttgens
>
> I would rather say
> "Proof: If it were false, there would be a counterexample. For
> theoretical reason, nobody can
> find one. So the theorem is proven."


And this is exactly where you are going to run into resistance. "For
theoretical reason, nobody can find one" is completely bogus. You made
that up, and you have no theoretical reason for that statement. To
date, nobody has found a counterexample, but nobody has found a proof,
either. So the conjecture is still open.



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.