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Re: Vindication of Goldbach's conjecture
Posted:
Jul 25, 2012 9:17 AM
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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."
Marcel Luttgens
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