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Goldbach/ Twin Prime stir
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 This is causing a bit of a stir:
http://unsolvedproblems.org/S13.pdf
If you allow for the erroneous "2j is a sum of two primes whose squares do not exceed 2j" [1.2] (cut the "do not")  it looks, as far as I can see, immaculate. Very elementary proof of Goldbach and Twin Prime Conjectures.
Dc   In primenumbers@yahoogroups.com, "descarthes" <mgullandt@...> wrote:
>
*That* is "every elementary"? My eyes glazed over looking at the symbols alone! It looked intelligent but I couldn't get past the symbol/terminology. Tom, is it possible for you to present here the gist of what you did in simple amateur/layman terms, without the rigour? Thanks,
> This is causing a bit of a stir:
>
> http://unsolvedproblems.org/S13.pdf
>
> If you allow for the erroneous "2j is a sum of two primes whose squares do not exceed 2j" [1.2] (cut the "do not")  it looks, as far as I can see, immaculate. Very elementary proof of Goldbach and Twin Prime Conjectures.
>
> Dc
>
Mark > >
Look at Figs 1 and 2 along with associated notes.
>
> *That* is "every elementary"? My eyes glazed over looking at the symbols alone! It looked intelligent but I couldn't get past the symbol/terminology. Tom, is it possible for you to present here the gist of what you did in simple amateur/layman terms, without the rigour? Thanks,
>
> Mark
>
Tom I got another PM of the same ilk.
Theorem 1 basically says that, given any interval of specified length, more (or an equal number of) members of it are divisible by at least one member of any set J of primes if the lefthand endpoint is 2, than if it's any higher; within an error margin of J.
That's quite simple to prove because 2 is the lowest prime, and for higher regions of the interval [1, prod(J)] it's more likely that the divisors of any given integer is a composite; there's a limit on the total number of divisors of integers we can possibly have in our interval of specified length.
If the consensus is that this is comprehensible, I'll move on.
Tom
 In primenumbers@yahoogroups.com, "Mark Underwood" <mark.underwood@...> wrote:
>
>  In primenumbers@yahoogroups.com, "descarthes" <mgullandt@> wrote:
> >
> > This is causing a bit of a stir:
> >
> > http://unsolvedproblems.org/S13.pdf
> >
> > If you allow for the erroneous "2j is a sum of two primes whose squares do not exceed 2j" [1.2] (cut the "do not")  it looks, as far as I can see, immaculate. Very elementary proof of Goldbach and Twin Prime Conjectures.
> >
> > Dc
> >
>
> *That* is "every elementary"? My eyes glazed over looking at the symbols alone! It looked intelligent but I couldn't get past the symbol/terminology. Tom, is it possible for you to present here the gist of what you did in simple amateur/layman terms, without the rigour? Thanks,
>
> Mark
>  Hi,
I had the 2 required courses in Modern Algebra to get a math degree (196667) but
I forgot a lot since then. I have books on it but have difficulty with the subject
now. Anyway, the paper states at the end:
"It follows that our proof of the Goldbach Conjecture implies the proof of the Twin
Prime Conjecture."
Is the converse true?
Sorry if this is answered elsewhere in the paper.
I guess I have a general question:
Which is harder, proving the Goldbach conjecture or the twin prime conjecture?
The above would imply Goldbach is easier but I am not sure. Putting it another
way, which is closer to undecidability?
Cino
To: primenumbers@yahoogroups.com
From: mgullandt@...
Date: Wed, 19 Aug 2009 20:11:51 +0000
Subject: [PrimeNumbers] Re: Goldbach/ Twin Prime stir
I got another PM of the same ilk.
Theorem 1 basically says that, given any interval of specified length, more (or an equal number of) members of it are divisible by at least one member of any set J of primes if the lefthand endpoint is 2, than if it's any higher; within an error margin of J.
That's quite simple to prove because 2 is the lowest prime, and for higher regions of the interval [1, prod(J)] it's more likely that the divisors of any given integer is a composite; there's a limit on the total number of divisors of integers we can possibly have in our interval of specified length.
If the consensus is that this is comprehensible, I'll move on.
Tom
 In primenumbers@yahoogroups.com, "Mark Underwood" <mark.underwood@...> wrote:
>
>  In primenumbers@yahoogroups.com, "descarthes" <mgullandt@> wrote:
> >
> > This is causing a bit of a stir:
> >
> > http://unsolvedproblems.org/S13.pdf
> >
> > If you allow for the erroneous "2j is a sum of two primes whose squares do not exceed 2j" [1.2] (cut the "do not")  it looks, as far as I can see, immaculate. Very elementary proof of Goldbach and Twin Prime Conjectures.
> >
> > Dc
> >
>
> *That* is "every elementary"? My eyes glazed over looking at the symbols alone! It looked intelligent but I couldn't get past the symbol/terminology. Tom, is it possible for you to present here the gist of what you did in simple amateur/layman terms, without the rigour? Thanks,
>
> Mark
>
[Nontext portions of this message have been removed]  hi !
cino hilliard wrote:> Hi,
good point.
> I had the 2 required courses in Modern Algebra to get a math degree (196667) but
> I forgot a lot since then. I have books on it but have difficulty with the subject
> now. Anyway, the paper states at the end:
> "It follows that our proof of the Goldbach Conjecture implies the proof of the Twin
> Prime Conjecture."
> Is the converse true?
anyway, as far as I know, they are "somewhat" related.
I would like Tom to explain this relation deeper,
as I'm sure there are interesting things to learn.
> Sorry if this is answered elsewhere in the paper.
"easier"... is relative. To Tom, it seems easier to tackle Goldbach,
> I guess I have a general question:
> Which is harder, proving the Goldbach conjecture or the twin prime conjecture?
> The above would imply Goldbach is easier but I am not sure.
but from my point of view and personal experience, the TPC
looks easier (and I have not found how to link Goldbach from TPC).
Also, "easier" could also mean "simplicity and concision of the proof".
Or "accessibility of the explanation to the newcomer".
But if Goldbach was "easy", it would have been solved for a long time, no ? :/
> Putting it another way, which is closer to undecidability?
no idea...
But we'll know it when they get "decided" by a definitive proof.
it's getting exciting.
> Cino
yg
 From: descarthes <mgullandt@...>
> This is causing a bit of a stir:
No it's not. At least not in the places where a stir would be meaningful. Nothing to see here, move along.
>
> http://unsolvedproblems.org/S13.pdf
  In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
>
So your identification of the (antistir) error is...
> From: descarthes <mgullandt@...>
> > This is causing a bit of a stir:
> >
> > http://unsolvedproblems.org/S13.pdf
>
> No it's not. At least not in the places where a stir would be meaningful. Nothing to see here, move along.
>
Give the line, eq. or what have you....  There is a little bit more to Theorem 1.
For any sets M and N of integers, let Mult(M, N) denote the members of M that are divisible by at least one member of N. Consider that all intervals are intervals of integers.
Let J be any set of primes.
For any interval I and any member j of J, Mult(I, {j}) is either floor(I / j) or floor(I / j) + 1. In other words, by crafty positioning of endpoints, you can squeeze into your interval one extra multiple for each member of J. But the total number of divisors in any interval of length I is always minimal, ie. is J fewer than the maximum, if the lefthand endpoint is one.
Now consider Mult(I, J \ {j}) /\ Mult(I, {j}). ( /\ = set interscetion; \ = set subtraction (relative complement)). I denote this by T(I, j, J). If for each j in J, T(I, j, J) is maximal, then the divisors are all crammed in the fewest possible number of integers in I, given that number of divisors and that set J \ {j} of divisors. So for that case, Mult(I, J) is minimal under those stated conditions.
Clearly this is not going to be the case if the lefthand endpoint of I is one, because, once more, fewer factors of the multiples of j, other than j, are composites that at almost all of the higher regions of [1, prod(J)]. Indeed, Mult([x,y], J) is maximal (subject to the error margin [discussed above] of J) if x = 1. But, by the same token, for any interval of length yx, T([x,y], j, J) is maximal if the lefthand endpoint is j*min J \ {j}.
And this is the point: when J is an initial segment of the sequence of primes, Mult([x, y], J) is maximal if x= 1, but in principle there is no reason why the single multiple of j in J that is found in [1, 2j) shouldn't, if you could rearrange the sequences of divisors (i.e. by using a different interval: [x+i, y+i]), be made to be a member of T([x+i, y+i], k, J) for some k in J. (BTW here we are strictly considering j in J \ {2}. Then 2j = j*min J.)
And, to continue to speak informally in terms of a 'rearrangement' of the sets of multiples of members of J, these two tricks  squeezing in an extra multiple of j for each j in J, or forcing a singleton (i.e. an integer whose sole factor in J is j) to move onto a multiple of some other member of J  give us the only two parameters that serve to vary the value of Mult([x+i,y+i], J) given the set of all i in N.
Tom   On Thu, 8/20/09, descarthes <mgullandt@...> wrote:
>  In primenumbers@yahoogroups.com, Phil Carmody wrote:
Maybe when you explain why you think you have a greater insight into sieve methods than Halberstam and Richert. I trust you've studied their peadagogical work /Sieve Methods/?
> >
> > From: descarthes <mgullandt@...>
> > > This is causing a bit of a stir:
> > >
> > > http://unsolvedproblems.org/S13.pdf
> >
> > No it's not. At least not in the places where a stir
> > would be meaningful. Nothing to see here, move along.
>
> So your identification of the (antistir) error is...
> Give the line, eq. or what have you....
From prior experience, though, my guess would be at about Theorem 1. Error terms do that. Even Hardy and Littlewood have been caught by them.
However, I will complement you on laying your ideas out neatly and understandably. A vastly better (well, worse from a bunthrower's perspective) show than the cranks lay on.
Phil  On a quick read I spotted that on page 9 n and n+1 are not twin primes. You meant p_n and p_(n+1)
Just before your statement of Theorem 1 on page 3 you say
"For any set J of primes, this will imply maxima and minima in regard to the number of integers k for which Div(J, k) = {k}, given the set of r for which ..."
What's the relevance of r? Presumably the clause "given the set of r for which" is meant to constrain something but it's not clear what, why or how.
Also according to your definition on the bottom of page 2 Div(J,k) is just the members of J dividing k. Given that J are primes, Div(J,k)={k} means that k is a prime in J and your statement becomes rather trivial.
Even if I consider that you got your definition of Div wrong I still can't make it make nontrivial sense.
The last paragraph on page 9 is important but I can't understand it. Could you explain what you are asserting and preferably some of the connecting logic more clearly.
At the top of page 12 shortly before you say QED to the Goldbach conjecture you mention two disjoint subsets of Q, K and K'. Don't you need to constrain the union of the subsets to be Q otherwise your following statement regarding an inequality of a product is likely to be false. Also it's not clear how the two statements you make result in Theorem 2 being proved.
What this "corollary" to theorem 1? A corollarly is a noteworthy result that follows easily from a previous theorem. It's not clear what your corollary to theorem 1 means, how it's relevant or how it follows.
The functions f(n) and a(n,r) are they seriously just introduced to help you prove that the product of (p2) is less than the product of (p1) for the elements p of K ?
It seems that your document details a rather longwinded proof of some very elementary results and then rushes over the interesting bits. I would hope to see building blocks for the proof of the Goldbach Conjecture to be set out explicitly as theorems and then for it to be made very clear how these blocks fit together to prove the main result.
I understand how difficult it is to write a good paper and how relatively easy it is to find mistakes. Keep plugging away at it!
Richard Heylen   In primenumbers@yahoogroups.com, "richard_in_reading" <richard_in_reading@...> wrote:
>[snip]
Thank you Richard. I'm glad you were able to penetrate it enough to see specific problems. I was able to understand Tom's 'colloquial' version, and realized that the core of his proof went from a very elementary result to a QED in two sentences flat. Tom told me that the missing link of proof was found in the full version, but I could not penetrate the set theory nomenclature. My eyes glazed over, even with  note  coffee *and* chocolate by my side.
> It seems that your document details a rather longwinded proof of some very elementary results and then rushes over the interesting bits. I would hope to see building blocks for the proof of the Goldbach Conjecture to be set out explicitly as theorems and then for it to be made very clear how these blocks fit together to prove the main result.
>
But the folding idea is an appealing picture, lending itself to a nice rephrasing Goldbach's conjecture: Every integer greater than one is the mean of two primes.
But here's a reality check for attempts at solving GB's conjecture.
We want to show that 2n is the sum of two primes. We draw a line from 0 to 2n, with n in the center:
0.......n.......2n
GB conjecture would have that for every n>1 there is a prime equidistant on either side of n.
On the way to prove such, of course it would have to be proven that there are indeed primes from n to 2n. Such a little thing. :)
Mark   In primenumbers@yahoogroups.com, "marku606"
<mark.underwood@...> wrote:
> On the way to prove such, of course it would have to be proven
Indeed :)
> that there are indeed primes from n to 2n.
> Such a little thing. :)
I sometimes wonder why purported provers of the G*ldb*ch conjecture
don't criticize Erdos for using central binomial coefficients in
his proof of Bertrand's postulate, nicely caught here
http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/bertrand.pdf
by Robin Chapman. The close reasoning in this proof of a result
immensely weaker than the G*ldb*ch conjecture ought to offer
some sort of warning, one might have thought? Of course
it is not beyond the bounds of reason that Chebyshev and Erdos
fooled themselves and the rest us into thinking that even
Betrand's postulate is this hard to prove.
David >> On the way to prove such, of course it would have to be proven
Nice point mark!
>> that there are indeed primes from n to 2n.
>> Such a little thing. :)
> I sometimes wonder why purported provers of the G*ldb*ch conjecture
Yes, and that is one of the simplest proofs. It would be kind for these
> don't criticize Erdos for using central binomial coefficients in
provers to offer us a halfpage proof of the Bertrand result which is so
much simpler than Goldbach. What a fine way to catch the attention of
mathematicians that would be... CC
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