Goldbach's Conjecture - Proof (better link)

[Andy Lee]

For reference, your first mistake is here:


In fact, out of N consecutive odd numbers, as few as (N - 2)/3 might remain.

For each iteration we could say as few as (N-M)/D might remain, where multiples of
D are eliminated and M is the maximum divisor in the process.

Then near the end of my proof instead of the expected number being
(1/sqrt(2N))*((N-1)/2)) I would have

(1/sqrt(2N))*((N-M)/2)) and M<=sqrt(2N) so N-M >= N-sqrt(2N) giving
an expected number of prime pairs greater than or equal to

(1/sqrt(2N))*((N-sqrt(2N)/2)

and this is >1 for N>=18.

So the proof is good for N>=18

 

[CRGreathouse]

M<=sqrt(2N)

Care to explain the derivation of that?

 

[CRGreathouse]

My previous counterexample with
6002108856728919
still applies. (47 - sqrt(2 * 47)) * 1/3 * 3/5 * ... * 45/47 > 1, so you've 'proved' that that interval contains at least 2 odd numbers with smallest prime factor > 47.

 

[Andy Lee]

My previous counterexample with
6002108856728919
still applies. (47 - sqrt(2 * 47)) * 1/3 * 3/5 * ... * 45/47 > 1, so you've 'proved' that that interval contains at least 2 odd numbers with smallest prime factor > 47.

Again, as I stated before, the counterexample is not relevant unless it was less than 47^2

 

[rofler]

Then you must prove that he cannot create a counterexample.

So first and foremost, if you are using the fact that his counterexample does not work because it is larger than n^2, then you must then prove the following statement for your proof to be valid:

We cannot have n consecutive numbers between 1 and n^2 so that each is divisible by at least one of 3,5,...,n.

After that, the other points of the proof can be discussed.

Cheers,

Rofler

 

[CRGreathouse]

Again, as I stated before, the counterexample is not relevant unless it was less than 47^2

Do you seriously think that my lack of a counterexample to Goldbach's Conjecture means that you have proved it?

Your proof is wrong for the reasons I have outlined. I have given a counterexample to your method, which is the best I can do since I don't have a counterexample to Goldbach's conjecture itself.

And you still haven't even suggested a reason that M <= sqrt(2N).

 

[Andy Lee]

I never suggested that your lack of a counter example means I proved Goldbach!

You attempted an example to counter a particular element of my proof, in an effort to demonstrate the proof as invalid. You were unable to do so. So it is that the proof was not destroyed by your example. Your example has not verified the proof true or false.

 

[CRGreathouse]

So first and foremost, if you are using the fact that his counterexample does not work because it is larger than n^2, then you must then prove the following statement for your proof to be valid:

We cannot have n consecutive numbers between 1 and n^2 so that each is divisible by at least one of 3,5,...,n.

Actually, this wouldn't be enough to make his proof work. But a proof of this statement (as quoted, or replacing "numbers" with "odd numbers" as Andy is won't to do) would be of interest to me.

Sloane's http://www.research.att.com/~njas/sequences/A049300 seems relevant, but sadly doesn't list any papers.

 

[CRGreathouse]

You attempted an example to counter a particular element of my proof, in an effort to demonstrate the proof as invalid. You were unable to do so.

Wrong. Reread what I wrote.

 

[rofler]

In the proof, you explicitly use the theorem that there is no string of n consecutive numbers below n^2 so that they are all divisible by at least one of 3,5,...,n, and you must prove this, or it will not be valid.

Furthermore, even if CRGreathouse cannot construct a counterexample (although I think he did construct a counterexample to your original proof), you have given no reason for one not to exist, so the proof is not "complete".

Cheers,

Rofler

 

[CRGreathouse]

even if CRGreathouse cannot construct a counterexample

A counterexample to that would be a counterexample to the full GC. I can't provide that.

 

[Andy Lee]

A counterexample to that would be a counterexample to the full GC. I can't provide that.

I know. So it will be necessary to show error in the proof by some other fashion (or accept it).

 

[Andy Lee]

In the proof, you explicitly use the theorem that there is no string of n consecutive numbers below n^2 so that they are all divisible by at least one of 3,5,...,n, and you must prove this, or it will not be valid.

Rofler

In fact, I did not use this conjecture as part of the proof. Rather I suggest it is a result that has been proved (though unintentionally) by my proof of GC.

 

[CRGreathouse]

I know. So it will be necessary to show error in the proof by some other fashion (or accept it).

I've already done that in posts #6 and #20.

In post #29 you agreed that I found a hole. In post #31 you attempt to fix it, but your fix doesn't work. I suggest carefully studying the transition from {3} to {3, 5} to {3, 5, 7} in order to see why this is.

For reference, the first mistake in that post is

For each iteration we could say as few as (N-M)/D might remain, where multiples of
D are eliminated and M is the maximum divisor in the process.

though this may perhaps be rescued. The first serious mistake is

Then near the end of my proof instead of the expected number being
(1/sqrt(2N))*((N-1)/2)) I would have

(1/sqrt(2N))*((N-M)/2)) and M<=sqrt(2N) so N-M >= N-sqrt(2N) giving
an expected number of prime pairs greater than or equal to

which, unsurprisingly, is not justified in any way.

 

[Andy Lee]

Care to explain the derivation of that?

I defined M as the maximum divisor in the process.

Take my example in post 26 where N=37. There is no reason to extend the
iteration beyond sqrt(74). Sundaram's sieve is illustrative...

9, 15, 21, ...
25, 35, 45, ...
49, 63, 77, ...
p^2+2px gives all odd composites.

 

[CRGreathouse]

I defined M as the maximum divisor in the process.

I didn't ask what it was -- though I'll admit that even after that explanation I don't know what you mean. I was asking how you knew that it was less than sqrt(2N).

Your claim is that there are at least (1/sqrt(2N))*((N-M)/2)) Goldbach pairs for 2N. This is really the only context in which I care about M. If M is defined in a way that makes it obvious that it is less than sqrt(2N), then I want to know how you came to this statement; if it's defined in a way that makes this statement obvious, I want to know how you can bound it below sqrt(2N).

 

[Andy Lee]

Sorry, I realize now from your post 27 in which you said:
"In fact, out of N consecutive odd numbers, as few as (N - 2)/3 might remain."

Your N was my (N-1)/2, so your (1/3)(N-2) becomes (1/3)((N-5)/2) for me.
It seems I was in fact too generous allowing for M to be the maximum divisor in the process, when 5 will do just fine.

I believe then my proof shows there are at least (1/sqrt(2N))*((N-5)/2)) Goldbach pairs for 2N. This is greater than 1 for N>=17.

 

[CRGreathouse]

Sorry, I realize now from your post 27 in which you said:
"In fact, out of N consecutive odd numbers, as few as (N - 2)/3 might remain."

Your N was my (N-1)/2, so your (1/3)(N-2) becomes (1/3)((N-5)/2) for me.
It seems I was in fact too generous allowing for M to be the maximum divisor in the process, when 5 will do just fine.

Ah. Then you entirely misunderstood. My bound (which is tight) was only for divisibility by 3 for n or 2N - n. It's worse for divisibility for 3 and 5, and worse yet for divisibility by 3, 5, and 7.

 

[Andy Lee]

Ah. Then you entirely misunderstood. My bound (which is tight) was only for divisibility by 3 for n or 2N - n. It's worse for divisibility for 3 and 5, and worse yet for divisibility by 3, 5, and 7.

Yes. Thanks for your time on this.