Talk:Goldbach implies Bertrand

This article should be deleted: It's just plain nonsense, the "proof" is absolutely wrong! See the diskussion at Matroids Matheplanet! --85.127.135.121 04:38, 25 August 2011 (CDT)

Restored the page content and added a proofread tag. Unfortunately my german isn't very good, so others will have to verify. --Joe (talk) 08:08, 25 August 2011 (CDT)

The Author has added an english translation, so you can check it in your language (I've done it). Both versions are wrong. Maik tries to convince us at Matroids Matheplanet since several years that he has proofed the famous Goldbach conjecture (although his lack of mathematical and logical basics is highly visible). My advice: Delete this nonsense!--Franz 08:56, 25 August 2011 (CDT)

Well the translation is absolutely diabolical, and is not good enough English to be understood. If you want anybody to make a proper comment, then:

a. Find somebody who is good enough at both English and German to translate it properly.

b. Learn how to typeset a page using MediaWiki and mathJax markup, so as to present it in a form that is pleasant to read.

Before that happens you may find it difficult to persuade anybody to take this page seriously, although the idea of it is interesting enough.

You may be quite correct in that the page is worthless and needs to be deleted. But until others have had a chance to investigate, please do not delete any pages without that page being discussed first. --prime mover 13:35, 25 August 2011 (CDT)

Another rollback

I have just rolled back another attempt to remove this page, by another anonymous contributor.

Clearly this has been a point of controversy on various German-speaking websites, and clearly this page contains material which may be controversially incorrect. We appreciate that if this is indeed the case, then the page needs to be amended appropriately (but probably not deleted - it may well be retained as an example of a fallacy).

But until somebody a) translates it into readable English and b) explains where the fallacies are, this page stays as it is. I would also require that such a person register with this community as a user, rather than act anonymously. If you contribute as a registered user, we will be able to enter into a dialogue, and as a result your views are more likely to be honoured. --prime mover 02:17, 26 August 2011 (CDT)

The persistent anonymous poster who keeps deleting this page and replacing it with a smug message as to its inaccuracy has been blocked after receiving a polite warning.

I have taken the time to study this page, removing the German (this is an English site) and replacing the English with something coherent enough to understand.

As a result, I have determined what is wrong with this page, and categorised it in the Fallacies and Mistakes page.

Anyone else who wishes to impose their ideas on the philosophy of this site without entering into rational discussion first will likewise be blocked. We understand the good intention of the anonymous poster who wished to clean up something dirty, and thank those who brought it to our attention. But please respect the philosophy of this site, and please do not try to impose your ideas of what you think we ought and ought not to include.

The block will last a week, during which time we are open to arguments as to why this page should be removed. We do not agree, as we believe it can be held up as a fine example of how not to write proofs. You may even like to link to it, or copy its intent, in your own disputes.

Again, thank you for your input on this, but as I say, please respect our philosophy, whether you like it or not. --prime mover 03:05, 26 August 2011 (CDT)

Goldbach => Bertrand

The new translation in mathematical speach is correct.

The proof is correct and not a fallacy. It speaks for itself.

Proof it!

http://www.uni-due.de/~gph120/wahrheitstafeln/

http://en.wikipedia.org/wiki/Truth_table

Sorry @ all for my bad english.

Maik Becker-Sievert

Verify it with :

H. Ricardo (2005). "Goldbach's Conjecture Implies Bertrand's Postulate". Amer. Math. Monthly 112: 492

http://www.math.hmc.edu/~benjamin/papers/monthly481-492.pdf

Yes, Goldbach implies Bertrand, agreed, but what I believe you were trying to prove was that Bertrand implies Goldbach, which does not happen. Correct me if I've got the wrong idea. --prime mover 06:39, 29 August 2011 (CDT)

The nice simple proof shows what it says Golbach => Bertrand, or Goldbach implies Bertrand.

The other direction Bertrand => Goldbach would be nice but this proof is not enough.

So you can sort it under elegant proof for: Goldbach implies Bertrand

The statement:

Goldbach is wrong because there exist an anti Golbach prime number gap!

Is proofed wrong.

Maik Becker-Sievert

Right okay I understand now. As long as no attempt is made to assert that the proof of Bertrand implies the truth of Goldbach, no problem.

This is the text from your original posting which made that assertion:

"Out of this follows that in the defined prime number gap which contradicts the Goldach Conjecture a prime number must lie, because n and 2n lie in this gap.

"This stands in the contradiction to the acceptance that this prime number gap exist

"Out of this follows that the Goldbach Conjectur is true and vice versa the sentence of Bertrand."

So I will remove that, and put this page into the number theory category.

Many thanks for providing the link to the source work. --prime mover 09:50, 29 August 2011 (CDT)

I thank you!

Goldbach => Bertrand is proofed but not Bertrand => Goldbach

OK

Maik Becker-Sievert

It's easy, but not so easy...

The proof ist still wrong, because in the negation of Bertrands postulate $p_1=n$ was omitted improperly (and this case would allow to satisfy Goldbach by $2n=p_1+p_1$). It could easily be repaired by weakening $n<p<2n$ in Bertrands postulat to $n\le\ p<2n$, but then the implication is so trivial that it is hardly worth to talk about it (Goldbach's $2n=p_1+p_2$ leads immediately to the fact that the two primes cannot both be less than $n$). Otherwise you must expand your proof to the case that $n$ is prime, which could e. g. be done by considering Goldbach's $2n+2=p_3+p_4$. (Besides that you should have considered only sufficiently great n to ensure the existence of a prime $p_1$ less than n - which doesn't hold for n=2.) Franz 22:19, 29 August 2011 (CDT)

Okay, I've implemented the proof as given in the originally cited paper. Any further inaccuracies feel free to correct yourself.--prime mover 00:36, 30 August 2011 (CDT)

The proof in this paper is not okay, because "n is prime" does not implicate "n+1 is composite" (take n=2 as a counterexample!). But it's corrected now.--Franz 02:05, 30 August 2011 (CDT)

Maik's "Alternative proof"...

...is - as every mathematician easily can see (and as I have described yet on this "talk") - wrong. I don't want to start an edit war with M-B-S (nor to start any mathematical discussion with him; I had too much of these on other sites), so I put the further ministration of this article in the hands of the admins.--Franz 07:12, 30 August 2011 (CDT)

No more work needed

I have once more deleted the original incomplete "alternative proof" and also the substandard "graphics" which accompanied the original non-proof.

There is no need to include any of the initial material which made this page up. Please do not make any further amendments, in particular, edits which attempt to go back to the original versions, because they are inadequate and therefore unnecessary.

The moderators (this one in particular) has a limited amount of time and patience, so if the above instructions are ignored, there exists the serious possibility of being blocked from making further contributions to this website. --prime mover 13:29, 30 August 2011 (CDT)