Jump to content

The Collatz Conjecture has been proved. What next??


Recommended Posts

Hi friends I am a high school student from india and have been trying to prove the collatz conjecture. Physics and maths have always been my favourite and recently I have been able to develop a proof for this conjecture. What should I do? Whom should I send it to? What is its value? Please do help me out.

Edited by Antony Howard Stark
....
Link to comment
Share on other sites

You would need to get it published in a mathematics journal. But that might be hard if you are not familiar with how to write a mathematics paper and a formal proof.

Alternatively, you could present your idea here and maybe someone can explain where you have gone wrong. :) 

That is probably a bit unfair; mathematics is one of the few areas where amateurs do occasionally make breakthroughs!

 

 

Link to comment
Share on other sites

I feel that I should say something here. But I do not know exactly what.

The Collatz conjecture is extremely difficult and hard to prove. Some people would say that it might be impossible. A proof or disproof of the conjecture would be extremely interesting.

If you have an idea for a proof, you should certainly discuss it with somebody, and find out whether there is a hole in your proof suggestion. 

If you cannot find a mistake, it is still not a disaster if you send a written up version to a journal and it turns out to be wrong. Of course if it turns out to be correct it is just great, but I think you are not worrying much about that case anyway. 

The problem is that if you submit a manuscript to a journal, and it is not written in formally coherent way, then it could be summarily rejected, especially if you are dealing with a high-end journal. Which you naturally should be, if you propose to solve this conjecture.

Is there some way that you can give more information about your work, without revealing enough to enable somebody to steal it?

If you have a finished manuscript, I can recommend to submit it to the Electronic Journal of Combinatorics, of which I am an editor: www.combinatorics.org - we will take a look at it and recommend what you should do with it further, possibly guiding you all the way to final publication.

 

 

Link to comment
Share on other sites

4 hours ago, Antony Howard Stark said:

Hi friends I am a high school student from india and have been trying to prove the collatz conjecture. Physics and maths have always been my favourite and recently I have been able to develop a proof for this conjecture. What should I do? Whom should I send it to? What is its value? Please do help me out.

First, you should show it to your local school mathematics teacher. Then he/she should show it to somebody on nearby university for verification..

Link to comment
Share on other sites

  • 2 months later...
2 minutes ago, Antony Howard Stark said:

Well the last thread didn't really turn up satisfactory answers

I thought there were some good answers. I particularly liked:

  • Discuss it with a local maths teacher
  • Post it here for discussion

What is unsatisfactory about those?

Link to comment
Share on other sites

On 9/12/2018 at 7:52 PM, Antony Howard Stark said:

Hey guys . Has the collatz conjecture been proved yet ? If yes then who and how and if no then I'd like to share a proof I've come up with. And also please say where can I submit this proof ?

No, it has not been proved yet in any published work.

If you have a proof, you should submit it to Annals of Mathematics.

Apologies if my answers do not live up to your stellar expectations.

Link to comment
Share on other sites

All right guys now you all are being a bit rude. You all have assumed many  a things about my expectations but all I really want is a reply.

1. I tried to contact AMS no reply

2. IJISRT no reply

3. Went to my maths teacher , didn't understand the goddamn proof. (Although I don't know why was it so hard to understand )

4. And As of your solution to post my proof here. What guarantee can you provide me with that my proof will still be mine ?

So guys I tried out all the solutions generated by the previous string. And this is what I meant when I said that it did not turn up satisfactory answers. Do I want people to know me ? Obviously yes who dosent ? But first I'm expecting someone to atleast understand what im trying to say.

I think I'll try the Annnals of mathematics. 

Then electronic journals of combinatronics.

I hope my answer has lived up to your rude comments and shows you that I'm trying to do what I can.

 

Link to comment
Share on other sites

If it is about priority, then why not post on vixra? Then you get submission acknowledgment with your name on it and a date stamp, everyone has to recognize that it is your work..

What you do not get there is any form of interest or feedback. After all I counted about 40 "proofs" of Collatz already on vixra, so it will be run of the mill there.

But after you are done with the above, you can send it here. There will be people telling you either that it seems fine, or pointing out possible mistakes.

Link to comment
Share on other sites

7 hours ago, Antony Howard Stark said:

All right guys now you all are being a bit rude.

That might be because you just ignored all the comments made. Some people get a bit iffy about that sort of thing.

7 hours ago, Antony Howard Stark said:

4. And As of your solution to post my proof here. What guarantee can you provide me with that my proof will still be mine ?

Is that a real problem? But I would have thought that posting it here (or vista, or a blog or ...) would give you as good a claim to ownership as anything else. With the benefit you might get some expert feedback. 

Have you looked at the other Collatz threads here? It might give you an idea of the sort of feedback you might get...

Link to comment
Share on other sites

3 hours ago, StringJunky said:

From what I've read, the conjecture has been proven to be undecideable under the current paradigm. It needs new maths that doesn't currently exist.

It certainly hasn't been proven independent of the usual axioms. That would have made the news and it would be quite surprising. I think there's a quote from Erdős that math isn't ready for such problems, but he's referring to techniques, not axioms. 

 

Link to comment
Share on other sites

13 hours ago, wtf said:

It certainly hasn't been proven independent of the usual axioms. That would have made the news and it would be quite surprising. I think there's a quote from Erdős that math isn't ready for such problems, but he's referring to techniques, not axioms.

It would be interesting if that is the case though, and not too difficult to imagine. I had already been thinking that I would look up if something is known about what happens in a non-standard model of the natural numbers. Maybe somebody reading this forum knows?

Link to comment
Share on other sites

On 9/18/2018 at 11:29 PM, taeto said:

It would be interesting if that is the case though, and not too difficult to imagine. I had already been thinking that I would look up if something is known about what happens in a non-standard model of the natural numbers. Maybe somebody reading this forum knows?

No direct knowledge but hard to believe. Any first-order statement has the same truth value in the standard or nonstandard integers, since they satisfy the exact same set of axioms. If the hyperreals or any nonstandard model of anything had produced any result of interest, we'd know about it. At best the proponents of the hyperreals claims pedagogical improvements in the teaching of calculus, but never a new research result.

Link to comment
Share on other sites

8 hours ago, wtf said:

No direct knowledge but hard to believe. Any first-order statement has the same truth value in the standard or nonstandard integers, since they satisfy the exact same set of axioms. If the hyperreals or any nonstandard model of anything had produced any result of interest, we'd know about it.

Indeed, same axioms and theorems. The point is that if there is a counterexample in a nonstandard model, then the conjecture cannot be proven. Even if it is true in the standard model, in which case it would be undecidable. But as you say, it seems that not much is known.

Link to comment
Share on other sites

21 hours ago, taeto said:

Indeed, same axioms and theorems. The point is that if there is a counterexample in a nonstandard model, then the conjecture cannot be proven. Even if it is true in the standard model, in which case it would be undecidable. But as you say, it seems that not much is known.

Can you say what aspect of this problem leads you to believe it might be true in some models and false in others, which means there could be no proof. 

As context let me give an example. Take the reals and the hyperreals. The hyperreals are a superset of the standard reals that include infinite and infinitesimal elements. It's a nonstandard model of the reals. Therefore it satisfies the same first-order properties and theorems tha the reals do. For instance. .999... = 1 in both the reals and the hyperreals, and the proof is identical in either case. This debunks a common theme in .999... crank threads, when people wrongly believe the hyperreals help them to falsify the equalty. They don't. 

But there IS an important property that the reals satisfy and that the hyperreals don't. And that is, metric completeness. The real line is complete. It has no "holes." The technical definition is that every Cauchy sequence converges. That means that every sequence that morally "should" converge, does converge. 

To illustrate this concept, consider the plain old rationals, the fractions n/m where n and m are integers. They are a field: You can add, subtract, multiply, and divide (except by zero of course). They are totally ordered: that is, for any two non-equal reals, one is larger than the other. The rationals are similar to the reals in the sense of being densely ordered fields. 

But the rationals are not complete. For example the Cauchy sequence 1, 1.4, 1.41, 1.412, ... consisting of more and more decimal places of the sqrt(2), "should" converge to sqrt(2). But it doesn't, because sqrt(2) isn't a rational number. So there's a "hole" in the rationals where the sqrt(2) should be. And we can express that without going outside the rationals by noting that there's a Cauchy sequence that doesn't converge.

However, the hyperreals are not complete! It's a theorem that any model of the reals that contains infinitesimal elements can not be complete. There are Cauchy sequences that don't converge. There are holes.

What happened? First I said that they share all first-order properties. But completeness is a second-order property. Another definition of completeness is the Least Upper Bound property, which says that a nonempty set of reals bounded above must have a least upper bound. This is the defining characteristic of the standard reals. And to express it we have to quantify not only individual real numbers, but over arbitrary nonempty SUBSETS of the real numbers. That makes the logic second-order; and the reals and the hyperreals may in fact differ on second-order properties.

So -- sorry if that's a lot of words but I hope it's clear -- why is it that you think the Collatz problem has some aspect that makes it behave differently in some nonstandard models of the integers? What aspect of the problem necessarily creeps into second-order logic?

Edited by wtf
Link to comment
Share on other sites

12 hours ago, wtf said:

Can you say what aspect of this problem leads you to believe it might be true in some models and false in others, which means there could be no proof. 

     I am not sure whether it is a reasonable requirement that there should be some such aspect. But you can feel free to educate me. The Gödel sentence itself is "self-referential", so that would lie within the scope of that requirement. But for undecidability of solutions of a Diophantine equation or something like Paris-Harrington, it does not obviously apply. Maybe you can mention some such "aspect" in those cases.

        

12 hours ago, wtf said:

As context let me give an example. Take the reals and the hyperreals. The hyperreals are a superset of the standard reals that include infinite and infinitesimal elements. It's a nonstandard model of the reals. Therefore it satisfies the same first-order properties and theorems tha the reals do. For instance. .999... = 1 in both the reals and the hyperreals, and the proof is identical in either case. This debunks a common theme in .999... crank threads, when people wrongly believe the hyperreals help them to falsify the equalty. They don't. 

But there IS an important property that the reals satisfy and that the hyperreals don't. And that is, metric completeness. The real line is complete. It has no "holes." The technical definition is that every Cauchy sequence converges. That means that every sequence that morally "should" converge, does converge. 

   Thank you for the interesting example! As I am not familiar with these kinds of extensions of the reals, I am quite interested to note that, whereas the first order theory of the ordered real field is provably consistent and complete, there are models within which the truths of second order statements differ.

    That does not seem relevant though to the case of the Collatz conjecture. This is a statement that can be formulated in first order Peano arithmetic, and it is possible that there are models for which this FOL statement is true and others for which it is false. It has not got to do with second order statements, like in the case of hyperreals.

Link to comment
Share on other sites

38 minutes ago, taeto said:

     I am not sure whether it is a reasonable requirement that there should be some such aspect. But you can feel free to educate me.

I was hoping you could educate me. You claimed that there might be a Collatz solution in a nonstandard model of the integers. I"m asking why you believe that. Trying to understand your earlier posts.

Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.