taeto

No it is not. Where on that page does one encounter such a definition which enables one to talk about "Actual Infinity" in mathematical language?

You assume that to prove P=NP you do it by demonstrating a polytime algorithm to an NP-complete (NPC) problem. And yes, that is the obvious way. And no, there is no restriction whatsoever on the degree of the polynomial which bounds the time demand of your solution algorithm, nor on the size of the constants involved in that polynomial. They could be hugely larger than in a polynomial which bounds the time demand of verifying a proposed certificate of a solution. So long as you get some polynomial bound on execution time, you are good.

Are you aware that in math you have to give definitions of the things that you want to talk about? Could *you* give me "a counter-example of Wombats from maths" without knowing what one might think of as being the meaning of "wombat" in math? Maybe if we are lucky there might actually be found a research paper which introduces the definition of a "wombat" as some kind of mathematical structure. Maybe the paper also presents a Theorem which states that a wombat exists if and only if no tabmow exists, where a "tabmow" is another structure with a precise mathematical definition. If *you* then were to actually produce a tabmow, then you would have "a counter-example of a wombat from math". Is that the kind of situation that you have in mind? You do not really know of any definition of a mathematical object called "Actual Infinity", am I right?

The possibilities, as far as I am presently aware, are that Collatz might be provably true in first order Peano arithmetic, or provably false, or undecidable. It was commented by someone that they thought the latter is the case. You replied, and we agree on this, that all options are still open. I did remark that the latter one would be interesting. There are some aspects of the conjecture that sets it apart from statements that have so far already been proved undecidable. So yes, if it is undecidable, since it is a first order statement it would be either false in the standard model and true in some nonstandard model, or vice versa, by Gödel's completeness theorem. I do not understand the reason for considering any second order properties; Collatz can be stated in first order logic, so the nonstandard model would have first order properties that differ from the standard model, if the truth value of Collatz differs between them. I think about the standard model being something like en.wikipedia.org/wiki/True_arithmetic And to me undecidable means that the truth value of a well-formed statement cannot be decided from axioms, so something like "G is abelian" in group theory. I haven't been aware of other uses until now.

That I doubt.

That wikipedia page is only created by mathematics cranks as a means of vandalizing wikipedia.

It was StringJunky who said that he thought that was the case, and then you said that this has not been proved, and I said that it would be interesting if it was true. Sure I believe that it might be the case, just as well as I can believe that it is not the case, so long as it has not been proved either way yet. Is there anything wrong, in your opinion. with the suggestion that it would be interesting if the conjecture is undecidable? You seem to resist this possibility. I said that it would be interesting. In comparison, there were people who found the fact interesting that it is possible to write up a multivariate polynomial equation with integer coefficients for which it is undecidable whether the equation has any integer solutions. Now for such an equation, if it did have a solution, then this would be straightforward to verify, by plugging in the suggested values of the variables and check that the solution is correct. Which gives an explanation as to why we can say that such an undecidable equation cannot possibly have a solution, since by the virtue of having a solution, it would be decidable. On the other hand, when the equation does not have a solution, you can go on forever trying to plug more and more values into the variables, without ever getting anywhere. Collatz would be interesting because it is different. If the conjecture is true, it is not obvious how to prove it true. If it is false, it is not obvious how to show this either. I cannot think of any other statement like that which is known to be undecidable. Which is the reason that if it could be proved undecidable, it would be quite interesting and new.

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. 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.

It is a wikipedia page maintained by crackpots, its purpose is to vandalize wikipedia. It has nothing at all to do with mathematics.

There is a small point to make though, about the case when you starting problem has all rational coefficients and you need to have an exact rational solution. If the programmer is not all too clever, because this is a lot less trivial than to solve for `real' numbers, then you might end up with only a crude approximation, or the execution time might go through the roof. If the programmer is not an idiot, the code preferably prevents either one from happening.

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.

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?

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.

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.

Did you mean to write that for any prime \(a\) and any natural number \(n\) you have \[a^n + a^{n-1} = 2^2 \cdot a^{n-1}?\] If + refers to usual addition of integers and \(\cdot\) means ordinary multiplication, then after cancellation of \(a^{n-1}\) this becomes \(a + 1 = 4,\) so \(a = 3,\) and the problem then is that this is not true for every prime \(a.\) Did you mistype something?

The existence of such functions has no logical impact whatsoever, since it is a trivial fact.

It appears the issue is with your phrase "there exists". What you stated, by accident means literally that there are some functions that reach powers of 2 after some iterations. What you intended to express was that *your* particular function G has this exact property (among others). I am not sure about what properties G is assumed to have exactly; at least if it follows from those assumed properties that G also has this property, then it has some similarity to the conjectured property of the Collatz function, that is granted.

Whether or not there is does not even matter. The function G that maps every N to 2 does the trick in 1 iteration.

For some reason, on this site you have to reload your post twice before the math shows correctly. If it is something else, then please be more specific about what is not working. Your code looks good.

Does such an argument not entail that gravity itself is just inferred and not directly observed?

Almost. You have to type "e^{i\phi}" in math mode to get the exponent right: e^{i\phi} = \cos \phi + i \sin \phi makes \(e^{i\phi} = \cos \phi + i \sin \phi \)

Nice.

I didn't want to implicate anything, just curious. In fact I just finished a research paper joint with two japanese and one korean colleague. I had to play grammar police quite a bit, so it is getting more easy for me to predict where the pitfalls lie.

Are you Asian, maybe Japanese or Korean? It might help to understand your explanations. I notice you are not using English grammar when you write "integers are infinite", because integers and any objects that are bijective with integers are pretty much the central examples of objects that are not infinite. Whereas in Japanese you might say 多い人がいます。Ooi hito ga imasu (people are many) to mean that in some place there are numerous people present. And not to mean that every person is numerous. In usual mathematical English language you would have to say "the set of integers is infinite", to express what you probably intended to express. Not that it would help your reasoning, since the set of even integers is infinite as well, and what you said is true would be false in that case.

I try with the "math" replaced by backslash square brackets: \[ \int_a^\infty f (x)dx = \mathop {\lim }\limits_{L \to \infty } \int_a^L f (x)dx \] As expected the integral signs extend a little further.