wtf

42 minutes ago, Zolar V said:

"As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7. "
- the next step is to extract either 7 or 5.  if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.)

I'm sorry i desperately looked for my notebooks and work has been real busy. 

No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure.

Still, just in case I want to try to read your most recent couple of posts, can you just tell me whether you are 

* Explaining how to handle 5 and 7 in the computation of G(139); or

* Trying to patch theorem 1 in some other way; or

* Introducing an entire new argument.

Also, are we still following the paper, or has that been abandoned? I just need to understand the context of the last few posts. Once I asked you about 5 and 7, your posts went off in a completely different direction without context, so I'm confused.

> if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.)

Still makes no sense to me. The sums are not discussed in theorem 1. So you have not established theorem 1. In your paper you talk about sums AFTER theorem 1. And you haven't provided an explanation I can understand of how 5 and 7 are handled.

What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?

 

 

Help me out here. The last thing I understood was theorem 1, right up until the recursion step. As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7. 

Now you do SOMETHING with 5 and 7 but for several posts you've failed to clearly explain it. Your two most recent posts seem to be veering off into some new direction. You have to put what you're doing in context. Is this a new idea, are you trying to patch theorem 1, are you trying to explain G(139), or what?

What you want to do here is say LESS and not MORE. I got to 140 = 2^2 x 5 x 7 and asked what happens to 5 and 7, and you have gone off with two long posts that are not anchored to anything I understand of your argument. Try just answering me with one- or two-sentence posts till I have some idea what you're talking about. 

I'm still at 2^2 x 5 x 7 and I've been there for two days now. Give me one clear sentence at a time as to what you are doing. Are you

* Explaining 5 and 7?

* Going off in a new direction with a new argument?

I just see no relation at all between what we agreed on a couple of days ago, and your most recent new direction.

> I think you are correct that the 3rd step isnt fully developed and it indeed cooks the goose. 

No, you have not even established theorem 1 yet. What happens with 5 and 7?

 

Sorry before I dive into this ... Theorem 1 is retracted? We're done with the paper now? Please be clear.

Natural number as sum of primes? Wow you're way beyond anything in your paper. You need the argument to become more grounded and focussed, not more wild. You're halfway to the famous partition problem, how many ways can you express a given number as a sum of natural numbers. You will not get any kind of similar argument under control.

What is the above symbology supposed to be about? What's your claim, what are you proving, what is your conclusion? I get that you're unhappy you can't find your original argument and that your paper's cooked. But this latest isn't possible to read without a sensible explanation.

To clarify: You are giving up on explaining G(139) and you have a new argument now?

2 minutes ago, studiot said:

It's good to see two members having such a cooperative discussion about a subject.

 

Here is a current discussion by someone else about much the same problems with latex that may help both of you.

 

https://www.scienceforums.net/topic/115767-test/

 

Evidently LaTeX does work on this site ... \[e^{i \pi} + 1 = 0\]

The delimiters are backslash-open-squarebracket and backslash-close-squarebracket.

 

On 7/22/2017 at 3:57 PM, Dave said:

After a recent update to our forum software, typesetting equations on SFN has changed a little bit. Although we are still using LaTeX, for a variety of reasons, we've elected to shift over from our custom-written LaTeX generator to the excellent MathJax library, which will take your equations from post text and render them in your browser.

Much as before, the idea is that in your post, you surround equations with special characters, and MathJax will convert the contained text into an equation for you. There's two types of equation that you can typeset:

• Inline math is displayed in the flow of a sentence, such as y=x2 . This example was produced by using the text \( y=x^2 \). Note that we do not support $ signs as most LaTeX users would be familiar with, since this occurs too frequently in text.
• Display math breaks up a paragraph and can be used for typesetting larger equations such as 
  y=∫f(x)dx.
  The text then picks up afterwards. This example was produced by using the text \[ y = \int f(x) dx \] , which we note is exactly what one would type in a usual LaTeX document.

For reference, the old guide is still available and has a number of useful examples for those getting started.

Finally, please note that for legacy posts, the old  [math] [/math] tags will still continue to work and these will display equations as inline. However it's likely that older posts may look different to the way that they did before.

\[ y = \int f(x) dx \]

10 hours ago, Zolar V said:

So here is where it gets a little tricky. You cant recurse 5,7 separately since they're multiplied together, but you can select which you want to recurse through.  
Once you select which prime you want to recurse, you manipulate the product such that you have a copy of that prime you selected.

 Suppose you select 7 then
5*7 = (5-1)7 + 7
and you can recurse through the first copy of 7 you have extracted.  Then recurse through all 5 copies of 7.    Unfortunately my gut is saying im missing something here or something isnt quite right.  I need to go find and look at all my notes again.    Something is a little off, but its also very late.    I'm gonna think about it and see if I can't remember whatever it is that's lurking in the shadows of my mind.
 

> Unfortunately my gut is saying im missing something here or something isnt quite right.

I'll stand by for clarification, because your exposition is pretty murky here. But what do you do if you're left with 3 primes or a million primes? What if you're left with 5x7x47x47x113 x 113 x 113? What now?

30 minutes ago, Zolar V said:

Ohhh! im sorry I read it wrong. I see now what you're doing.
You have defined 2^n as the largest power of 2 within p+1.  You then move it over and leave yourself the non-distinct product of prime factors. 
I got mixed up on your step 4 and assumed that you were further along, forgetting entirely that we are only trying to define G.
In that case, yes I think you are on the right path.

 Hey man!  Glad to see you're still around here too!

 

> You have defined 2^n as the largest power of 2 within p+1.

Yes.

> You then move it over and leave yourself the non-distinct product of prime factors. 

Yes! Just like you told me to!

> I got mixed up on your step 4 and assumed that you were further along, forgetting entirely that we are only trying to define G.
In that case, yes I think you are on the right path.

Ok good. But I'm still confused about G(139). We have 139 -> 140 -> 2^2 x 5 x 7. Now do you recurse each of 5 and 7 separately? Or recurse five copes of 7 separately? I found your earlier remark extremely confusing.

> Hey man!  Glad to see you're still around here too, 

Oh cool. Still, I should be clear about something. I don't think you have a proof. First, no amateur has a proof. A lot of really smart people have looked at the problem. If there's an elementary proof along the lines of doing arithmetic on prime factorizations, it would have been found.

Secondly, even if it is logically possible that you have a valid proof in your head, that is not enough. What you wrote isn't a proof. And your expository skills aren't sufficient to explain it to me. Perhaps you're right, maybe you need a real mathematician. If you're Ramanujan I'll perfectly well stipulate that I'm no Hardy!

Given that, I'll stick with this for a while more. I'm pleased I understood your G function. In fact I agree that your theorem 1 is correct, although I haven't worked out the detailed induction. So I want to push on a little more and see if I can understand the next step. 

56 minutes ago, Zolar V said:

Maybe I'm reading yours wrong, but it appears you're doing a modified form of the collatz conjecture.
if it is odd then add 1 
if it is even divide by 2

 

Nothing of the sort. I'm not even at the second half of your paper. All I'm doing is trying to define G as in your first theorem. I'm not even at the part where you introduce the additive term. I"m just trying to make sense of theorem 1.

ps -- I'm just trying to understand section 2, the prime reducing function. Lines 16 through 34. That's the sole focus of my attention. Are you saying I'm not understanding that correctly? Can you clarify then? The business with adding a term to a product of primes isn't discussed in section 2 so I'm not up to that yet. 

33 minutes ago, Zolar V said:

I actually was about to use LaTex, but decided to use MS Word's insert equation function instead.  Its more clunky but a bit faster for a mockup.

I would say that you're generally right.  I don't divide mine by 2ⁿ  for the sole purpose of saving them all for the end. 
I think the tricky detail you're missing is converting the product of primes into a summation of the primes. That conversion lets you work with one prime at a time and also gives you some detail about how many times you need to do step 3
Maybe a bit better of a way to think about it would be to convert all products back into basic addition.
5*7 = 7 + 7 + 7 + 7 + 7 

Well then now I'm perfectly well dispirited. Not factoring out 2^n till the end I can certainly accommodate. But the way you described it sounds like I've got something substantively wrong. Wow, 5*7 written out as five 7's? I confess, I gave it my best shot and cannot make sense of this. I'll read what you wrote but I'm afraid I'm at the end of my tether on this problem.

ps -- Let me get this straight. You get, what's a prime that works with 5 and 7? 2 x 35 is 70 and 69's not prime. 4 x 35 is 140 and 139 is prime, isn't it? Yes 139 is prime. So 139 -> 140 -> 4 x 5 x 7.

Now I thought you are going to recurse down both subnodes 5 and 7 separately, and note that they both end up at a power of 2. I believe that.

But now you're saying you write that as 7 + 7 + 7 + 7 + 7 and ... then what? Please be clear and take it slow. I see that the 7's go to 8 = 2^3 and done. So you want to add them? Why? What is the point of all this? 

10 minutes ago, StringJunky said:

I think the omission is an Invision software thing, not to do with  the site owners.

I'd like to say up front that I'm mindful that I'm a guest here, and there's a Feedback section, so my comments regarding the site are off-topic in this forum, and a breach of etiquette. I've just got a bug up my butt about this because it's the second time in two weeks that someone asked a question I'm interested in, and I wanted to write out some mathematical exposition, and had to resort to posting screenshots. Having said my piece I'm going to stop talking about it here.

However, just to reply to what you said ... I was not around when the software change happened. But I'm perfectly well aware that the owner of a site chooses the software. So if the owners went from brand X forum software to brand Y, that's a choice. And if brand Y is a significant downgrade in capability, that's on the owners. Someone please straighten me out if I've got this wrong and there are circumstances that let the owners off the hook in this instance.

Thing is, implementing MathJax amounts to pasting a short block of Javascript into the HTML headers. So I am concluding that the site owners don't have access to the HTML headers to add CSS or Javascript. It's hard for me to understand why anyone would install software like that but I'm not privy to the details. 

So as I say I'll just put a sock in it regarding this topic from now on. 

29 minutes ago, Zolar V said:

maybe the definition you're trying to get at looks like this

 

 

 

I didn't look at what you wrote yet, I've been working on my version. Just tell me if I've got this right. 

I must say that now TWO of us are screenshotting LaTeX and I think this is a very inefficient way to work. The problem is we can't quote each other's markup so everything has to be done from scratch. I'm not going to do this much more if at all. I hate giving the site owners the satisfaction of knowing they can delete LaTeX from the site and still have people discuss math. I'm just going to stick to ASCII from now on. This is dumb.

 

1 minute ago, Zolar V said:

I do recall the notation; unfortunately, I only briefly used the notation in a few of my classes.  So I am familiar but probably not much past a familiarity.

"I note that your paper is marked up nicely, which is good." 
- Thank you,  i wrote it over the course of a few weeks as I learned how to use LaTex.

I posted it here because I re-found the forum and remembered that this was a great place to have a good discussion.  I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor. I also don't really know where I would have posted this anyways.   I did show my paper to a few math major friends of mine and we did have some discussion about it.  They didn't seem to have the problems we are having here,  so I wonder if I did some explanation to them verbally that I just entirely missed.

"Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?" 
- yes, every prime will eventually reduce to a power of 2.

I am going to try to really define our function, or functions here so that it makes sense.

 

>  I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor.

Yes they broke it. Also made it so you can't intersperse responses with quotes. No idea what the powers that be are thinking. Reason this is on my mind is that I'm formalizing your G process (not really a function, more of an algorithm) and the most expressive way to do that is with LaTeX. As before I'll have to write my response offline and post a screenshot. And nobody else can quote my post to leverage my markup. It's extremely annoying. 

Anyway I just want to make sure I've got this. You input a prime and add 1. If it's a power of 2 you're done. Otherwise divide out the highest power of 2 from p+1 and factor it into a product of primes 2 < p_i < p, each prime written as many times as needed. Then you recursively apply G to each of the primes and eventually everything drills down to a power of 2. I have no idea what that proves or why it's important, but we'll get to that later. Do I have the basic algorithm right?

> They didn't seem to have the problems we are having here

I never realized it at the time, but a lot of people get degrees in math without learning any math. 

 

 

6 hours ago, Zolar V said:

That is the purpose of wanting to work with a real mathematician.

This part I don't understand. No professional mathematician is going to read a paper claiming to have solved a famous open problem in which the author doesn't know basic function notation and can't write a proof. It's like applying for a job as a piano player and admitting that you can't read music or play the piano, but you have a lot of great tunes in your head. They'll tell you to come back when you can play. "I don't want to practice my scales, I just want to bang the groupies." 

Anyway, this is what I'm getting about G. We have G(3) = 4, which is a power of 2 so we're done. G(11) = 12, and after factoring out the highest power of 2 we're left with 3, and G(3) = 4. So now what do you do with G(29)? G(29) = 30, factor out the highest power of 2 leaves 3 x 5. Then what? Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?

> G^n (p) = 2^m

So what you mean is this:

 

Do you understand the notation? It says that for all p, there exist natural numbers n and m such that G^n(p) = 2^m. That's how you express that. Of course you still haven't defined G, but at least this expression tells us what n and m are. They're bound to the existential quantifier. You're not saying this is true for all n and m, but rather that it's true for SOME n and m. 

I note that your paper is marked up nicely, which is good. Why did you choose to post your proof on a site that doesn't allow math markup? It makes mathematical conversation difficult. One can copy/paste most of the necessary symbols, but they're hard to read and they don't look good. I had the same problem here a couple of weeks ago. 

 

Ok this is way too long but what the hell. Hope you find a nugget of wisdom in here. If you reply, no point replying paragraph-by-paragraph since that will make it even longer. Bottom line let's figure out how to define G. Your exposition is seriously stuck there. 

And at my end I'll make another run at your paper now that you've explained the prime products. 

 

> Great critique!  I didn't design the proof with any sort of programing language in mind.  However, I typically think in that sort of nested fashion. 

I have to say I was charmed by the idea. Almost like indenting a block of code. 

Theorem 1: blah blah blah
    Lemma 1.1: such and so
    Proof: blah blah
Proof: blah blah blah blah.

It really makes perfect sense. Unless you want to use the same lemma more than once I guess. For what it's worth, math isn't written this way so it's better to do things the standard way. 

By the way as I go I'm making stylistic remarks. One of the problems you're having is that you don't yet know how to write a clear mathematical argument. I'm doing my best to engage with your ideas and mentally filling in the blanks in your notation. But at some point the notation gets so garbled that it expresses no idea at all. You have something in your head, but I assure you that you have NOT expressed the idea in your paper. 

Reading ahead, at the end of your post you write: "I don't believe your questions yet disprove that this isn't the solution to the collatz conjecture."

I quite agree. It's perfectly possible that you have a valid proof in your head. I can only tell you that you have not committed one to paper. 

And that more importantly, it's not that your ideas are clearly wrong. It's that they are so unintelligible as to be, in the famous words of Wolfgang Pauli, "Not even wrong!"

https://en.wikipedia.org/wiki/Not_even_wrong

I hope you will take these comments in the spirit of sharp but constructive criticism. You need to make an effort to write a much clearer and cleaner mathematical exposition. 

Not on my account, I'm just some guy on the Internet. But your paper says it's been sent to the Journal of Number Theory. How are they expected to react when they skim down the page and see that (P_i)^(n_i) is "prime by definition?" Your case is lost at that moment. 

So for your own interest in promoting your proof to the world, I urge you to rewrite your paper. Define every term and variable, use consistent notation throughout, and pretend you're some hapless journal editor who has a pile of papers from established mathematicans, and another pile from people he's never heard and who don't have Ph.D.'s in math. The papers in that latter file will get very short shrift. Bad exposition will get you roundfiled very quickly. 

So. Onward. Looking forward, I did understand your clarififaction of the prime products, and I'll make another run at your paper with that understanding. But at some point I'm going to run out of steam. I really think you need to go to version two on this paper. What you have isn't enough to express an actual mathematical idea, no matter what is in your head. Which as far as anyone knows, is a valid proof of Collatz that will make you famous tomorrow morning. I can't disprove the possibility; for the simple reason that your argument as written is incomprehensible. 

> " On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof.  " 
-Though the result is obvious and trivial, I think it is worth mentioning since it is the underlying concept to why adding 1 to a prime causes the resulting primes contained in the composite number to be less.  

You're pointing out that if p is a prime, then p+1 is divisible only by primes strictly smaller than p. And that if you factor out all the 2's, then every remaining prime factor of p+1 is greater than 2.

This is perfectly true. It's so trivial as to not require proof. The fact that you wrote it out as a lemma and then laboriously proved it is a credibility point against you. Someone reads that and thinks, "This person believes this needs proof. There is no way they proved Collatz." 

Just leave it out. 

> " On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what. " 
- This is the proof of the G(p) = p +1,  but I believe you figured that out.

I truly thank you for reading my post in its entirety instead of snapping back line by line. I wish everyone would do that everywhere. I'm often guilty myself.

> " Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general.  " 
-I think a note here that explains why we can still apply the function to a composite number would work.    The reason why we can iterate G(p) on any integer is because every integer is either prime or a composite of primes.  The latter needs a little bit of work, but we can extract a prime out such that we have a number + a prime.  Like stated above, E.G. 7*3 = (3-1)*7+ 7 

Every time I ask you what G is, you respond with a convoluted paragraph and you never define G. You still have not defined G. 

Regarding notation, I think I have the right to expect you to know and follow the standard notation for functions taught at the high school level everywhere in the world. 

OUTPUT = f(INPUT).

If you say f inputs a prime, and p is prime, then f is not defined for p+1. Like I say I'm willing to cut you a lot of notational slack in order to understand your idea. But on this we really have a problem. G(p) = p + 1 has the same meaning for every math student and physical scientist in the world. G(7) = 8, G(8) = 9. 

Someone earlier mentioned that you may be in need of the concept of a recurrence relation. This is a way of defining a function of the sort you have. For example here is how we can define the factorial function:

f(0) = 1

f(n) = n x f(n-1)

You can see that if you plug in 1 for n, then 2, and so forth, you always come down to the "base of the recursion" of n = 0, and the computation terminates. The idea is intimately related to how loops and recursion work in programming languages, and also the principle of mathematical induction in number theory.

You might have a go at defining G in this manner. 

The paragraph you wrote is simply a massive red flag for anyone honestly trying to read your paper. You say G is defined on primes, and then you define G(Gp)) by waving your hands at the prime factors of p+1, but you never actually tell us how G is defined. I really hope you will take this to heart because you simply have not defined G and until you do, your paper can never get anywhere. 

> " You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define. "
-G(p) = p+1 is what I called the prime reducing function.

I truly do get that you call G the prime reducting function. In fact I've figured that much out. What I don't know is how G is defined. And in particular, why you repeatedly say G inputs only primes, then immediately claim to compute G(G(p)). That's a logical contradiction. Another HALT instruction for the poor reader of your paper. 

> "I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n. "  
-I suppose a better way to write that would be G^n (p) = 2^m  , since after n iterations you get a 2^m number.  I mean you're really only adding 1 a number of times until you reach a 2^n.  its rather trivial if you think about it like that.

G^n (p) = 2^m? Well the right side is not only independed of n, it's now indepent of p as well. So it's a constant function. G^3(47) = 2^m, and G^154(p) = 2^m. G^anything(p) = 2^m. And you didn't say what m is!!

I simply have to draw a line. You either know how to use function notation, or else we need to talk about that before we talk about anything else. I can cut you a lot of slack on notation but not on this. What you wrote is meaningless and demonstrates a lack of familiarity with high school math past algebra I. That is not a good look for someone claiming to have solved a famous open problem.

Again I hope I'm not piling on. I'm being emphatic in the hopes of being helpful. Your notation is killing your argument because it makes no sense. 

> "On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case. " 
-Fair enough,  but if p+1 happens to be a power of 2 then the problem is done.  We're only trying to create a string of 2's, so then we can divide by 2 however many times to get 1.  We only do this because that is part of the collatz cojecture.

Yes perfectly well agreed, once we get to a power of 2 we're done. Yet what you originally wrote was wrong as stated. You need to explicitly call out this case so that we know you're paying attention to your own work!

> " On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k. "  
-You are both correct and wrong.  I shouldn't have written the notation with arbitrary powers of P_k,  I should have written it as a fully expanded string of powers.  3^3 should be 3*3*3...   We are attempting to handle individual primes, one at a time.  Not some grouping of primes in a shorthand notation.

" You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right? "
- I believe the prior answer to your prior question also solves this question.   by definition of the fundamental theorem of arithmetic each P_k is prime.

Ok I accept your explanation. It's on me now to reinterpret your paper with this new understanding of what you are trying to say. 

On your part, have you considered rewriting the paper from scratch in the light of some of these comments from me and others?

> " ppps -- No, you don't have an argument left. On line 32 you apply the "prime reducing function" to a list of POWERS of primes, VERY FEW OF WHICH are prime in the general case. Your proof is busted right there. "  
- I believe rewriting the composite number as a series of primes with a power of 1 solves your question here.  Very similar to the above two questions.

Yes I agree. However, that's why I originally mentioned the following point ...

> " I'll also mention that line 33 doesn't follow from anything that came before EVEN IF we allow that prime powers are prime (which of course they are not)! And what is the 'x' on the right side of line 33?  " 
- Wrong notation here.  In my mind "x" was the first prime we punched into the problem.  so it should have been written as "p" not "x"  
Line 33 is supposed to show that for each iteration of G(p), the resulting primes are less than p and also less than the last iteration of G(p). 

What I saw was that you established your facts about the non-2 prime factors of p+1, then waved your hands that since each factor is prime you can recurse your way down to your conclusion. But I did not see a proof, just a vague intution that there's a recursion to be had. Without a clear definintion of G, probably a recursive one, you won't have a proof.

That's why I mentioned that. I anticipated that even if I ALLOWED that p^n is a prime (!!!!) your argument STILL doesn't work. You glossed over every bit of detail you'd need to make your case.

> E.G  for the first iteration, let our p = 5 . 
G(5) = 6 
rewrite 6 as 2*3, then rewrite it as (2-1)*3+3.  (again the purpose of the rewrite is to show that we can get a prime number to apply G(p) to.
So G(5) = 3 + 3
Then the next iteration is G(3) = 4 .
so G(5) = G(3) + G(3)   (there are two iterations here, since there are two 3's we're reducing)
So after 3 iterations we reach a 2^n number and we are done with G(p).
So   2 < 3 <5

I still don't get it. Let's nail down what G is, maybe we can figure out the proper recursive definition needed to make the rest of this work. 

"... we can get a prime number to apply G(p) to." Oh my. G(p) is the value of the function G at the input value p. It's not some other function. Or is G(p) some other function? That's legal too, functions can output other functions, for example the differntiation operator in calculus. High school function notation, please! 

I don't mean to be impolite but I need to ask this. Are you misusing function notation because you don't know it? Or because you aren't familiar with the recursive definition of G you'd need to give and are hoping we'll understand what you mean? If the latter, let's fix that. If the former, let's talk about function notation before we do anything else.

> Note:  I don't believe your questions yet disprove that this isn't the solution to the collatz conjecture.  

I have certainly proved that your exposition doesn't prove Collatz, since it's not a mathematical argument at all. On the other hand, I don't know what's in your head, and can't definitively say you haven't proved Collatz. If the latter, once you rewrite your paper and become world famous, maybe you'll give me credit for helping you fix your notation!

> They do help enormously to clarify the logical jumps I made.

Good to know. Now I don't feel so bad being so critical. I'm curious to understand your idea and your exposition is in the way so I'm trying to help you clarify the exposition.

> In a simple sense, adding 1 to a number an arbitrary number of times such that you get a 2^m number is exceptionally trivial and intuitive.

(Off topic) It's nothing of the sort. It's very deep. Suppose I give you the Peano axioms that say whenever you have p you can make p+1. I define addition as repeated "plus-one"-ing, multiplication as repeated addition, and exponentiation as repeted multiplication. I give you X = 2^2^2^2^2^2^2^2 where the operations bind right to left, so that this is a very very big number. If all you have is the plus-one operation, how do you know when you're reached X? It's a deep question. 

>   of course if I add 1 enough times I will encounter an integer that is a power of 2.  But this result is why the collatz conjecture itself converges to 1.

Well that's a handwavy argument that doesn't convince me. 
 

4 hours ago, NortonH said:

No. I'm stuck. I missed something else and now i think i am in a state beyond repair.

 

Congratulations on being able to recognize that. It's the beginning of learning.

ppps -- No, you don't have an argument left. On line 32 you apply the "prime reducing function" to a list of POWERS of primes, VERY FEW OF WHICH are prime in the general case. Your proof is busted right there.

I'll also mention that line 33 doesn't follow from anything that came before EVEN IF we allow that prime powers are prime (which of course they are not)! And what is the 'x' on the right side of line 33? 

So that's the error. Your Lemma 1 shows that any P_k < P (once you've handled the case where p + 1 is a power of 2 so that there are no P_k's); but the rest of your proof requires that all POWERS of p_k are less than P, and prime to boot. Both those assertions are false.

1 hour ago, Zolar V said:

 

On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof. 

On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what.

Can you please supply the missing claim you're trying to prove?

Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general. 

You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define.

Out of curiosity, are you aware that the function notation G(p)  = p + 1 means exactly what it says, and NOT some convoluted function you have in your mind but have never clearly described?

After all, @studiot was charitable enough to suggest that you are accustomed to the use of '=' as an assignment operator in some programming languages. That's a sensible interpretation in general, but it hardly explains how the notation G(p) = p + 1 is being used to denote some function G whose definition is not possible to discern from your exposition.

I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n.

 

ps -- Oh I see what the second "proof" is. It's the proof of the OUTER assertion. You nested the lemma inside the main thing you're trying to proof. Not a bad idea actually but really not how it's done. A little like an inner class in Java, is that where you got the idea? Way too confusing for readers. Next time do Lemma 1, Proof of Lemma 1. Main claim, Proof of main claim. Your readers will thank you. Take that as constructive criticism. The main issue is to untangle the definition of G.

 

pps -- Ok more. I'm slogging through your exposition trying to understand what G is, and you have a little problem on line 29 and a BIG BIG problem on line 30.

On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case.

On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k.

You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right?

You need to fix these errors and see if you still have an argument left.

3 minutes ago, Zolar V said:

That would be all from the first equation.  the next part is to take the even equation from the collatz conjecture and apply it however many times it takes to reduce each 2 into a 1.
The purpose of G(p) = p + 1 is to reduce each prime to 2. 

Please continue working one of your examples to completion. You had G(7) = 8, and 8 = 2^4. Ok what now? I hope there's more to this than noting that 2 is always a factor of 1 plus an odd prime.

21 minutes ago, Zolar V said:

7  ->  7+1 = 8 = 2*2*2       |  2,2,2 are  < 7
 

> 7  ->  7+1 = 8 = 2*2*2       |  2,2,2 are  < 7

Perfectly clear and I appreciate specific examples. But what happens next after breaking down 7 + 1 into 2^4?

18 minutes ago, Zolar V said:

I edited it  for  -> instead of equals.

to me "->" means a sort of action.  G(p) = p + 1 is the function and  G^n(p) = 2^n is the result of n iterations of the function. 
I tend to think in a very step oriented logical (if then) fashion. unfortunately I dont express myself very well.
 

Yes but if G(p) = p + 1 is a function, then G(5) = 6 and G(11111) = 11112, and G^n(p) = p + n + 1, right? Not 2^n. Using the convention that G^0 = G.

25 minutes ago, Zolar V said:

Or differently stated:

Consider a function f(x) = x + 1  where x is prime.  then  x+1 is a composite of primes where each individual prime is less than x.  
Suppose you applied f(x) to that resulting number for some m iterations.  Then each resulting prime eventually tends to 2.  That is to say the first iteration of f(x) produces a number whos primes are less than x,  then each subsequent iteration produces a number who's primes are less than the first prime x and each subsequent prime that is plugged into the function.  such that after a number of iterations the composite number is 2^m.

of course after dividing 2^m by the even function m times that results in 1. 

> Consider a function f(x) = x + 1  where x is prime.  then  x+1 is a composite of primes where each individual prime is less than x.  
Suppose you applied f(x) to that resulting number for some m iterations.

But if x is prime then x + 1 is NOT in general prime, hence G(x+1) is NOT DEFINED. So you can't iterate it. Someone else already pointed this out to you.

15 minutes ago, studiot said:

Isn't this 'equals' a computing assignment statement?

 

Perfectly sensible explanation, but not in a math paper. Equals means equals in math. But now that you mention it, yes that's a good interpretation ... except that he wrote "G(p) = p + 1   =  G^n(p) =  2^n". How should I understand that?

16 minutes ago, Zolar V said:

"That's utter nonsense. If you claim a proof of an open problem you can't make up all your own terminology."  - precisely why I need to work with a true mathematician.

lol

 

> G(p) = p + 1   =  G^n(p) =  2^n

You seem to be deeply confused about the equal sign. 

8 minutes ago, Zolar V said:

yes typo.  It should read  3*2*11 +11  .   The point of the rewrite is to show that any integer can be rewritten such that it is a number + a prime.   Of course the whole purpose is that when you have a number + a prime you can then use the prime reduction theorem upon the (+ prime) part.

 

Lines 17 and 17 are nonsense. Perhaps it's just worded awkwardly but your nonstandard terminology makes your paper difficult to read. You say that the function G(p) = p + 1 "converges to 2." That's utter nonsense. If you claim a proof of an open problem you can't make up all your own terminology. Convergence has a very specific meaning in math. G(p) = p + 1 means that G(3) = 4, G(47) = 48, etc. In what way does G(p) "converge to 2?"

I certainly believe that G(1) = 2 converges to 2. Past that, you're just making up your own funny meanings for standard mathematical terms. And this leads me to believe that your more complicated expressions contain similar errors and obfuscations. 

10 minutes ago, Zolar V said:

I meant to say that using the prime reduction theorem, the odd collatz function can be rewritten such that it is a 2^m number and thus divided by the even function m times results in 1.

What do you mean limiting?  I didn't limit the collatz conjecture by any means.

 

Oh I see you're using the word "convergence" in a funny way. Regardless, 33 is not equal to 99 even though you claim it is on line 41. Is that a typo? 

10 minutes ago, Zolar V said:

Well here, let me just post my pdf.   Im quite serious about my proof.

Draft2.pdf

"The problem itself converges to 1?" What on earth can that mean? in any event as I pointed out in the OTHER Collatz thread, no limiting argument works. 

ps -- Line 41 is wrong. 99 does not equal 33.