Zolar V 70 Posted August 18 Is anyone here interested in the Collatz Conjecture? If so I believe I have the solution, seriously, and I need to work with someone who knows how to write a formal proof better than I. There are some errors in my proof, but the underlying principle is right. I only have my minors in mathematics, so my proof writing skills are subpar. 1 Share this post Link to post Share on other sites

Achilles 30 Posted August 18 Yes I am a layman just hand it over as a pdf file 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 I feel like I've been here before ... 1 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 (edited) Well here, let me just post my pdf. Im quite serious about my proof. Rip it to pieces and tell me im an idiot please. Draft2.pdf Edited August 18 by Zolar V 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 (edited) 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. Edited August 18 by wtf 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 (edited) 12 minutes ago, wtf said: "The problem itself converges to 1?" What on earth can that mean? in any event as a pointed out in the OTHER Collatz thread, no limiting argument works. 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. "ps -- Line 41 is wrong. 99 does not equal 33. " correct, the line should read 3 * 2 *11 + 11 Edited August 18 by Zolar V 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 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? 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 (edited) 3 minutes ago, wtf said: 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? 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. "Oh I see you're using the word "convergence" in a funny way." - yes, my abilities to write proofs are limited and inexperienced. Edited August 18 by Zolar V 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 (edited) 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. Edited August 18 by wtf 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 (edited) 50 minutes ago, wtf said: 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?" "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. G(p) = p + 1 -> G^n(p) = 2^n G^n(p) / 2^n = 1 What I was trying to say was that after n iterations, G(p) = 2+ 2+ 2.... = 2^m AND G(p) / 2^m = 1 50 minutes ago, wtf said: 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. Here is a simple online calculator that shows G^m (p) = 1 else multiply by 3 <- replace 3 with 1. https://www.dcode.fr/collatz-conjecture Edited August 18 by Zolar V 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 (edited) 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. Edited August 18 by wtf 1 Share this post Link to post Share on other sites

studiot 1478 Posted August 18 7 minutes ago, wtf said: lol > G(p) = p + 1 = G^n(p) = 2^n You seem to be deeply confused about the equal sign. Isn't this 'equals' a computing assignment statement? ZV did say he worked in some sort of IT environment. 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 I would like you guys to take a serious look at the mathematics, instead of attacking the semantics. Is it really that difficult for you to comprehend what the paper is trying to state? -1 Share this post Link to post Share on other sites

taeto 37 Posted August 18 If you really want to write a text on a mathematical topic, you should try to become familiar with concepts, read a fair amount of well written expositions, and then practice a lot on formulating your own thoughts so that they become accessible to someone else when you put them into writing. I actually believe that you did read some of Lagarias's papers and enjoyed it, and you would want to emulate the style of exposition if you could. As wtf also pointed out already, your draft here is of very limited expressive power. Could you try yourself to look at line 6 together with the initial bit of line 7, and think about what a reader will see there. There is an "n", an "f", an "x" and an "m", none of which is explained as being connected to anything else. Yes, "n" is explained as being a positive integer, which is good, but n is not relevant to anything that follows. The name "f" in mathematics is quite different from the name "F". We do not know what x is. And the conjecture does not fix any particular number m to be the number of iterations before reaching the end. In particular the conjecture itself is badly misstated. Since you have read Lagarias's treatments, then why not stay on the safe road and simply state the conjecture in the same way as he does? Apart from your use of "converge", which I take to mostly mean "reach in a number of iterations", the lines 17-18 make no sense, because you cannot have a function G defined only on primes so that you can iterate and compute from a prime p the sequence p,G(p),G(G(p)),G(G(G(p))),... unless G(p) is also always a prime. If p is a prime, then you can compute G(p), that is fine, but if G(p) is not a prime then there is no G(G(p)). In particular, the function G for which G(x)=x+1 does not cut it, since G(p)=p+1 is only a prime in a very limited circumstance, that is, when p is equal to 2. Maybe for now you should just be content with explaining in a simple language, and using no mathematical terminology which you are not sure that you understand completely, how you intend to go about your plan for a proof. 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 2 minutes ago, taeto said: If you really want to write a text on a mathematical topic, you should try to become familiar with concepts, read a fair amount of well written expositions, and then practice a lot on formulating your own thoughts so that they become accessible to someone else when you put them into writing. I actually believe that you did read some of Lagarias's papers and enjoyed it, and you would want to emulate the style of exposition if you could. As wtf also pointed out already, your draft here is of very limited expressive power. Could you try yourself to look at line 6 together with the initial bit of line 7, and think about what a reader will see there. There is an "n", an "f", an "x" and an "m", none of which is explained as being connected to anything else. Yes, "n" is explained as being a positive integer, which is good, but n is not relevant to anything that follows. The name "f" in mathematics is quite different from the name "F". We do not know what x is. And the conjecture does not fix any particular number m to be the number of iterations before reaching the end. In particular the conjecture itself is badly misstated. Since you have read Lagarias's treatments, then why not stay on the safe road and simply state the conjecture in the same way as he does? Apart from your use of "converge", which I take to mostly mean "reach in a number of iterations", the lines 17-18 make no sense, because you cannot have a function G defined only on primes so that you can iterate and compute from a prime p the sequence p,G(p),G(G(p)),G(G(G(p))),... unless G(p) is also always a prime. If p is a prime, then you can compute G(p), that is fine, but if G(p) is not a prime then there is no G(G(p)). In particular, the function G for which G(x)=x+1 does not cut it, since G(p)=p+1 is only a prime in a very limited circumstance, that is, when p is equal to 2. Maybe for now you should just be content with explaining in a simple language, and using no mathematical terminology which you are not sure that you understand completely, how you intend to go about your plan for a proof. Thank you for your insight: Here is the prose version:Since the most important function within the collatz conjecture is F(x) = 3x + 1 , I really tried to understand what it was this function did to the problem. Two facts are present, any number is either prime or a composite of primes. Therefore I broke it down and looked at a different equation: G(y) = K + 1. For y prime. Notice the composite number K +1 is less than 2K, therefore all of the primes within the number are also less than K. Now since K + 1 is a composite number it can be rewritten as a multiplicative sets of primes. Moreso, you can reduce one set by 1 and extract a prime. E.G. 9 +1 = 10 = 2*5 → (2-1)5 + 5 Therefore you can extract a prime (E.G: 5) and plug it back into G(y). For each iteration you reduce the resulting composite number and after some M iteration you can reduce it to 2. Now given the original collatz function F(x) = 3x + 1 , you can rewrite this as F(x) = 2x + x+1 . Notice that we have an x+1 and that we can also rewrite any composite number in such a way as to extract a single prime to work with. Ergo F(x) = 2x + G(y). Since G(y) = K +1 for K prime and converges to 2 after M iterations. Then F(x) = 2x + G(y) → F(x) = 2 (x + 1) and again (x +1) is our G(y) so F(x) = 2(G(y)) → F(x) = 2*2 for some M,N iterations. Then of course you can apply the even collatz function and it converges naturally to 1. Side note: Any odd function of the form ax+1 also converges to 1 following the exact same logic, and the collatz odd function is just a case where a = 3. I also hope it’s well enough written that you can see my thoughts on it. 0 Share this post Link to post Share on other sites

studiot 1478 Posted August 18 6 minutes ago, Zolar V said: I would like you guys to take a serious look at the mathematics, instead of attacking the semantics. Is it really that difficult for you to comprehend what the paper is trying to state? If you would be so kind as to answer my question (as required by the rules of this forum) instead of being rude, help might be more forthcoming. 0 Share this post Link to post Share on other sites

taeto 37 Posted August 18 6 minutes ago, Zolar V said: I would like you guys to take a serious look at the mathematics, instead of attacking the semantics. Is it really that difficult for you to comprehend what the paper is trying to state? Well in that case the thing is just not worth dealing with: what the paper states just does not make sense mathematically. Sorry to have bothered you. 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 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. 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 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? 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 2 minutes ago, wtf said: 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? 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. 0 Share this post Link to post Share on other sites

studiot 1478 Posted August 18 5 minutes ago, wtf said: 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? Coupled with the reference to 'limits' ZV may be trying to reinvent recurrence relations. I can't tell, he won't discuss it. 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 (edited) 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. Edited August 18 by wtf 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 Just now, wtf said: 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. You need to convert a non prime number into a composite of primes then extract a prime from that composite. p+1 is a composite number. p+1 = a*b*c*d*e.. where each of those are primes. What you need to do is then take (a*b*...*(d -1) ) *e + e E.G: http://www.wolframalpha.com/input/?i=(5*3*7)+%3D+(5*3+-1)*7+%2B+7 once you do that you can then utilize the function to reduce each prime E.G: G(5)= 5 +1 = 6 6 = 2*3 = (2-1)*3 +3 = 3 + 3 so 3+ 3 -> G(3+3) = G(3) + G(3) 10 minutes ago, wtf said: 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. > 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. yes, x+1 is a composite number. you can then follow the rules of math to give you a expression that has a prime that you can use. any composite number can be rewritten as a product of primes. as such you can rewrite that expression of primes to be an even product of primes multiplied by a certain prime and added to that prime. E.G 486485 + 1 = 2*3*3*3*3*3*7*11*13 = (2*3^5*7*11-1)*13 + 13 Since you have a composite part and a prime part you can then apply the function on the prime part. Maybe you can see more of what I am trying to explain by taking a look at a few primes. G(p) = p + 1 where each prime in the composite number p+1 is less than p 7 -> 7+1 = 8 = 2*2*2 | 2,2,2 are < 7 11 -> 11+1 = 12 = 2*2*3 | 2,2,3 are <11 13 -> 13+1 = 14 = 2*7 | 2,7 are <13 17 -> 17+1 = 18 = 2*3*3 | 2,3,3 are < 17 19 -> 19+1 = 20 = 2*2*5 | 2,2,5 are <19 23- > 23+1 = 24 = 2*2*2*3 | 2,2,2,3 are < 23 ... and so on. 0 Share this post Link to post Share on other sites

wtf 99 Posted August 18 (edited) 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? Edited August 18 by wtf 0 Share this post Link to post Share on other sites

Zolar V 70 Posted August 18 1 minute ago, wtf said: > 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? 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. 0 Share this post Link to post Share on other sites