On Collatz Problem

Collatz problem can be found here: http://mathworld.wolfram.com/CollatzProblem.html

Please look at the attached paper:

http://www.geocities.com/complementa...y/3n1proof.pdf

Thank you,

Organic

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 Recognitions: Homework Help Science Advisor Comments1. k must be fixed one presumes or many things make no sense as we will see. 2. The second line, starting 'A direct convergence', has an extraneous comma that makes it unclear what you mean 3. the 4th paragraph implies that k is not fixed, since if it were 2^{k+1} would contradict that statement. 4.of the 4 options you give one must conclude that k is fixed otherwise the assertion j is an even number >2^k is nonsense. It is bad maths anyway, as one should not use symbols when words are required. 5. similarly don't use XOR like that, neither of the 'inputs' is a statement than can be true or false. are those contradictions about k being fixed or not enough for you? probably not. edit: there are also these: what does it mean 'to be out of the range'? why are you misusing decidability like this? godel states that something is undecidable if both it and its negation are consistent with the other axioms. i don't even want to touch the von neumann heirarchy stuff.
 Recognitions: Homework Help Science Advisor thinking about this for a while it just seems that you're saying: because of a couple of iterations of the rules of Collatz we see it can never be decided what happens. but that is obviously wrong.

On Collatz Problem

Hi Matt,

 k must be fixed one presumes or many things make no sense as we will see.
k is any positive integer ( 0 included).
 2. The second line, starting 'A direct convergence', has an extraneous comma that makes it unclear what you mean
Thank you, you right (my English problems again), A direct convergence to 1 must start in one of the even numbers that belong to 2^k sequence.
 3. the 4th paragraph implies that k is not fixed, since if it were 2^{k+1} would contradict that statement.
As i wrote, k is any positive integer ( 0 included).
 4.of the 4 options you give one must conclude that k is fixed otherwise the assertion j is an even number >2^k is nonsense. It is bad maths anyway, as one should not use symbols when words are required.
Please explain what do you mean by "k is fixed"?, again k is any positive integer ( 0 included).
 5. similarly don't use XOR like that, neither of the 'inputs' is a statement than can be true or false.
If i write "or" instead of "XOR" is it ok?
 what does it mean 'to be out of the range'?
If you look at the Binary Tree that stands in the base of Von Neumann Hierarchy, then you can see that this structure is the invariant symmetry of N members, therefore can be used to define rigorous proofs about N members.

Collatz problem is also based on this invariant symmetry of the Binary Tree, and for any k we can find N members that are out of the range of 2^k (case 4 in page 1).

Therefore it is non-decidable for any Binary Tree, which means non-decidable within N (for any n).

 Recognitions: Homework Help Science Advisor I don't think you understand the subtlety of the word fixed. As you compare j to 2^k it cannot be that when you say 'examine 2^k' that you mean 'examine the set {2^k | k in N u{0}}' you are confusing sets and numbers again and the comparison j>2^k is meaningless without explanation. I really don't see what you are getting at, to be honest, at least beyond stating in a very confused fashion the obvious about what might happen when you apply the Collatz iteration. Anyway, that can be readily rectified, however you don't explain what you mean by 'out of the range' no matter what you may think. A simple statement will do, you know. Nor do you explain how your facile observations imply anything to do with decidability, which as I pointed out is quite an intricate issue - the continuum hypothesis is undecidable in ZFC for instance. or is better than xor, but you shouldnt write 'either >x or
 Dear Matt, Thank you very much for your help. I fixed some of the problems. Please read it again: http://www.geocities.com/complementa...y/3n1proof.pdf and please tell me more about "fixed" if needed. Please be aware that i am using here a non-standard point of view on N members, which based on its fundamental symmetry of the Binary Tree. Yours, Organic
 Recognitions: Homework Help Science Advisor changing the word range to the word scope doesn't make it any clearer. just state what you mean to be out of the scope of, or out of the range of
 Well, I used scope because "range" is already used by Math. By out of the scope I mean that we shall always find n's which are beyond any Binary Tree. Because this symmetry stands in the basis of N members, we can conclude that Collatz Problem is non-decidable for any n in N, because any n is nothing but a part of the Binary Tree invariant symmetry. I'm going to sleep, see you tomorrow.
 Recognitions: Homework Help Science Advisor so you assertion is that given any binary tree there is a number 'beyond' it. Care to explain what that means at any point? anyway, you've not proved that at no point in the future iteration of 47 does it not return to one. I'll put money on it doing so. Putting it in your language, you've omitted to consider things in the correct order, which is that given an n there is some k with n in the scope of the binary tree 2^k, and in fact in the scope of infinitely many of them. You're taking things in the wrong order again. Of course if you could prove that there is some n, such that for any k, there is some point in the Collatz iteration with i_r(n)>2^k, where i_r stands for the r'th iterate in the sequence, but that isn't what you've shown. The assertion that given any k there is an even integer, n, not a power of 2, with n>2^k does not prove anything interesting here. Let's repeat the main point. There is no number beyond the scope of any binary tree. Given any binary tree there is a number beyond its scope, but that is different for each tree. There is no number that is beyond the scope of EVERY tree. Not that I see what that would prove anyway.
 Hi Matt, I made major update in my proof, please tell me what do you think: http://www.geocities.com/complementa...y/3n1proof.pdf Thank you, Organic
 Recognitions: Homework Help Science Advisor The assertion that, given a k, 3*(2^k+1)+1 is not in *any* binary tree is wrong. It is in an infinite number of them. It is in the one corresponding to 2^r where r is any number greater than 3*(2^k+1)+1. When you say the first number after 2^k you are fixing k at some (arbitrary) value. because, once you've fixed this k, you can find a number iterated to some larger number is not important because it is dependent on k, only if it were independent would you even start to have something, but even then that is moot as you'd have to prove that no iteration at any point in the future would return you to 1 example k=3, so we start with 9, the iterations are 9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1 It still wouldn't state it was undecidable: a statement is undecidable in some axiomatic theory if either the assumption of it or its negation is consistent with the axioms, example, the continuum hypothesis and ZFC.
 Hi Matt, But what I show is based on the invariant symmetry that stands in the base of any n in N. (2^k)+1 has a 1-1 and onto with {} that exists in the internal structure (The Von Neumann Hierarchy Binary Tree) of any n in N. Let current n be 2^(k+1). Any 3*((2^k)+1)+1(=3n+1) > 2^(k+1) where 3n+1 is beyond the internal Binary Tree structure of current n(=2^(k+1)) in N, which means that we always need to get n+1 for general and rigorous proof. Shortly speaking, if n then n+1 (the ZF axiom of infinity), therefore it is non-decidable for any n in N. Q.E.D Please see my updates: http://www.geocities.com/complementa...y/3n1proof.pdf
 This self similarity of N members, which constructed by The Von Neumann Hierarchy Binary Tree can't be ignored. Maybe someone knows what is the Mathematical brunch that researches the symmetry of Mathematical objects like numbers?

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Homework Help
 Originally posted by Organic Hi Matt, But what I show is based on the invariant symmetry that stands in the base of any n in N. (2^k)+1 has a 1-1 and onto with {} that exists in the internal structure (The Von Neumann Hierarchy Binary Tree) of any n in N. Let current n be 2^(k+1). Any 3*((2^k)+1)+1(=3n+1) > 2^(k+1) where 3n+1 is beyond the internal Binary Tree structure of current n(=2^(k+1)) in N, which means that we always need to get n+1 for general and rigorous proof. Shortly speaking, if n then n+1 (the ZF axiom of infinity), therefore it is non-decidable for any n in N. Q.E.D Please see my updates: http://www.geocities.com/complementa...y/3n1proof.pdf

Look, the number you pick for each k that is out of the 'scope' of the tree you are looking at is different for each tree, there is not universal number in those that you pick. Got it? Evidently not. You fix the k, then you say take 3*(2^k+1)+1, note that is different for each k and there is not one of them you can pick that works for all k.

For pity's sake don't start using the axoim of infinity again, as this isn't what it states. It merely states the the set of integers must be a set in the model you are using.

You are evidently using your own private defintion of undecidable which is not the mathematical one (again).

In particular all it appears you are saying is that there is an odd number freater than 2^k. That number you chose is different for each k. The axiom of infinity that you seem obsessed with doesn't all ow you take the limit of this sequence and claim it is a meaningful number. This is just the same as the fundamental error in your cantor argument.

What you've written down does not support any conclusions you've made.

Matt,

Again you ignore the invariant fundamental symmetry that stands in the basis of N.

My proof is not about any fixed k value. It is about the "universal" (your word) structure that stands in the basis of N.

What I show is that there is self similarity between this basic structure and the Binary Tree used in Collatz problem.

Therefore if there is a general and rigorous proof to Collatz problem, it means that you also prove the ZF axiom of infinity ("if n exists then n+1 exists" is an axiom, therefore cannot be proved, otherwise it is not an axiom).

Shortly speaking, a general and rigorous proof to Collatz problem is a contradiction of ZF axiom of infinity existence as an axiom.

Because no proof of any Math system contradicts its axioms, Collatz problem is undecidable within N.

Also sets R and C can't be used here because their existence depends on set N existence.

My proof is "structured oriented", not "fixed k oriented".

If we look at this tree http://michael.cleverly.com/funstuff/3x+1/collatz2.jpg we can see that there are some numbers which are the results of both 3n+1 and n/2, for example:

Number 22 is the result of 3*7+1 and also the result of 44/2.

Let K be the set off all natural numbers, which are a common result to both 3n+1 and n/2.

|K|=|N| therefore transfinite system is too strong for Collatz problem.

Arithmetic

The arithmetic of Collatz problem is not importent, because we can use another arithmetic and get the same "Collatz tree" structure with different values instead of N members.