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Topic: Proof of the Collatz Conjecture
Replies: 12   Last Post: Jan 12, 2013 11:19 PM

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Posts: 9
Registered: 12/29/12
Proof of the Collatz Conjecture
Posted: Dec 29, 2012 2:52 AM
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December 29, 2012

I have a nine-page rough-draft of a proof of the Collatz Conjecture posted at:

I would appreciate any questions, feedback, or suggestions you may have concerning the proof. I will provide a rough outline below.

Let me state up front that there is very little new or interesting mathematics in the proof. My proof is limited to the positive integers of the original conjecture. The Generalized Collatz Conjecture, which deals with all integers, retains its great mysteries.

Outline of the Proof
1) I partition the positive integers into ordered sets such that each set begins with an odd number and each subsequent member of the set is twice the value of the previous member. These sets will become the branches of the Collatz tree.

2) I link the initial members of each set (the odd integers) to other nodes using the function f(x) = (3x+1)/2, and I link each member of a set to the next member of the set.

3) I show that the only possible intra-branch cycle is the trivial (1,2) cycle.

4) I show that there cannot be any inter-branch cycles for any branches connected to the root branch ("trunk"?) {1, 2, 4, 8, 16, ... 2^n ...} of the Collatz tree.

5) I show that there cannot be any trees or graphs that begin with a positive odd integer that are not connected to the Collatz tree.

5a) I assume that there is an infinite tree constructed as part of steps 1) and 2) above that is not connected to the Collatz tree.

5b) I show that all such trees can be described by the simple pattern (s+(de+)+), where s, d, and e are each unique classes of positive integers and the "+" indicates one or more occurences of the preceding item.

5c) I show that, no matter what finite value one chooses as the root of the infinite tree, following the above pattern eventually excludes the value, and therefore such an infinite tree cannot exist.

5d) I show that for any cyclic graph I can construct an infinite tree that is its equivalent. Having already proven that there can be no infinite tree, there therefore can be no cyclic graph.

That is all there is to it. The messy details of the proof are in steps 5b) and 5c). Thank you for your time and assistance.

Mark A. Durham

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