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Topic: Where is the flaw in this proof of the Collatz Conjecture?
Replies: 8   Last Post: Aug 1, 2013 11:58 PM

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 raycb@live.com Posts: 41 Registered: 10/25/08
Re: Where is the flaw in this proof of the Collatz Conjecture?
Posted: Jul 25, 2013 10:20 AM

On Thursday, 25 July 2013 11:06:40 UTC-3, ra...@live.com wrote:
> The conjecture states that:
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> Given a positive integer n,
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> If n is even then divide by 2.
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> If n is odd then multiply by 3 and add 1
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> Conjecture: by repeating these operations you will eventually reach 1.
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> Proof:
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> Let n be the smallest positive integer that is a counterexample to the conjecture.
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> If n is even then it can be divided by two to give a smaller number, leading to a contradiction.
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> Assume n = 4k + 1.
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> Multiply it by 3, add 1, and divide by 2 twice.
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> The result is 3k + 1, a number smaller than n, leading to a contradiction. Therefore n has the form
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> n = 4k - 1.
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> Multiply by 3, add 1, and divide by 2.
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> The result is 6k - 1. If k is odd, then 6k - 1 is one more than a multiple of 4, which is impossible, therefore k is even, and n has the form
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> n = 8k - 1
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> Multiply by 3, add 1, and divide by 2.
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> The result is 12k -1, with k necessarily even. In this manner it can be proved that n must have the form 16k - 1, 32k -1, 64k -1, and so on, requiring n to be infinitely large, which is impossible.

A poster on another site questioned my statement that 12k -1 requires k to be even. Here's my explanation:

Well, I'm trying to say that n has the form of a number with a remainder of -1 for some power of 2.

The number 12k - 1 can be written as 8k + 4k -1, and the number has to have a remainder of -1 mod 8, therefore 4k is a multiple of 8, and k is even. My argument is that these operations always require k to be even, that there is a repeating pattern.

Date Subject Author
7/25/13 raycb@live.com
7/25/13 raycb@live.com
7/25/13 trj
7/25/13 Bart Goddard
7/25/13 raycb@live.com
7/25/13 raycb@live.com
7/25/13 Thomas Nordhaus
7/25/13 Gottfried Helms
8/1/13 b92057@yahoo.com