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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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Gottfried Helms

Posts: 1,902
Registered: 12/6/04
Re: Where is the flaw in this proof of the Collatz Conjecture?
Posted: Jul 25, 2013 12:12 PM
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Am 25.07.2013 16:06 schrieb raycb@live.com:
> The conjecture states that:
>
> Given a positive integer n,
>
> If n is even then divide by 2.
>
> If n is odd then multiply by 3 and add 1
>
> Conjecture: by repeating these operations you will eventually reach 1.
>
>
>
> Proof:
>
>
>
> Let n be the smallest positive integer that is a counterexample to the conjecture.
>
> If n is even then it can be divided by two to give a smaller number, leading to a contradiction.
>
> Assume n = 4k + 1.
>
> Multiply it by 3, add 1, and divide by 2 twice.
>
> The result is 3k + 1, a number smaller than n, leading to a contradiction. Therefore n has the form
>
> n = 4k - 1.
>
> Multiply by 3, add 1, and divide by 2.
>
> 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
>
> n = 8k - 1
>
> Multiply by 3, add 1, and divide by 2.
>
> 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.
>

Odd numbers of the infinitely following forms with k>=1 transform to 6k-1
4k - 1 -> 6k -1
16k - 3 -> 6k -1
64k - 11 -> 6k -1
256k - 43 -> 6k -1
...

Odd numbers of the infinitely following forms with k>=0 transform to 6k+1
8k + 1 -> 6k +1 (k=0 gives 1 -> 1 the only known cycle in odd positive numbers)
32k + 5 -> 6k +1
128k + 21 -> 6k +1
512k + 85 -> 6k +1
...

That two groups of odd numbers cover the whole set of odd positive numbers.
But you must prove, that iterations form a tree connecting all that numbers
via the root 1.

Gottfried Helms




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