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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 10:51 AM
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On Thursday, 25 July 2013 11:32:48 UTC-3, Bart Goddard wrote:
> raycb@live.com wrote in
>
> news:c78723d3-def0-478d-83bb-8ae787abffb7@googlegroups.com:
>
>
>

> > 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
>
>
>
> Why it is impossible? You've shown that the _smallest_ counterexample
>
> can't be of the form 4k+1, but that doesn't apply to any other numbers.
>
>
>
> You're confusing 'n' with its image under the operations.
>
>
>

> > 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.
>
>
>
> Same problem. n is 4k-1, not 8k-1 or 64k-1 or anything else.
>
> You're confusing using k as a parameter and using it to describe
>
> the form of the number.
>
>
>
> B.


I'm using 8k -1 to say that n has a remainder of -1 (mod 8).

Suppose n has a remainder of 3 (mod 8).

3*3 = 9

9 + 1 = 10

10/2 = 5 = 1 (mod 4).

That is impossible, and eliminates all numbers with a residue of 3 (mod 8).

Yes, I'm jumping back and forth from the size of numbers to their remainders. The size of numbers eliminates the first two possibilities, the rest is done with remainders.




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