
Re: Where is the flaw in this proof of the Collatz Conjecture?
Posted:
Jul 25, 2013 10:32 AM


raycb@live.com wrote in news:c78723d3def0478d83bb8ae787abffb7@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 4k1, not 8k1 or 64k1 or anything else. You're confusing using k as a parameter and using it to describe the form of the number.
B.

