
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 UTC3, ra...@live.com wrote: > 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.
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.

