
Re: Where is the flaw in this proof of the Collatz Conjecture?
Posted:
Jul 25, 2013 12:12 PM


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 6k1 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

