
Re: Where is the flaw in this proof of the Collatz Conjecture?
Posted:
Jul 25, 2013 11:09 AM


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.
So far OK.
> > 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 is this impossible? Let k=2m+1, m odd. Then n = 8m+3. Furthermore 6k1 = 12m+5 > 36m+16 > 9m+4 (odd) > 8m+3 = n. So this is not a contradiction to n being the smallest counterexample. So m has to be even. But why should it?
 Thomas Nordhaus

