
Where is the flaw in this proof of the Collatz Conjecture?
Posted:
Jul 25, 2013 10:06 AM


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.

