1
$\begingroup$

I recently stumbled across this self-proclaimed proof of the Collatz Conjecture. It seems very irritating to me that this very hard conjecture is supposedly proven by using very basic counting techniques. Can someone maybe point out some of the mistakes?

Clearly the author didn’t bother to go into any formalisms or to even make a complete proof.

$\endgroup$
  • $\begingroup$ There are very many "so-called" proofs. I am not sure it is even worth to look at it, if the methods are as you said. Usually someone finds an "error" soon, but this doesn't change anything. The proof will just be "modified" a little bit, and then we restart the whole procedure. Or, as Max says very poetically, it is an "endless flow". $\endgroup$ – Dietrich Burde Apr 19 '19 at 10:10
4
$\begingroup$

The first unjustified passage is

"Since it is repeatedly dividing a even number by two, if the series converges, it will sure reach 1. Therefore it is enough to demonstrate that the series converge"

Here "converges" is of course to be taken in a modifed sense of "cycle" (because even if the Collatz conjecture is true, the sequence obviously does not converge in the usual sense). But then it hasn't even been shown yet that $1-4-2$ is the only possible cycle.

Then

"For any starting number N, the series will converge if z is ON AVERAGE greater than the number given by the equation above. In this way the sequence of operations will subtract more than add to the series."

This is not justified either. For it to have any chance of being true, "on average" would have to be quantified way more than just this way. What if, every time it's not greater, then it just becomes way bigger and you can't control it that way ?

I won't go further, because it seems to be again just an endless flow of unjustified claims.

Let me take this opportunity to make a statement about proofs in maths : the point of proofs is that it's not the reader's job to try to correct the proof; the author of the proof has to justify every single one of their claims, if they can't and the reader has an objection to which the "prover" can't answer, then it's not a valid proof.

Of course, the amount to which one has to justify the claims also depends on the public you're aiming at : professional mathematicians are used to certain types of reasoning and so can fill some gaps by themselves when reading paper (for a stupid basic example, if you're proving something to a mathematicall educated audience, you don't have to say "and then by modus ponens we have ..."); but it is still the case that in the prover vs reader "game", the reader is always right, and when a part of a proof is not clear to them, they are right and the prover has to justify that part of the proof.

$\endgroup$
  • $\begingroup$ Hi @Max, can you let me know what you think about the following math.stackexchange.com/questions/3489883 it is an attempt, I'm assuming irritating, hopefully slightly less than what is being discussed here. $\endgroup$ – Francis Laclé Dec 28 '19 at 20:01
2
$\begingroup$

As has already been pointed out by Max, it is not a proof as it based on unproven claims: in particular, the claim that if you start the sequence with any given number, the number of odd and even numbers in the sequence will in the long run balance out in the ratio 1:2, which would have been 1:1 if you replace the rule for odd numbers with $n\mapsto (3n+1)/2$.

The argument is a heuristic argument of why you might expect the Collatz sequence of some arbitrary number to reach 1. In other words, if the numbers of the sequence behave as if they are random numbers, with nothing special or systematic going on, that is what would happen.

However, the numbers of the sequences are not random. There could be something going on which makes the number of odd and even numbers in some Collatz sequence skewed. How do you prove that doesn't happen?

It could even be the case that the heuristic argument would be correct for most cases, but with a few exceptions. So even from a heuristic point of view, it is not enough to just argue what happens on average: you have to argue why that would be the case for every possible starting point, of which there are an infinite number of possibilities.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy