Where is my Collatz conjecture proof wrong?

Question

I am amateur and don't have very good understanding of mathematical proving. My proof is so simple i don't believe noone thought of this before. But I am so blinded by the hope it is correct, that I can't find any errors in it.

Here are all the conclusions I made in the process:

1. You can always make smaller odd number from even, so we only need to prove that every odd number works.
2. Out of the two ways (3n+1 and n/2) only the n/2 can be used to create n ≡ 0 (mod 3) from n_-1, where n_-1 ≡ 0 (mod 3) has to hold true
3. n ≡ 0 (mod 3) always creates n ≡ 1 (mod 3) (EDIT: every odd n ≡ 0 (mod 3))
4. therefore once we abandon n ≡ 0 (mod 3), we can never find any other n ≡ 0 (mod 3) later in the sequence
5. in other words we can't compose any n ≡ 0 (mod 3) from smaller number
6. But, we can "compose" any n !≡ 0 (mod 3) _{(sorry, if I invented the notation. !≡ means the remainder after division is not 0)} from smaller number.
   • reverse Collatz Example: 5 => 10 => either 3 or 20
   • The only way to reverse n ≡ 2 (mod 3) is n*2, after which every n_-1 ≡ 1 (mod 3)
   • n_-1 ≡ 1 (mod 3) can be reversed in both ways, but the (n-1)/3 yielding n_-2 will always be smaller then n
7. If n !≡ 0 (mod 3) can be "composed" from smaller number, we need only to prove the Collatz conjecture works for the smaller number.
8. which can be either n !≡ 0 (mod 3) and can be "compsed" from smaller number again or be n ≡ 0 (mod 3)
9. So the only thing I have to prove now is that the Collatz conjecture works for every odd n ≡ 0 (mod 3)
10. But since every odd n ≡ 0 (mod 3) will yield n₊₁ ≡ 1 (mod 3) and I already concluded that once we abandon n ≡ 0 (mod 3), no later number will be n ≡ 0 (mod 3), the Collatz conjecture works also for n ≡ 0 (mod 3)

I've already looked at this for two days and can't find any errors and it drives me crazy. Is it cyclic or something? Am I depending on my own proves that also depend on my own proves? Did I make algebraic mistake? Hopefully, some of you will be willing to help.

P. S. Please be mindful of my experience and don't use very mathematical language if not neccessary. I've tried to search for this and didn't understand many papers on this matter.

EXPLANATIONS:

• "be composed of" means "is preceded by"

EDIT:

• don't overlook EDIT in point 3
• made the points ordered so we can reference them more easily.
• added explanations

Answer 1

EDIT after OP edited the question.

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Your reasoning fails after point $6$ (included). Indeed, you talk about a certain number in the chain being "smaller" or "bigger" than another, and then mistakenly apply this rule to multiple elements of the chain (for instance in your point $6c$).

This is not strong enough, as $n/2$ might be followed by $3(n/2)+1$, which is always bigger than your original $n$. And then the same behaviour might happen again, etc.

So you have not proved that every $n ≡ 1\ ( mod \ 3)$ eventually becomes smaller than $n$.

Nor have you proved that there does not exist a cycle other than $4\to2\to1\to4$, a problem that your reasoning did not address.

Answer 2

But, we can "compose" any $n \not≡ 0\,(\text{mod}\space 3)$ from smaller number.

Not true. How do we get $11$ from a smaller number? You've only proven that you don't need to consider odd cases of the form $3k+1$, and you've also proven that even $3k+2$ can be 'composed' from a smaller term. You still have $n ≡ 2\,(\text{mod}\space 3)$ (where $n$ is odd) to think about.

Answer 3

Alright, some people post answers to the comments and I'd like them to be seen. Also to have all the errors of my proof at one place. I will edit this as someone finds new errors.

1. (point 6) Not every n !≡ 0 (mod 3) can be preceded by smaller elements (credit to Arnaud D.). More precisely every odd n !≡ 1 (mod 3) DOESN'T HAVE TO. Example (7, 13, 19 (19 will be eventually preceded number 5, but not after just two steps as i thought before)).

Change in my phrasing, because it might have caused confusion: - "composed of" -> "be preceded by"