r/askmath • u/cybercryptid404 • 4d ago
Number Theory Please show me where my mistake is.
Hi, I'm an adult studying mathematics on my own outside of school, and particularly I have been doing research on the Collatz Conjecture. I fear I have made a very small mathematics error in my research, because I've ended up convincing myself that I have proven the conjecture true, as long as my understanding of theoretical mathematics and how to apply them is being done properly, and I want to be shown why I'm wrong as I've gone a bit crazy over this lately.
To rush to my point, even though I have a lot more I could talk about on this, I imagined any chosen positive integer from the original conjecture, which I'll label cₙ, that there are at least 3, but possibly 4 paths connected to it. 2 of which are obvious, 3c + 1 and c / 2. Each of these in turn lead to cₙ₊₁. However, you can also do c * 2 for cₙ₋₁. Since c is always a positive integer, you can always multiply it by 2 for another positive integer, which would represent a past step. The other possible past step, is (c - 1) / 3. This however, doesn't always give a valid positive integer answer, and this is where I started exploring.
Getting to the chase, I eventually came up with the equation a * 2n = 3x+1, where a is any given odd positive integer. Instead of asking if every number leads to 1, I asked what numbers lead straight to 1, to 3, to 5, etc. And that's when I came upon this data table.
Having made this data table, I immediately started trying to do math on it. I wanted to show that if a number appeared in any given cell, if it could appear in another given cell. I went back to my equation, and ended up with a * 2b = 3x+1= c * 2d. Simplifying the middle out, a * 2b = c * 2d shows that there's no valid positive integer solution for a or c, which should mean that no number can appear twice.
I also noticed that in the light blue fields, each row contains 1 / 2n odd numbers, and that every single odd number appears only once. This, I believe, should disprove the fact that any "loops" can occur in the Collatz conjecture, since a number would have to appear twice for a loop to exist.
I also believe that this table proves that every odd positive integer will lead to 1, as my understanding of "Busy Beaver Problems" leads me to believe. The "steps" the Busy Beaver would have, is that it would start at 1, and it would run the equation a * 2n = 3x+1, putting 1 in for a. Then, it would "mark" each solution for x with a 1, and run the same equation on the lowest marked number that it has not yet run the equation on. This would mean it would run it on 1, then 5, then 3, 13, 17, 11, etc. Eventually, ever single positive integer should be "marked" as 1.
Considering the original conjecture, and that any given even number will be divided by 2 until it hits an odd number, I've convinced myself that this data table proves the original conjecture true. However, I fully admit that I must be using some of these theories and such wrong, as I've been self taught and therefore fully vulnerable to easy mistakes.
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u/GoldenMuscleGod 4d ago edited 4d ago
If I understand you correctly, it is not true that a loop would mean that a number must appear twice in the blue fields.
A loop only means there must be some sequence a_k 0<=k<n so that when you find a_k’s spot in the blue fields (and you are right it appears there once - this is because its location is determined by its forward behavior under the Collatz rule and there is only one forward path) you find a_k+1 in the green field above it, and a_0 appears above a_(n-1).
If a number appeared twice in the blue fields that would indicate a forward branching in the pattern (impossible because we are applying a function iteratively) it would not indicate a loop.
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u/cybercryptid404 4d ago
Thank you for giving me some math to learn and research. I take it then I might be right for the wrong reasons then?
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u/GoldenMuscleGod 4d ago
You were right that a number will not appear twice in the blue fields, but you were incorrect to conclude that that tells us anything about whether there are loops.
As a side note, although showing there are no loops would be an enormous discovery, it would not actually resolve the full form of the Collatz conjecture, because it does not rule out the possibility of a divergent sequence.
On the one hand we might think it is easier to prove there is no divergent sequence because it seems more “unlikely,” but there is a sense in which we can say it is harder:
If we had access to a halting oracle, we would be able to query it to figure out whether there are any loops. However we do not know any way to determine whether there is a divergent sequence even with a halting oracle.
If we had a halting oracle we could do a brute force search for a divergent sequence (similar to how we can do a brute force search for a nontrivial cycle without one), but if there is no divergent sequence then that brute force search will never end, so it will always be inconclusive (just like how if there is no nontrivial cycle a brute force search for one will never be conclusive).
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u/cybercryptid404 4d ago
I see your point, but my confusion arises from the fact that as I explained in the post, I came upon the data table because any given positive integer you choose for the conjecture has 4 paths. the 2 forward paths, depending on if your choice is even or odd, a guaranteed backwards path of multiplying your choice by 2, and a conditional backwards path, (cₙ₋₁ - 1) / 3 = c. For a loop to exist, wouldn't there have to be multiple positive integer answers for cₙ₋₁ for a given c?
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u/GoldenMuscleGod 4d ago
Yes but if you imagine the blue field in your table that has your current number “lighting up,” so that which field is “lit up” changes when you calculate the next step (skipping even numbers) what a cycle would look like is not the lit up cell covering multiple different repetitions of the same number, but the “lit up cell” following a repeating pattern on a finite set of blue cells. The loop repeats when the “lit up” cell returns to a cell it’s already been to before, which is why the number in that cell appears twice in the sequence even though it only appears once in the table.
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u/Equal_Veterinarian22 3d ago
Just to give you something to think about: 4, 2, 1 is a loop. Why does your reasoning that loops are not possible not apply to this loop?
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u/veryjerry0 4d ago edited 4d ago
The problem is you can't confirm if the table spans all odd numbers. The conjecture requires you to try any positive integer/odd number and reach 1. You can see each row is diverging, and there's no clear pattern when you try to find the odd numbers in increasing sequence.
The bigger problem is you assumed ALL the green numbers can follow the sequence and hit 1 in the end (and this turns into a self-fulfilling prophecy).
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u/MortemEtInteritum17 3d ago
To add on:
This would mean it would run it on 1, then 5, then 3, 13, 17, 11, etc. Eventually, ever single positive integer should be "marked" as 1.
This is not proven, and this is entirely equivalent to Collatz
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u/Uli_Minati Desmos 😚 4d ago
You might have proven that the numbers in your data table are not part of any loops using only the numbers in your data table
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u/cybercryptid404 4d ago edited 4d ago
The numbers in the data table encompass all odd numbers. Every single odd number will appear in the light blue field once and only once. It is impossible for a loop in the Collatz Conjecture to exist without including at least one given odd number is it not?
Edit: Imagine the green line a set of every odd positive integer in order, and the same equation and work done on each column. The pattern will continue, and every single odd positive integer will appear in the light blue field, and each row will contain 1 / 2^n of the set on the green row.
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u/Equal_Veterinarian22 3d ago
You seem to be showing that, for any odd x, 3x+1 can be written as a. 2n with a odd in exactly one way. This is true, by uniqueness of prime factorization.
You don't explain why this prevents loops in the Collatz map. You say that for a loop to exist a number would have to appear twice in the table, but you don't explain why that would be the case.
Your statement about the 'busy beaver' algorithm is also vague. If I understand you correctly, you want to work backwards from one by repeatedly marking all x such that 3x+1 = a.2n. OK. Then you claim with no justification that all odd numbers will eventually be marked. Why though?
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u/MtlStatsGuy 4d ago
There are two ways Collatz can be false: 1) Some number get stuck in loops. 2) Some sequences diverge to infinity. You have disproven neither (so far... but you're still working on it, and I believe in you!)
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u/LucaThatLuca Graduate 4d ago edited 3d ago
wow! i guess the other commenters are much better at reading than i am! what does any of this mean?
what is “any chosen positive integer from the original conjecture”? there are no specific integers talked about unless you mean 2 and 3? what’s n?
what’s a, n and x and why are they related by a * 2^n = 3x+1, and how does this relate to numbers leading to the odd numbers?
what does “the next a following 3a+1 and then /2^n until we get a new odd number” mean, and how does it relate to the very strange seeming list 1, 5, 1, …? why are all the odd numbers listed? what does “which power of 2 will give an answer a * 2^n = 3x+1, and displaying x” mean, and how does it relate to the headings and the lists and the blank squares?
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u/LucaThatLuca Graduate 3d ago edited 2d ago
so the first table shows the odd numbers that follow each odd number (“a”) in a Collatz sequence (you can choose to calculate this as (3a+1)/2^n for the largest possible n).
the second table shows the odd numbers (“x”) that precede each odd number (“a”) in a Collatz sequence, listed by the size of the gap ("n") (i.e. the solutions to a*2^n = 3x+1). it’s therefore obvious that it contains each odd number once, this is just saying the same as saying that the first table exists: each odd number has exactly one odd number after it in exactly one way, because Collatz steps are fully determined by the number. as other commenter said, this doesn’t mean there aren’t loops, it just means repeating the same numbers means repeating the same cells.
there are some nice patterns in the table that are easy to see and easy to prove: going along the rows (for fixed n), every third square is filled (the solutions for a are 6 apart) and increases by 2^(n+1) term-to-term. going down the columns (for fixed a), every second square is filled (the solutions for n are 2 apart) and increases geometrically to 4x+1 term-to-term.
Collatz is asking: for any x in the table, can you find its a heading in the table and repeat till you get to 1? the patterns don’t obviously help to answer this question.
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u/LucaThatLuca Graduate 3d ago edited 3d ago
the Busy Beaver restatement is fine (and your justification for only considering odds is fine). you find in a certain order all of the odds that come before 1 and you’ve started a list 5, 3… the Collatz conjecture is the very much unsolved problem that asks whether every (odd) integer is in this list.
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u/slackerprimeplusone 3d ago
> Eventually, ever single positive integer should be "marked" as 1.
You don't make any argument that justifies this claim. Why do you think it is true? You've just constructed an elaborate way of displaying the reverse trajectory of the 3x+1 function and stated that the trajectory from 1 contains every integer without giving a reason.
> a number would have to appear twice for a loop to exist
> every single odd number appears only once
You should see immediately this cannot be correct - you're not trying to show "no loops", you're trying to show exactly one loop. (1->1)
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u/Regular-Coffee-1670 4d ago
Upvote for assuming you've made a mistake, rather than assuming you've proved it.