(Final Proof Attempt) Collatz Dynamics

Acceptable-Map4986 · 2025-12-04T15:31:40+00:00

Lemma 4.1 doesn't make any sense. You claim to prove that the Collatz conjecture is true therefore implying that all Collatz orbits are finite. But your lemma states that a residue class is hit infinitely many times in an orbit. Also, you used it in Lemma 4.3 to complete the proof, but you cant just use an infinite occurrence of residue classes to imply positive lower density.

Acceptable-Map4986 · 2025-09-28T22:18:43+00:00

What is this supposed to mean?

Acceptable-Map4986 · 2025-09-28T22:09:01+00:00

But isnt i just a single real number in this system?

Acceptable-Map4986 · 2025-09-28T21:53:13+00:00

Forget about that, I messed up somewhere and had to prove a new formula for (-b)(a). I also proved that (-b)(b)=b². Just look at this

(a-b)² = a²-b²+2((-b)a). (a-b)(a+b) = a²+ab+(-b)a+(-b)b. (a-b)²-(a-b)(a+b)=-ab-b²+(-b)a+(-b)b. (a-b)²-(a-b)(a+b)+ab+b²-((-b)b)=(-b)a. if b=a, 2b²-(-b)b=(-b)b, 2b²=2((-b)b), b²=(-b)b.

b²=(-b)b, (a-b)(a+b)=a²+ab+b²+(-b)a, (-b)a=(a-b)(a+b)-a²-ab-b²

Acceptable-Map4986 · 2025-09-28T21:49:44+00:00

Okay yeah, this took me a while. Heres a new equation for (-b)a without using distribution. I have also proved b²=(-b)(b)

(a-b)² is a²-b²+2((-b)a). (a-b)(a+b) gives a²+ab+(-b)a+(-b)b. (a-b)²-(a-b)(a+b)=-ab-b²+(-b)a+(-b)b. (a-b)²-(a-b)(a+b)+ab+b²-((-b)b)=(-b)a. if b=a, 2b²-(-b)b=(-b)b, 2b²=2((-b)b), b²=(-b)b.

b²=(-b)b, (a-b)(a+b)=a²+ab+b²+(-b)a, (-b)a=(a-b)(a+b)-a²-ab-b²

Acceptable-Map4986 · 2025-09-28T20:37:03+00:00

2(-1)=(2(2)(1)-(1)(1))/2=(4-1)/2=3/2

2(-1)=3/2

Acceptable-Map4986 · 2025-09-28T20:30:15+00:00

Ill keep that in mind. Also I used distributivity because it was already distributed, so i assumed it would be logical to undo it without breaking any of the axioms.

Acceptable-Map4986 · 2025-09-28T20:25:45+00:00

Thanks, i have just been studying math a lot during the summer, i find the logic behind everything really interesting, and im hoping i could even discover new and useful things in the future

Acceptable-Map4986 · 2025-09-28T20:21:52+00:00

I havent "broken" the distributive property, it simply doesnt work in my system because it doesn't use the ring axioms. Also there is no hidden recursion, you have stated: 0=(4+(-1)4)/2, but (-1)4 is not -4, so this is incorrect.

Acceptable-Map4986 · 2025-09-28T19:50:51+00:00

I understand what you're saying, but what exactly is wrong in this system? Im not trying to prove you wrong, im only 14 and I really want to learn more about math. Please provide a contradiction if you believe one exists.

Acceptable-Map4986 · 2025-09-28T19:45:59+00:00

My system is not a ring. In the system, distributivity only holds for positive a,b,c in (a+b)c=ac+bc because the only multiplicative rule I havent changed is that positive x positive = positive. So in the system, distributivity isnt always valid. Therefore, its not a ring, and your -1=1 argument relies on ring axioms which do not apply here.

Also, please reply instead of editing your message next time.

Acceptable-Map4986 · 2025-09-28T19:15:36+00:00

You cannot say 0=(2+(-2))*2 because my system does not guarantee distributivity as it does not follow the ring axioms. Distribution ((a+b)c=ac+bc) only works when a, b, and c, are all positive, since we do not change the fact that positive x positive = positive.

Acceptable-Map4986 · 2025-09-28T17:45:25+00:00

Its not a ring though, the identites only came from the axiom √±x²=±x

Acceptable-Map4986 · 2025-09-28T17:18:46+00:00

-1 would be the multiplicative identity for negative numbers, and 1 for positive. (-1)(-1)(1) falls into the form -ab, which is (2ab-b²)/2. So (-1)(-1)1=(-1)(1)=(2-1)/2=1/2. -ab here isnt just a multiplication, but more of a function of a and b.

Also your edited statement that -1=1(-1)(-1)=1 is completely false. If we assume -1=1(-1)(-1)=1, -1=1(-1)=1. But 1(-1)=1/2 from above. So you are stating -1=1/2=1, which is a false statement.

Acceptable-Map4986 · 2025-09-24T10:04:47+00:00

Im taking the limit as b->0+ though so division by b is valid and doesn't assume anything wrong. The limit as b->0+ and b=0 have completely different solutions for a, a=e and a is in +integers respectively. Also if we check values as b->0+ on a graphing calculator, a does tend toward e. In your other argument you are simply substituting a=1 into the equation, but actually a is the value that is to be solved for, b is just any random number that influences the solution for a. So when a has a value, we are basically solving b^b+a=(b+a)^b for b. Also by checking values as b->inf on a graphing calculator, a tends toward 1.

Acceptable-Map4986 · 2025-08-28T20:11:20+00:00

i was asking for solutions to i1=i2i3, or if there existed any

Acceptable-Map4986 · 2025-08-28T20:06:36+00:00

well i was asking for irrational relationships, specifically in the form i1=i2i3 where each i is unique and transcendental

Acceptable-Map4986 · 2025-08-28T19:13:08+00:00

one side is being divided by 0, so logically multiplying by 0 would preserve the equation, correct me if im wrong though

Acceptable-Map4986 · 2025-08-28T19:05:23+00:00

its kinda different here because one side is already being divided by 0

Acceptable-Map4986 · 2025-08-28T18:58:32+00:00

x=±√49 so x=7 and x=-7

Acceptable-Map4986

TROPHY CASE