# Collatz Conjecture proof for review. [closed]

This is a simple attempt as a proof of the Collatz Conjecture, as defined in Wikipedia. I must tell you that I'm not a math expert and I did it just for fun. Surelly it have serious flaws I just curious to know what could they be.

Introduction

The conjecture holds true if all natural numbers will make the succession to converge to 1 after a finite number of applications of the recursive function $f$, computing $n/2$ if $n$ is even, and $3n+1$ if $n$ is odd.

A useful observation

One useful observation is that if it holds true for $n \geq 2$ an even number, then must hold true for $f(n)=n/2$, since it is the next number in the succession. Analogously if it holds true for $n \geq 3$ an odd number, then it must hold true for $f(n)=3n+1$ too, since it is the next in the chain. In both cases, also the conjecture remains true for $2n$ the previous number in the chain, that is if $a_i=n$, then $f(a_{i-1})=f(2n)=a_i=n$, since $2n$ will always be even, and is in the same chain that $n$.

Proof

The Collatz conjecture holds for all natural numbers, and that the succession is finite (the convergence to reach 1 requires a finite number of steps).

I will proceed by induction on $a_0=n$, for natural $n \geq 1$.

Base case $a_0=1$: The succession goes: $a_0=1; a_1=3.1+1=4; a_2=2; a_3=1$. Also the succession converged in 3 steps.

Inductive hypothesis: The succession is finite an always converge to 1.

The inductive step is divided in two sub cases: the $n$ is even or the $n$ is odd.

Inductive case: $n \geq 2$ is an even number: This means that exist a natural number $q < n$ such that $a_0=n=2q$. Then the next number in the succession is $a_1=f(a_0)=f(2q)=2q/2=q$. As $q < n$ then by inductive hypothesis lets assume that for $q$ the succession converged in $s$ steps, then for $n$ must converge is $s+1$ steps.

Inductive case: $n \geq 3$ is an odd number: This means that the initial number in the succession is $a_0=n$, an odd number. We can construct a virtual succession by using the number $a_{-1}=2n$, which is an even number, and it is in the same chain on the succession, since $f(a_{-1})=a_0$. As shown previously, the conjecture apply for even numbers, so it holds true for the number $2n$, and therefore for $n$ since it is one step further in the succession.

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## closed as too localized by Grigory M, lhf, Asaf Karagila, Henning Makholm, Zev ChonolesFeb 9 '12 at 2:12

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Dear Gabriel, I have not read the details of your argument, but: this being a rather famous open problem at which probably almost every mathematician in the last 50 years has taken a stab, it is rather unlikely that such a short argument would solve it! –  Mariano Suárez-Alvarez Feb 8 '12 at 5:44
In your odd case, you assume the result for $2n$, which you can't do since you don't know it for $2n$; you just know it for $1,...,n-1$ by your induction hypothesis. –  Zarrax Feb 8 '12 at 5:50
Numerous incorrect proofs of famous open problems is not what Math.SE is for -- voted to close as "too localized". –  Grigory M Feb 8 '12 at 7:56
(+1) at @Grigory. One might assume good faith anyway and modify the question to set the focus on the question of "is my approach to the method of induction valid here? (example 3x+1-problem)". Otherwise I also would like to see the question being closed. –  Gottfried Helms Feb 8 '12 at 11:24
chat somewhere, anyone? some new ideas on collatz –  vzn Mar 4 '13 at 16:57

You handle the case of odd $n$ by considering instead $2n$ (which indeed has $n$ in its succession) and appealing to your even case. However, the even case is proven via strong induction. So, to prove $2n$ converges, you must know already that $n$ converges. Your logic here is circular.