# A Possible Proof of Collatz Conjecture

Our reader **Eswar Chellappa** has sent his work on the solution of **‘3X+1’ problem**, also called **Collatz Conjecture**. He had been working on the proof of *Collatz Conjecture *off and on for almost ten years. The *Collatz Conjecture *can be quoted as follow:

Let $\phi : \mathbb{N} \to \mathbb{N}^+$ be a function defined such that:

$$\phi(x):= \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even } \\ 3x+1, & \text{ if } x \text{ is odd} \end{cases}$$Then the iterates of $\phi(x)$ will eventually reach $1$ for any initial value of $x$.

*See this post about Collatz Conjecture for more details.*

Plenty of proof attempts were made by various mathematicians. But none of those could flawlessly prove the statement. Mr. Chellappa’s attempt is based upon the famous ** Sieve of Eratosthenes. **Despite of his experience & confidence, I can not guarantee if this work is perfect. I invite readers to cross check the flawlessness and tell what they think.

## Introduction

Let, $f(x) = x/2$ if $x$ is even and $g(x) = 3x + 1$ if $x$ is odd. Since $3x+1$ is an even number for any odd $x$, we can replace any odd number by an even number which equals to $3x+1$. And when, $3x+1$ is an even number, we can successfully halve it according to first step of the function defined in the conjecture.

## The method proposed below is similar to the famous Sieve of Eratosthenes

To start with the proof, choose the natural numbers from 1 to 100.

Since we have $$f(x) = x/2$$ for even $x$. Let’s strike out all even numbers for each of them reduces to another odd or even number (less than x) by f(x), by exactly one iteration. This we do from 100 to 2, in the decreasing order.

- We strike out 100 as $f(x)$ reduces it to 50.
- Then we strike out 98 keeping 49 as the result of one iteration.
- 50 is also struck out giving 25 as result of one iteration.

Now all it remains, the odd numbers and natural number $1$. As $1$ automatically verifies the conjecture, we start from 3 and end to 99.

- We have $g(3) := 10$. Since $10$ has already been struck out, we can simply say that $3$ satisfies the conjecture. Hence we strike out 3.
- $g(5) := 16$, so 5 can also be struck out.

By this process we can strike out all odd numbers up to 33 as we have taken only numbers up to 100 and $g(35) >100$.

By suitably increasing the range of numbers we can conclude that ultimately $\phi (x)$ has to reach 1 irrespective of the starting natural number $x$.

## Any possible worries?

The only possibility that might worry us was falling in to a loop. This happens only if $\phi_i(x) = \phi_j(x) \neq 1$ where $i\neq j$. But that is not possible since $\phi (x)=x/2$ is one to one.

## Remark

- By striking out the numbers, we verify that they have images under $f(x)$. That is to say, more iterations are possible until the sequence reaches $1$.
- Concerned about $\mathbb{N}^+$ notation?

PS: This isn’t a **paper. **