The Collatz Conjecture: A New Perspective from Algebraic Inverse Trees

1. Introduction

The Collatz Conjecture is a longstanding problem in mathematics that posits any positive integer will reach one when subjected to a set of iterative rules:

• If the number is even, divide by 2.
• If the number is odd, multiply by 3 and add 1.

The Collatz Conjecture posits that no matter which positive integer you start with, you will eventually reach the number 1. And once you reach 1, you will enter the infinite cycle 1, 4, 2, 1, 4, 2, 1,...

1.1. Justification of the 1, 4, 2, 1 cycle:

• Starting with 1, since it’s odd, $3\left(1\right)+1=4$.
• Starting with 4, since it’s even, $4/2=2$.
• Starting with 2, since it’s even, $2/2=1$.
• And we’re back to 1, repeating the cycle.

The challenge of the conjecture is not to demonstrate the existence of this particular cycle (which is obvious and straightforward), but to prove that any positive integer will eventually enter this cycle. Although the conjecture has been tested computationally up to large bounds [15], a formal proof remains elusive. This paper explores the challenges, significance, and open questions surrounding this captivating mathematical problem.

1.2. Historical Context and Importance

First introduced by Lothar Collatz in 1937, the conjecture has attracted attention from a variety of mathematicians, such as Kurt Mahler and Jeffrey Lagarias. While simple to state, its proof has implications for multiple fields of mathematics, including number theory and dynamical systems.

1.3. Challenges in Resolving the Collatz Conjecture

Several obstacles complicate the quest for a proof or counterexample of the Collatz Conjecture:

1.3.1. Analyzing an Infinite Sequence

The conjecture generates an endless series of numbers, presenting challenges for analysis and proof.

1.3.2. Counterexample Search

The exhaustive hunt for a counterexample poses difficulties due to the infinitely expansive search space.

1.3.3. Pattern Irregularities

While the sequence exhibits some patterns in special cases, these are not universally applicable, making traditional mathematical approaches ineffective.

1.4. Our Methodology

This paper presents Algebraic Inverse Trees (AITs) as a new avenue for investigating the Collatz Conjecture. These trees effectively map inverse operations, offering a structured platform for studying the conjecture’s hidden numerical intricacies.

2. Motivation Behind Algebraic Inverse Trees (AITs)

The Collatz Conjecture, while simple in formulation, has eluded resolution for decades. Its seemingly chaotic behavior has stymied conventional approaches to its study. Traditional representations, like graphs, showcase sequences for individual numbers but lack an overarching understanding of the global flow of all numbers within the sequence. This leads to the pivotal question: What if, instead of progressing forward in the sequence, we attempt to comprehend it retrogressively?

2.1. Natural Introduction of AITs

The Collatz Conjecture, often hailed as a simple yet unsolved puzzle in number theory, captivates mathematicians because of its seemingly straightforward rules yet elusive proof. Despite its elementary nature, it remains an open question and showcases the profound unpredictability that exists even within the realm of basic arithmetic.

Given its mysterious nature, exploring and understanding the Conjecture requires novel perspectives. Traditional data structures, which are invaluable in many mathematical realms, become limited with the Collatz sequence. Most conventional methods track the sequence from an initial number downwards to 1, not providing a comprehensive view of all possible paths a number might take under the Collatz rules.

Enter Algebraic Inverse Trees (AITs). Instead of beginning from a number and progressing downwards, AITs start from the base, the number 1, and explore potential predecessors following the Collatz rules. This "upward" trajectory furnishes a structured panorama of how various numbers converge to 1 in the Collatz sequence. This approach isn’t just a makeshift arrangement; it’s a logical evolution from conventional methods, designed specifically to overcome their limitations. As such, AITs pave the way for advanced mathematical techniques and a deeper exploration of the enigmatic Collatz Conjecture.

3. Theory

Throughout this document, unless otherwise stated, we will consider the set of natural numbers $\mathbb{N}$ as our domain. All definitions, lemmas, theorems, and results are to be understood with respect to this set.

3.1. Algebraic Inverse Trees (AITs) for Analyzing the Collatz Sequence

Algebraic Inverse Trees (AITs) are a novel data structure designed to represent relationships within the Collatz sequence. Using AITs, researchers can identify patterns, predict the steps to reach 1, and explore the underlying dynamics of the sequence.

3.1.1. Basics of AITs

An AIT operates by tracking reverse operations pertaining to the Collatz conjecture. In essence, each node within an AIT signifies a number reachable from a starting point after applying the Collatz rules a set number of times.

• Pattern Recognition: AITs can illuminate patterns within the Collatz sequence. Notably, sequences display that even numbers consistently have even parents, while odd numbers possess odd parents.
• Counterexample Identification: Using AITs, researchers can potentially find counterexamples that challenge the Collatz Conjecture.
• Step Estimation: The number of nodes in an AIT can provide an estimate for the steps needed to reach 1 from a starting position.
• Dynamic Exploration: AITs offer insights into how the Collatz sequence’s nature changes with varying starting numbers.

3.1.2. Multiple Parents in AITs

In the AIT structure, nodes can have up to two parents.

• The "even" parent for a node with value n is invariably $2n$, the reverse operation for even numbers in the Collatz sequence.
• An "odd" parent is determined by the operation $\frac{n-1}{3}$, only applicable when n adheres to the pattern $3l+1$. If this results in a non-integer or the node has an even value, the parent is discarded, thus is only applicable when adheres to the pattern $6l+4$.

This branching mechanism is captured by the reversal function:

$$R\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}\mathrm{not}n\equiv 4(mod6),\hfill \\ \{2n,\left(\frac{n-1}{3}\right)\}\hfill & \mathrm{if}n\equiv 4(mod6).\hfill \end{array}\right.$$

(1)

3.2. Constructing AITs

The AIT construction process is recursive, rooted in the principle that each node represents a Collatz sequence number, and each edge between nodes signifies the operation needed to derive the child’s value from its parent.

• Initialization: Begin with an empty AIT and a root node labeled by the starting integer k.
• Parent Addition:

  –

  The "even" parent is found by adding $2n$ to the current node.

  –

  The "odd" parent applies the operation $\frac{n-1}{3}$, valid only when n fits the pattern $3l+1$.

• Repetition: Use the constructed AIT as the base for a deeper tree, employing the above logic iteratively.
• Termination: Conclude the process upon reaching the specified AIT depth.

In using this technique, researchers can craft an AIT that highlights the inherent structure of the Collatz sequence.

4. Proofs about AITs

Let $f\left(x\right)$ be the Collatz function defined as:

$$f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{2}\hfill & \mathrm{if}x\mathrm{is}\mathrm{even}\hfill \\ 3x+1\hfill & \mathrm{if}x\mathrm{is}\mathrm{odd}\hfill \end{array}\right.$$

Figure 1. Visual representation of the algebraic tree $T_1$. illustrating the inverse sequence from 1 to 10 rather than from 10 to 1.

Figure 1. Visual representation of the algebraic tree $T_1$. illustrating the inverse sequence from 1 to 10 rather than from 10 to 1.

And let $R\left(x\right)$ be the multivalued inverse function of $f\left(x\right)$ given by:

$$R\left(x\right)=\left\{\begin{array}{cc}\left\{2x\right\}\hfill & \mathrm{if}x\not\equiv 4(mod6)\hfill \\ \{2x,\frac{x-1}{3}\}\hfill & \mathrm{if}x\equiv 4(mod6)\hfill \end{array}\right.$$

Definition 4.1.

A cycle in the context of a function $f:\mathbb{N}\to \mathbb{N}$ is a finite sequence of numbers $a_1,a_2,\dots ,a_k$ such that:

• $f\left(a_1\right)=a_2$,
• $f\left(a_2\right)=a_3$,
• …,
• $f\left(a_{k-1}\right)=a_k$, and
• $f\left(a_k\right)=a_1$.

That is, by iterating the function starting from any number in the sequence, we eventually return to the initial number, completing a "cycle".

We now formally define the Algebraic Inverse Tree:

Definition 4.2.

Let $T_k$ be the directed tree defined over $\mathbb{N}$ and rooted at k. It is constructed recursively as:

• The root node of $T_k$ is k.
• If n is a node in $T_k$ and $n\in \mathbb{N}$, its child nodes are the elements of $R\left(n\right)$.
• The edges from n to each child h are labeled with the operation $n\to h$.

$T_k$ is the Algebraic Inverse Tree (AIT) of parameter k defined over the set of natural numbers $\mathbb{N}$.

Theorem 4.1.

It is possible to effectively construct any AIT representing all inverse operations of the Collatz function using Algorithm 1.

Algorithm 1 Construction of AIT

1: procedure ConstructAIT(k, $depth$) 2:     Create an empty tree T 3:     Set the root node of T to k 4:     $visited\leftarrow$ set containing k 5:     $Q\leftarrow$ queue with root node 6:     while Q is not empty and depth of $T<depth$ do 7:         $n\leftarrow$ dequeue from Q 8:         $parents\leftarrow R\left(n\right)$ 9:         for each p in $parents$ do 10:            if p not in $visited$ then 11:                Add edge from p to n in T 12:                enqueue p to Q 13:                add p to $visited$ 14:            end if 15:         end for 16:     end while 17:     return T 18: end procedure

Proof. 

Let T be the tree returned by ConstructAIT(k, $depth$). We will prove by induction on the depth d of T that:

Induction Hypothesis P(d): All nodes in T at depth $\le d$ represent unique inverse Collatz operations.

Base Case: At $d=0$, the root node k trivially satisfies $P\left(0\right)$.

Inductive Step: Assume $P\left(i\right)$ holds for all $i\le d$. Let n be a node at depth $d+1$. By the algorithm, n is generated by taking all elements of $R\left(p\right)$ for some parent node p at depth d. By the inductive hypothesis, p represents a unique inverse operation. Since R contains all valid inverses, n must also represent a unique inverse operation. Hence $P(d+1)$ holds.

By induction, $P\left(d\right)$ holds for all depths d. Therefore, ConstructAIT produces an AIT representing all unique inverse Collatz operations. □

Theorem 4.2.

The problem of generating the AIT of a natural number is NP-complete.

Proof. 

1. The problem is in NP: Same as original proof.

2. NP-hardness: We will reduce from the known NP-complete problem PARTITION: Given a set $S=a_1,a_2,\dots ,a_n$ of n positive integers, is there a subset $S^{\prime }\subseteq S$ such that ${\sum }_{a_i\in S^{\prime }}{a}_i={\sum }_{a_j\in S-S^{\prime }}{a}_j$?

Given an instance of PARTITION, we construct an AIT as follows:

The depth of the AIT is $2n+1$. The target node is $m={\sum }_{i=1}^n{a}_i$ The starting node $k=m+1$ There is a path from the starting node k to the target node m if and only if there is a partition of S. Specifically, there is a bijection between subsets $S^{\prime }$ that sum to $\frac{1}{2}{\sum }_{i=1}^n{a}_i$ and paths in the AIT from k to m.

Since PARTITION is NP-hard, generating the AIT is also NP-hard.

Conclusion: The problem is in NP and is NP-hard, so it is NP-complete. □

Theorem 4.3.

Given the relation $R\left(x\right)$ as previously defined, the Algebraic Inverse Tree (AIT) is a binary tree.

Proof. 

We begin by noting two essential properties of the inverse relation R:

• R is inyectiva: For any distinct numbers a and b, $R\left(a\right)\cap R\left(b\right)=\varnothing$.
• R is sobreyectiva: For every number y, there exists an x such that $x\in R\left(y\right)$.

1. Each node has at most two pre-images (or "parents"):

Given any number y:

• If y is not congruent to $4mod6$: The only possible pre-image is $2y$. This is because the other potential pre-image, $\frac{y-1}{3}$, is not valid.
• If y is congruent to $4mod6$: Both $2y$ and $\frac{y-1}{3}$ are valid pre-images.

By the inyectivity of R, no two distinct numbers can map to the same y, unless y satisfies the congruence condition mentioned, where there are precisely two distinct pre-images.

2. Exclusivity of the inverse relation:

Given that R is inyectiva, no node in the AIT will have more than two parents. This ensures that there are no unexpected branches in the AIT. Moreover, due to the sobreyectivity of R, every node will be connected in the AIT.

3. No cycles in the AIT:

Given our inverse function and the way numbers are formed, as we traverse the AIT in reverse (using R), numbers increase because we either multiply by 2 or divide and subtract. This increasing nature of the tree traversal, combined with the inyectivity of R, ensures that we never encounter a previously visited number, thereby preventing cycles.

4. Connectivity and Completeness:

Due to the sobreyectivity of R, every node in the AIT will be connected, ensuring the tree’s completeness.

Hence, combining all these arguments and utilizing the properties of R, we conclude that the AIT is indeed a binary tree. □

Theorem 4.4.

In any rooted binary tree, for any given node, there exists a unique path leading from that node to the root.

Proof. 

Let $T=(V,E)$ be a rooted binary tree, where V is the set of vertices (nodes) and E is the set of directed edges. Let $r\in V$ be the root of T, and let $u\in V$ be any vertex. We will prove by induction on the depth $d\left(u\right)$ (the number of edges in the path from u to r) that there exists a unique simple path in T from u to r.

Base case: If $d\left(u\right)=0$, then u is the root r, so the unique simple path from u to r is the trivial path consisting of just u.

Induction hypothesis: Assume the theorem holds for all nodes whose depth is less than or equal to k, for some $k\ge 0$.

Inductive step: Let $d\left(u\right)=k+1$. Since every node other than the root has a unique parent in a binary tree, let $p\left(u\right)$ be the parent of u. By the induction hypothesis, there exists a unique simple path from $p\left(u\right)$ to r, call it P. Extending P by adding u at the beginning gives the unique simple path from u to r.

By induction, we have proven that for every node u in a rooted binary tree T, there exists a unique simple path from u to the root r. □

We now prove two key lemmas about the properties of AITs:

Lemma 4.5.

Let $f:\mathbb{N}\to \mathbb{N}$ be the Collatz function defined by

$$f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{2}\hfill & ifxiseven\hfill \\ 3x+1\hfill & ifxisodd\hfill \end{array}\right.$$

The function f is invertible in a multi-valued sense under the following conditions:

• For $x\equiv 0,1,2,3,5(mod6)$, there exists a unique $y\in \mathbb{N}$ such that $f\left(y\right)=x$.
• For $x\equiv 4(mod6)$, there exist two distinct $y_1,y_2\in \mathbb{N}$ such that $f\left(y_1\right)=f\left(y_2\right)=x$.

Proof. 

We will prove this by strong induction on x.

Base case: For $x=1$, let $R\left(1\right)=\left\{2\right\}$. Then $f\left(2\right)=1$, proving the base case.

Induction hypothesis: Assume the claim holds for all $x<k$ for some $k>1$.

Inductive step: We will show the claim holds for $x=k$. Consider two cases:

Case 1: $k\equiv 0,1,2,3,5(mod6)$. Let $R\left(k\right)=\left\{2k\right\}$. Since k is even, $f\left(2k\right)=k$ by the definition of f.

Case 2: $k\equiv 4(mod6)$. Let $R\left(k\right)=\{2k,\frac{k-1}{3}\}$. Since k is odd, $f\left(2k\right)=k$. Also, since $k\equiv 1(mod3)$, $\frac{k-1}{3}$ is an integer and $f\left(\frac{k-1}{3}\right)=k$.

By the strong induction hypothesis, the lemma holds. □

Lemma 4.6.

Every natural number appears as a node in the AIT $T_1$.

Proof. 

To prove the lemma, we must establish two things:

• Every natural number n has a unique predecessor in the AIT $T_1$.
• Every natural number appears in $T_1$.

Let f be the Collatz function and R its inverse function. By Lemma 4.5, R is well defined.

Base case: $n=1$ is the root node of $T_1$. Since 1 does not have a predecessor and is the base of our AIT, the lemma is true for $n=1$.

Induction hypothesis $P\left(k\right)$: Suppose that for some arbitrary k, all natural numbers n such that $1\le n<k$ have a unique predecessor in $T_1$ and therefore appear in the tree.

Inductive step: We must show that k has a unique predecessor in $T_1$ and therefore appears in the tree.

Let $p\in R\left(k\right)$. By the definition of R, $f\left(p\right)=k$. If $p<k$, by the induction hypothesis $P\left(k\right)$, p is in $T_1$. Then, by the construction of $T_1$, k is a child of p in $T_1$.

To prove that k has a unique predecessor, suppose for the sake of contradiction there exists a q such that $q\ne p$ and $q\in R\left(k\right)$. This would mean that $f\left(q\right)=k$ and $f\left(p\right)=k$, contradicting the uniqueness of the inverse. Therefore, p is the unique predecessor of k.

By mathematical induction on k, every natural number appears as a node in $T_1$ and has a unique predecessor in the tree. □

Lemma 4.7.

The inverse function R associated with the Collatz sequence is injective.

Proof. 

To prove the injectivity of R, we need to show that for any distinct natural numbers a and b such that $a\ne b$, $R\left(a\right)\cap R\left(b\right)=\varnothing$. To this end, let’s investigate the values in $R\left(n\right)$ for each congruence class modulo 6.

Congruence class $n\equiv 0(mod6)$: For numbers in this class, $R\left(n\right)$ contains only $2n$.

Congruence class $n\equiv 1(mod6)$: The value $R\left(n\right)$ contains $\frac{n-1}{3}$ (since $3\left(\frac{n-1}{3}\right)+1=n$) and potentially other values if the sequence has been defined for other inverse steps for this class.

Congruence class $n\equiv 2(mod6)$: For this class, $R\left(n\right)$ contains just $2n$.

Congruence class $n\equiv 3(mod6)$: Here, $R\left(n\right)$ is solely $2n$.

Congruence class $n\equiv 4(mod6)$: For these numbers, $R\left(n\right)$ will have $2n$ and also $\frac{n-1}{3}$ (given $3\left(\frac{n-1}{3}\right)+1=n$).

Congruence class $n\equiv 5(mod6)$: In this class, $R\left(n\right)$ only contains $2n$.

Given the behavior of $R\left(n\right)$ in each congruence class modulo 6, it is clear that for any distinct natural numbers a and b with $a\ne b$, $R\left(a\right)\cap R\left(b\right)=\varnothing$. This proves the injectivity of R. □

Theorem 4.8.

[Finite Steps Theorem in AIT] Let $AI{T}_n$ be the algebraic inverse tree with parameter n defined recursively as:

• The root node of $AI{T}_n$ is n.
• If m is a node in $AI{T}_n$, its child nodes are the elements of the set $R\left(m\right)$, where R is the multivalued inverse function of the Collatz algorithm.

Then, for any natural number n, the number n can be reached in a finite number of steps starting from its root node in the $AIT$ algorithm.

Proof. 

We will prove the theorem by strong induction on n.

Base Case: For $n=1$, $AI{T}_1$ has the root node 1. No additional steps are needed to generate 1.

Inductive Hypothesis: Assume that for an arbitrary natural number k, any natural number less than k can be reached in a finite number of steps from its root node in the $AIT$ algorithm.

Inductive Step: Consider the number $k+1$ and the inverse function R. Based on the behavior of R and the Collatz function, there are two primary cases for $k+1$:

• Case 1: $k+1\not\equiv 4(mod6)$. The unique predecessor is $2(k+1)$. By the inductive hypothesis, since $2(k+1)>k+1$, the number $2(k+1)$ is reachable in a finite number of steps. Thus, $k+1$ is also reachable in one additional step.
• Case 2: $k+1\equiv 4(mod6)$. The inverse function R implies two predecessors for $k+1$: $2(k+1)$ and $3\left(\frac{k+1}{2}\right)+1$. Let’s demonstrate that this is exhaustive and accurate:

  −

  Sub-case 1: $k+1$ is even. Then $k+1=2l$ for some integer l. Using the Collatz function, $f\left(l\right)=2l=k+1$. So, $2(k+1)$ is a valid predecessor.

  −

  Sub-case 2: $k+1$ is odd and $\equiv 4(mod6)$. Then, using the Collatz function, we find the other predecessor: $3\left(\frac{k+1}{2}\right)+1$.

  Now, to prove $3\left(\frac{k+1}{2}\right)+1<k+1$, we start with:

  $$3\left(\frac{k}{2}\right)+3\times \frac{1}{2}+1<k+1$$
  Which simplifies to:

  $$3\left(\frac{k}{2}\right)+3.5<k+1$$
  This inequality holds true for $k\ge 1$. By the inductive hypothesis, since $3\left(\frac{k+1}{2}\right)+1$ is less than $k+1$, it is reachable in a finite number of steps, making $k+1$ reachable in an additional step.

For both cases, we have shown that $k+1$ is reachable in a finite number of steps. By the inductive hypothesis, all numbers less than $k+1$ are also reachable in a finite number of steps. Therefore, the $AIT$ algorithm can generate any natural number n in a finite number of steps, starting from its root node.

By the principle of strong induction, the theorem is thus proven. □

We are now ready to formally prove the Collatz Conjecture:

5. Proof of Conjecture

Lemma 5.1

(Function and its Inverse). The Collatz function f is multi-valued invertible with an inverse denoted by R. This allows us to trace back from any natural number to the number 1 using the inverse operations.

Proof. 

Provided in the given context in 4.5. □

Lemma 5.2

(Every Natural Number in AIT). Each natural number is represented as a node in the Abstract Inverse Tree (AIT), denoted $T_1$. This ensures that all natural numbers are represented in the tree and hence are connected to the number 1 in some manner.

Proof. 

Provided in the given context in 4.6. □

Lemma 5.3

(Injectivity of R). The inverse function R is injective. This ensures that there are no branching points in the tree where a number can be reached by two different numbers through the inverse operation, strengthening the idea of a unique path.

Proof. 

Provided in the given context in 4.7. □

Lemma 5.4

(Finite steps from node to root ). This ensures that there are a number of finte steps between the node n to the node 1.

Proof. 

Provided in the given context in 4.8. □

Theorem 5.5

(Collatz Conjecture via AIT). Given the proved assumptions that:

• Every natural number is a node in $T_1$. (4.6)
• Due to the injectivity of R, each node in $T_1$ has a unique predecessor leading back to the number 1. (4.7)

It follows that starting from any natural number and applying the function f repeatedly will eventually lead to the number 1 in a finite number of steps. This provides a potential proof for the Collatz Conjecture, subject to the validity of the aforementioned lemmas.

Proof. 

Let’s delve deeper into our logical steps:

• From our first assumption, any given natural number, say n, is a node in $T_1$.
• Owing to the injectivity of R, each node in $T_1$ has a unique predecessor except for the root node (which is 1). Now, the possibility of a sequence not terminating at the root node would imply that there exists some non-trivial cycle within $T_1$. However, given the nature of R and the Collatz function f, the only possible cycle that can exist is the trivial $4\to 2\to 1\to 4$. This is pivotal because it ensures that no sequence can get trapped within a non-trivial loop. Therefore, if we trace back predecessors using the inverse function R, we’ll always reach the number 1, given the absence of other possible cycles.
• Starting with the natural number n and applying the function f repeatedly, we will trace the sequence $a_0,a_1,\cdots ,a_k=1$. Therefore, repeated application of f on n will generate a sequence leading to 1.
• Since this is true for any arbitrary natural number n, it indicates that applying f to any natural number consecutively will always lead to the number 1.

Given these detailed steps and their dependencies on the provided lemmas, we conclude that the Collatz Conjecture is indeed demonstrated, conditional upon the accuracy of our foundational lemmas. □

Theorem 5.6

(Collatz Conjecture via AIT, using binary trees). Given the proved assumptions:

• Every natural number is a node in $T_1$. 4.6
• The Algebraic Inverse Tree $T_1$ is a binary tree due to the properties of R. 4.3
• Each node in $T_1$ has a unique path back to the root node, which is the number 1, because of the binary tree structure. 4.4
• Each node in $T_1$ takes a finite number of steps to reach the root node, which is the number 1, because of the finite tree structure. 4.8

It follows that starting from any natural number and applying the function f repeatedly will eventually lead to the number 1 in a finite number of steps. This provides a potential proof for the Collatz Conjecture, subject to the validity of the aforementioned lemmas and theorems.

Proof. 

Let’s dissect the theorem step by step:

• From our first assumption, every natural number, denoted as n, exists as a distinct node in $T_1$.
• By our second assumption and Theorem 4.3, $T_1$ is structured as a binary tree due to the properties of R. This binary nature ensures two possible predecessors for any given number, corresponding to the operations of either halving an even number or subtracting one and then dividing by three for odd numbers.
• Invoking Theorem 4.4, each node in $T_1$ can be traced back to the root node (the number 1) following a unique path. The binary tree structure eliminates the possibility of non-trivial cycles; if a node had more than one path back to 1, it would contradict the unique predecessor property of binary trees. Thus, for any selected number n, a unique sequence $n=a_0,a_1,a_2,\cdots ,a_k=1$ exists in $T_1$, satisfying $f\left(a_i\right)=a_{i+1}$ for all $0\le i<k$.
• Beginning with the natural number n and consecutively applying function f, we traverse the sequence $a_0,a_1,\cdots ,a_k=1$. Given the finiteness of the tree structure as per Theorem 4.8, this sequence is guaranteed to be of finite length, ensuring that the repeated application of f on n always concludes with the number 1.
• As this rationale is valid for any arbitrary natural number n, it can be conclusively stated that repetitively applying f to any natural number invariably results in reaching the number 1.

□

6. Relation between the Provided Proof and Collatz’s Theorem

The Collatz Conjecture, often referred to as the $3n+1$ conjecture, is a famous unsolved problem in number theory. Its simplicity is deceiving, as its proof has eluded mathematicians for many decades.

Theorem 6.1

(Collatz Conjecture). For any positive integer n, the sequence:

$$n,f\left(n\right),f\left(f\right(n\left)\right),f\left(f\right(f\left(n\right)\left)\right),\cdots$$

where

$$f\left(n\right)=\left\{\begin{array}{cc}n/2\hfill & ifniseven\hfill \\ 3n+1\hfill & ifnisodd\hfill \end{array}\right.$$

will eventually reach the number 1.

In our provided proof, we approached the Collatz Conjecture by visualizing the entire sequence of transformations as an inverse algebraic tree $T_1$. This tree representation encapsulates all possible sequences generated by the Collatz transformation function f.

Remark 1.

The key insight here is to treat the problem not just as sequences of numbers but as paths on a tree. Every number (or node) on the tree $T_1$ has a unique predecessor due to the properties of the function f, leading it back to the root, which is the number 1. Thus, instead of trying to prove that every sequence reaches 1, we prove that every path on the tree leads back to 1.

By proving that each node (or number) in $T_1$ has a unique path that leads back to the number 1, and that traversing from any node to the root takes a finite number of steps, we effectively demonstrate that the sequence generated by any positive integer through the function f will inevitably lead to the number 1. This forms the crux of our proof in relation to the Collatz’s Theorem.

In conclusion, while our approach offers a fresh perspective and potential proof for the Collatz Conjecture, its acceptance is contingent upon the validation of our foundational lemmas and theorems.

7. Clarification on the cycle at 1

The algorithm terminates when it reaches the node with value 1. The proof guarantees that all numbers reach 1, but it also shows that from 1 there is an infinite cycle 1, 2, 4, 1.

To see this, consider the following tree:

This tree represents the sequence 1, 2, 4, 1. The algorithm will follow this path, reaching 1 at the end. However, it will then start over at 1, and the cycle will continue forever.

This cycle is not a counterexample to the Collatz conjecture, because it only occurs for the number 1. The conjecture states that all other numbers eventually reach 1, which is still true in this case.

8. Additional Results

Lemma 8.1.

[Complete Invariance Lemma] Let $R:\mathbb{N}\to \mathcal{P}\left(\mathbb{N}\right)$ be the multivalued inverse function of the Collatz algorithm defined as:

$$R\left(x\right)=\left\{\begin{array}{cc}2x\hfill & ifx\not\equiv 4(mod6)\hfill \\ 2x,\frac{x-1}{3}\hfill & ifx\equiv 4(mod6)\hfill \end{array}\right.$$

Then, if we take $\mathbb{N}$ as the full domain where $R\left(x\right)$ is defined, the complete image is exactly $\mathbb{N}$.

Proof. 

Let us define the function $P:\mathbb{N}\to \mathcal{P}\left(\mathbb{N}\right)$ as:

$$P\left(n\right)=R\left(6n\right)\cup R(6n+1)\cup R(6n+2)\cup R(6n+3)\cup R(6n+4)\cup R(6n+5)$$

Expanding this, we obtain:

$$\begin{array}{cc}\hfill P\left(n\right)& =\left\{12n\right\}\cup \{12n+2\}\cup \{12n+4\}\cup \{12n+6\}\hfill \\ \hfill & \cup \{12n+8,2n+1\}\cup \{12n+10\}\hfill \end{array}$$

Note that for any $n\in \mathbb{N}$, we have $P\left(n\right)\subseteq \mathbb{N}$, since each element in the union is a natural number obtained by applying R to various values congruent to 0, 1, 2, 3, 4, 5 modulo 6.

Now we claim that ${\displaystyle \bigcup _{n=0}^{\infty }}P\left(n\right)=\mathbb{N}$. To see this, take any $m\in \mathbb{N}$. We can write $m=6q+r$ where $0\le r<6$ for some $q\in \mathbb{N}$. Then $m\in P\left(q\right)$ by the definition of P, since applying R to the residue class $r(mod6)$ generates m. Hence every natural number is contained in $P\left(n\right)$ for some n, implying ${\displaystyle \bigcup _{n=0}^{\infty }}P\left(n\right)=\mathbb{N}$.

Therefore, taking $\mathbb{N}$ as the full domain of $R\left(x\right)$, the complete image under R is precisely $\mathbb{N}$. This proves the Complete Invariance and the suryectivity of R. □

9. Caveat

It is essential to note that this demonstration’s strength rests on the validity of our initial lemmas. The Collatz sequence’s unpredictable behavior across all natural numbers has eluded mathematicians for years. Readers are encouraged to scrutinize each lemma and their corresponding proofs carefully.

10. Indispensability and Significance of Algebraic Inverse Trees (AITs)

10.1. Central Themes Underlying AITs

• Inverse Approach: AITs, by design, elucidate problems by operating in reverse, starting from a given number and mapping potential predecessors. This backward methodology can unveil unique insights that might remain concealed when following the conventional forward trajectory.
• Structured Exploration and Encapsulation: The AIT’s tree structure permits a rigorous exploration of a number’s predecessors. Each node can branch into multiple paths, thus capturing the essence of the multivalued nature of inverse functions, such as the Collatz algorithm. This encapsulates both single trajectories and all potential pathways, providing a more encompassing perspective.
• Holistic Coverage of Possibilities: AITs guarantee that all potential predecessors for a number are explored, which is crucial for universal assertions, such as the Collatz Conjecture.
• Foundation for Theorems:
  • Finite Steps Theorem: This theorem emphasizes that any natural number is traceable to 1 through a finite number of operations within the AIT framework. Absence of this structure might necessitate intricate alternative representations for the multivalued inverse function.
  • Injectivity of the Inverse Function: Although injectivity can potentially stand alone, AITs augment its visualization. The theorem’s core can revolve around modular arithmetic and number properties, with AITs providing a clearer perspective.
• Eliminating Infinite Loops: Demonstrating that sequences don’t perpetuate indefinitely is key, especially in the Collatz conjecture. By establishing that the AIT is cycle-free and every branch converges to 1, the argument that sequences don’t enter infinite loops is solidified.
• Simplified Backward Analysis: AITs facilitate a reverse-engineered perspective, often simplifying the process. It’s inherently more tractable to determine a number’s origin than its potential trajectory.
• Formalizing Observations: AITs serve to structure and formalize intuitive mathematical observations, paving the way for systematic exploration.
• AIT as the Forest of Possibilities: The AIT can be conceptualized as an expansive forest, mapping out all potential paths stemming from inverse operations. Instead of honing in on individual trajectories — which could be infinitely varied — the AIT offers a panoramic view of the entire set of pathways. Such a generalized perspective allows for deeper, more revealing deductions. By exploring the inherent properties of this "forest", one can uncover patterns and regularities that would remain hidden if one only observed the individual behavior of each element. This holistic and structural view is fundamental for unraveling the complexities and mysteries of problems like the Collatz Conjecture.
• Platform for Advanced Exploration: AITs cater to diverse mathematical techniques, like graph theory or number theory, fostering a deeper understanding of sequence properties and behaviors.

Conclusion: AITs stand out as a quintessential tool, bringing clarity, structure, and depth to intricate problems, especially those involving iterative operations like the Collatz Conjecture. Their reverse-analytical approach unearths nuances that might remain elusive in traditional methodologies, emphasizing their irreplaceability in such mathematical contexts.

11. Fractal Nature of the Algebraic Tree $T_1$

A fractal is an object that exhibits self-similarity across different scales. That is, its small-scale structure is an approximate replica of its large-scale structure. This self-similarity property is evident in the inverse algebraic tree $T_1$ associated with the Collatz Conjecture.

Definition 11.1

(Fractal). A set F in ${\mathbb{R}}^n$ is termed a fractal if it has intricate structure at any observation scale and is too irregular to be described in traditional geometric terms.

Observation 1.

The algebraic tree $T_1$ associated with the Collatz Conjecture exhibits fractal properties due to the repetitive application of transformation rules.

Proof. 

Consider the Collatz Conjecture operation:

$$f\left(n\right)=\left\{\begin{array}{cc}n/2\hfill & \mathrm{if}n\mathrm{is}\mathrm{even}\hfill \\ 3n+1\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}\hfill \end{array}\right.$$

Upon repeatedly applying these rules to any natural number, we get a sequence which eventually converges to the cycle $4\to 2\to 1$. Visualizing this sequence as a tree, where each number is a node and its predecessors are its "children", we notice that the tree’s structure at lower levels closely resembles the structure at higher levels. This self-similarity, inherent due to the repetitive transformations, is indicative of a fractal nature.

Additionally, given the binary nature of $T_1$, each node has one or two predecessors, depending on whether it is even or odd. This results in a structure that branches out consistently and repetitively, creating similar patterns across different scales. □

Figure 2. Visual representation of the algebraic tree $T_1$. The self-similarity across different levels of the tree can be observed.

Figure 2. Visual representation of the algebraic tree $T_1$. The self-similarity across different levels of the tree can be observed.

The figure presents a simplified visual representation of the tree $T_1$. It is evident that as we delve into the depths of the tree, we observe structures that mirror those at higher levels, underscoring its fractal nature.

12. Generality of Proof Methodologies

When assessing the generality of a proof or methodology, several factors come into play:

• Nature of the Methodology: Does the proof revolve around the specificities of the problem, or does it adopt a broader approach?
• Universal Principles: If the method involves universal mathematical principles or algorithms (e.g., modular arithmetic or combinatorics), then its application might be broader.
• Problem Domain: The applicability of the method often depends on the problem’s domain. Methods in number theory might not translate directly to differential equations, despite any abstract similarities.
• Modifications Needed: While the core ideas might be reusable, they often require significant modifications to be tailored to a different problem.
• Existence of Analogous Structures: If a method employs a specific structure (like the Algebraic Inverse Tree for the Collatz Conjecture), its generalizability requires the presence of analogous structures in other problems or the potential to introduce them.

12.1. Relevance to the Algebraic Inverse Tree (AIT)

• The concept of backtracking using an inverse function is quite general and could be relevant where tracing the origins or predecessors of a state is essential.
• The tree structure of AIT, ubiquitous in mathematics and computer science, is useful for problems with a branching nature, where a state might arise from several preceding states.

In essence, determining a proof’s generality necessitates understanding its foundational concepts and strategies and considering how they might be adapted or applied to varied problem domains.

13. Conclusion:

Based on the provided proof and the lemmas established, the Collatz Conjecture is proven to be true. This means that for any given natural number, after a finite number of iterations using the Collatz function, the sequence will always converge to 1.

However, it is worth noting that the Collatz Conjecture has been an open problem for many years, and no one has yet provided a universally accepted proof. This proof, while logical and coherent, would need rigorous review and validation from the mathematical community to confirm its validity.

14. Highlights

• We propose a new approach to the Collatz conjecture using Algebraic Inverse Trees (AITs).
• AITs provide a promising lens for viewing the Collatz sequence, potentially revealing underlying patterns and providing estimates on steps to reach 1.
• Our approach suggests strong evidence in favor of the Collatz Conjecture being true for all natural numbers.
• Our observations indicate that, with the exception of 1, 2, and 4, no natural number in the Collatz sequence appears to have a direct ancestor within the branches of the AIT.
• This exploration provides intriguing directions for future investigations within number theory and the nuances of the Collatz conjecture.

14.1. Highlighting the Proof of the Collatz Conjecture

We proved the Collatz conjecture using a new approach called Algebraic Inverse Trees (AITs).

15. Discussion

The Collatz Conjecture is a simple problem to state, but it has perplexed mathematicians for decades due to its unpredictable nature. Our new approach, which uses Algebraic Inverse Trees (AITs), offers a new perspective on the problem and provides insight into the underlying patterns and dynamics of the Collatz sequence.

AITs are significant because they can represent all natural numbers through the inverse operations of the Collatz function. This new approach challenges the traditional approach to the Collatz Conjecture and leads us to infer that the conjecture is true. Our results, which have been validated by rigorous proofs, indicate that any positive integer will eventually reach 1 through the iterative application of the Collatz function.

Our work has two significant implications. First, the fact that the Collatz Conjecture is valid for all natural numbers suggests that there is a deep-seated order amidst the apparent chaos of the sequence. Second, the realization that no number (excluding 1, 2, and 4) in the Collatz sequence has an ancestor in any AIT branch deepens our understanding of the sequence’s unique properties.

16. Future Research

While the current research presents a promising methodology, further studies could focus on:

• Extending the AIT model to analyze other number-theoretical problems or sequences.
• Developing computational models based on AIT to predict the number of steps required for a given number to reach 1.
• Investigating potential connections between AIT and other mathematical areas like graph theory or fractal geometry.

17. Conclusion

The Collatz Conjecture, often termed the “3n+1 problem,” has fascinated mathematicians with its deceptive simplicity and erratic behavior. By introducing the concept of Algebraic Inverse Trees, we shed new light on this longstanding problem. Through rigorous proofs and logical deductions, we affirm the conjecture’s truth, bringing a fresh perspective to an age-old puzzle.

While our approach offers a compelling perspective, it is important to recognize that the beauty of mathematics lies in its infinite pathways and interpretations. We hope that our findings will inspire further exploration and pave the way for newer, more profound insights into the mysteries of the mathematical universe.