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发表于 2025-9-2 23:30
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Proof of the Collatz Conjecture
Author:Yang Yanhong,Yulong County,Lijiang City,Yunnan Province
Email:13312690681m@Sina.cn Lei Zhigang
Abstract:The Collatz Conjecture involves an iterative operation on a number:if the number is even,divide it by 2;if it is odd,multiply it by 3 and add 1.Will this process eventually reach 1?This proof expresses a number in binary form,folds it,and analyzes the structure of the number in the form of 0 and 1,combined with the I Ching(Book of Changes)and the Bagua(Eight Trigrams),to conclude that the Collatz rule is equivalent to a Möbius strip.
Keywords:Collatz Conjecture,Collatz Rule,Life Formula,Equal Transformation;I Ching,Bagua,Qian(Heaven),Möbius Strip
Starting from the operation rule of the Collatz Conjecture,suppose there are numbers\(S+1\)and\(S-1\)that follow the same Collatz operation rule(\(S\)is an integer),then:
\[A=3(S+1)+1\]
\[T=3(S-1)+1\]
\[A+T=6S+2\]
According to the Collatz operation rule,since\(A+T\)is even,it should be divided by 2.This is denoted as the Collatz operation rule:
\[L(S)=\frac{A+T}{2}=3S+1\]
​Starting from the 3X+1(Collatz operation rule)when solving-5,-7,-17,the calculation will enter a loop when repeatedly executed.Starting from the negative number operation rule,modify the Collatz operation rule:for negative odd numbers,repeatedly execute 3X-1;for even numbers,divide by 2.
\[G=3(S''+1)-1\]
\[C=3(S''-1)-1\]
\[G+C=6S''-2\]
Denote the rule as:
\[F(S'')=\frac{G+C}{2}=3S''-1\]
Then:
\[A+G+T+C=2[L(S)+F(S'')]\]
Represent an integer Y as:
\[Y=\log(N\cdot 1/N\cdot X)\]
Then:
\[-Y=-\log(N\cdot 1/N\cdot X)\]
\[A+T=6Y+2\]
\[G+C=6(-Y)-2\]
\[L(S)+F(S'')=3\log N+3\log(X/N)+1+3\log N+3\log[1/(NX)]-1\]
\[=6\log N+3\log(1/N\cdot 1/N)\]
\[=6\log N-6\log N=0\]
Thus:
\[A+T+G+C=L(S)+F(S'')\]
That is,Theorem 1:The property remains unchanged after folding.
Write the number S in binary and then fold it.There are four forms corresponding to the starting number:0,1,10,11.By writing the Bagua in a left-right structure instead of the traditional top-bottom structure,the correspondence between the 64 hexagrams and the AGCT genetic code can be derived.
​○○A
●○C
●●G
○●T
AAA corresponds to Qian in the 64 hexagrams.
AAA
○○
○○
○○
GGG corresponds to Kun in the 64 hexagrams.
GGG
●●
●●
●●
In the horizontal row,there are the eight trigrams:
○○○
A○
○○●
A1
●○○
C○
●○●
C1
○●○
T○
○●●
T1
●●○
G○
●●●
G1
Qian represents Heaven
AAA
○○
○○
○○
Equivalent to
AOAO
oooooo
Qian represents Heaven
AAA=AOAO
Starting from the equation\(\text{AAA}=\text{AOAO}\),
we derive the following:
\[x^3=2x^2\quad(1)\]
\[3x=2x+2\quad(2)\]
From the life formula(equivalent transformation),
\[A=T\quad\text{and}\quad G=C\]
Thus,
\[A+G=T+C\]
which implies
\[x^3+3x=2x^2+2x+2\]
Rearranging,we get
\[3x+1=2x^2+2x+3-x^3\]
When\(x=2\),the equation holds true.
Substituting\(x=3\)into the equation,
we get
\[10=0\]
Checking 10,we have the sequence:
\[10\rightarrow 5\rightarrow 16\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1\]
Thus,10,when checked through the Collatz conjecture,also eventually reaches 1.
If\(x=3\),we can calculate:
\[10=0\]
Checking out 10,we get:
\[10\rightarrow 5\rightarrow 16\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1\]
Therefore,10 will also fall to 1.
For negative numbers,consider the sequence starting from-10:
\[-10\div 2=-5\]
\[-5\times 3+1=-14\]
\[-14\div 2=-7\]
\[-7\times 3+1=-20\]
\[-20\div 2=-10\]
\[-10\div 2=-5\]
For an infinitely large number,\(\infty\),take the reciprocal:
\[S=\frac{1}{\infty}\]
which is denoted as
\[S=000000\]
The I Ching trigrams of Qian with an upper-lower structure and those with a left-right structure are equivalent.
It can be known that:
000000
○○
○○
○○
Thus,AOAO=AAA
That is,\(2A^2=A^3\)①
\(2A+2=3A\)②
Adding ① and ②:
\(A^3+3A=2A^2+2A+2\)
\(3A+1=2A^2+2A+3-A^3\)③
When\(A=2\),equation ③ holds true.
When\(A=3\):
\(3A+1=2A^2+2A+3-A^3\)
\(10=0\)
This is denoted as\([ten=O]\).
That is to say,when using a computer to calculate the Collatz conjecture,when the number reaches infinity,the memory overflows,and it cannot be verified by a computer.
For example,20 written in binary is 10100.When folded,there are two possible starting digits:1 or 0.
The number 20 written in binary can be mathematically expressed as\(S=E(s)\).
• When starting with 1,\(S=E(s)\)where\(s=2\mod(3)\).
• When starting with 0,\(s/2\)where\(s=0\mod(2)\).
By substituting\(s=0\),\(s-1\),and\(s+1\)into the Collatz rule,we can determine that\(A=3(0+1)+1\)and\(T=3(0-1)+1\).Therefore,\(A+T=2\).
It follows that there exists a Collatz rule\(L(0)=1\).This means that if you write 0 on one side of a strip of paper and 1 on the other side,and then twist and join the ends,you will create a Möbius strip,which illustrates the Collatz operation rule.
The Collatz conjecture involves converting a number to binary and folding it.There are four possible starting digits in binary:1,0,10,and 11.Assuming that the Collatz operation iteration eventually returns to the starting number,and given\(L(0)=1\)(i.e.,0 and 1 are equivalent),and since binary 10 equals 2,we can conclude that\(A=2\),thus establishing Equation ③.
Given that binary 11 equals 3,which is denoted as\(G\),the binary representation of-5 should be in two's complement form,which is 1111 1011(the original code is 1000 0101,the one's complement is 1111 1010,and the two's complement is 1111 1011).
When-5 is folded in binary and represented symbolically,it is\(GCGG\).This indicates that\(X=-5\)has entered a loop of-10→-5→-7→-20→-10.
References:
[1]I Ching
[2]Asimov's New Guide to Science-The Formula of Life
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