Theoretics On Hailstone Numbers

Theoretics On Hailstone Numbers

A Lesson by cyphertext
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This is an unsolved number theory problem called the Collatz Conjecture, if my hypothesis is true, (evey odd and even positive integer greater than one will eventually become a power of two under the conditions of the Collatz Conjecture), then every positive integer is a hailstone number.

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Residue Theory and the Collatz Conjecture (Hailstone Numbers Problem)

By: Steven Zheng

15/02/2011

Introduction

            In 1937, mathematician Lothar Collatz asked whether any positive integer will always reach the number one after a succession of operations are applied, one being to divide the number when it is even, two being to triple the number then add one when it is odd. Until this day, the conjecture is still unsolved but most mathematicians expect the answer to be affirmative. Current computer assisted research has verified the first 5.48×10^18 positive integers reach one. (2009).

Purpose

 To prove the Collatz conjecture by verifying the following:

1)Any integer will not diverge into the infinite under the operations of the Collatz conjecture.

2)Digit loops do not exist before reaching the integer one.

Common Notions

1) An even number divided by two will result an even number if it is not composed of an even number followed by two (ex. 62, 122, and 4082).

2) An odd multiplied by an odd number then summed by an odd number is even.

3) Powers of two are hailstone numbers under the conditions of the Collatz Conjecture.

Assumptions

1) Any integer will not diverge into the infinite under the operations of the Collatz conjecture.

2) Digit loops do not exist before reaching the integer one.

Axioms

1) One is the smallest positive integer.

2) Every positive odd number can be described by the equation 2n+1.

            Every positive even number can be described by the equation 2n+2.

Parameters

1) The Collatz conjecture is restricted to the positive integers.

2) The two operations are n/2 for every even number and 3n+1 for any odd number.

3) When the number sequence reaches one, stop.

4) The order type of the positive integers is ω (a definite minimum but no definite maximum).

5) The cardinality of the positive integers is _0.

6) The number one is not accounted in any set of positive integers because it is the theorized “last number” in all sequences.

Hypothesis

If every odd and even positive integer greater than one will become a power of two under the conditions of the Collatz Conjecture, then every positive integer will be a hailstone number.

Part 1: Turbulent Operations

Definition: Turbulent operations are a set of operations that prevent a counting set to diverge to the infinite.

 

Let there be a counting set of integers {Z} and an operation *.

{Z}* will never approach ±∞.

Postulates

1) Turbulent operations can be made not turbulent by a change of the set and vice versa. (Obvious/ Strong)

Proof- Let a set of real numbers {R}. When {R is placed under the operation * = n/2, where n is an element of {R}, {R} will approach the infinitely small. When the set {R} changes to a set of positive integers {Z+}, the minimum element is restricted to the number one, therefore the operation terminates when it reaches one.

2) None turbulent operations can be made turbulent by introducing a complementary non-turbulent operation. (Weak)

Proof-In the Collatz conjecture, there exists two operations: n/2 for every even positive integer and 3n+1 for every odd positive integer.

            3n+1 is a non-turbulent operation in the set {Z+}

            n/2 is a turbulent operation in the set {Z+}

For every sufficiently large odd number in the set {Z+}, the operation 3n+1 will force n to become an even number. For every even number, it is guaranteed that the number can be divided by two at least once.

 

Complementary Argument

Let n be a positive odd integer.

(3n+1)/2>n

(3n+1)/4<n

(3n+1)/2^(m+1) <n,where m is a positive integer

There are an infinite number of odd numbers that when the Collatz operation 3n+1 is applied, the new even number can be divided by two more than once. Thus, the number n will be reduced to a value less than itself, preventing it from reaching infinity.

Part 2

2^n numbers - positive odd integer continuum

Statement 1: If all positive odd integers eventually reach a 2^n number, then the sequence will not diverge into the infinite because a 2^n number will reach the number one in (n – 1) steps.

Statement 2: If all positive odd integers eventually reach a 2^n number, then a digit loop will not occur until it reaches one because a 2^n will become a smaller 2^n number each step.

The new focus is now the universal link between all positive odd numbers and powers of two.

Part 3: Theory of Residues and the continuum

Definition

1) Residue numbers are numbers that any value within the realms of a set can be evaluated by all turbulent operations.

2) The residue operation is a reduction (division) operation that differs from the modular operation where a separate operation is first applied to a value.

Division: N/ m

Modular: Na(mod m)

Residue: (N)* R.N.(mod m)

Let the set of R.N. be defined by the function R(x).

Derivation of R(x) = 6x+4 for the Collatz conjecture

It is known that any odd number multiplied by three then added to one will result an even product.

Fact: The number four is the smallest residue number because 8/ 2 = 4 and 3(1) + 1 = 4.

The operation * is 3n+1 for all odd numbers.

All odd integers can be defined by the equation 2(x) +1

3(odd number)+14 (mod m),where m is 3

3(2x+1)+14 (mod 3),       x{Z+}

R(x)=6x+4

1          3          5          7          9          11        13        15        17        [R(x)-1]/3

4          10        16        22        28        34        40        46        52        R(x)

Notice every positive odd number is directly linked to a residue number under the operation 3n+1.

Questions

Question 1: Do all residue numbers lead to a different residue number?

Answer: Yes

Proof- 1§ [6(even number) + 4]/2 = [6(2n+2) + 4]/2 = 6n + 8 ,6n+8 Even numbers

(6n + 8)/2 = 3n + 4: n = even number (2k + 2) or odd number (2k + 1)

Even number: 3(2k + 2) + 4 = 6k + 10, 6k+104 (mod 6)

Odd number: 3(2k + 1) + 4 = 6k + 7,6n+7 Odd numbers

3(6k + 7) + 1 = 18k + 22, 18k+224 (mod 6)

2§ [6(odd number) + 4]/2 = [6(2n+1) + 4]/2 = 6n + 5 ,6n+5 Odd numbers

3(6n+5) + 1 = 18n + 16, 18n+164 (mod 6)

                                                                                                                        QED

Question 2: Do all residue numbers lead to a power of two?

Answer: Indeterminate

Some powers of two are residue numbers.

Let 2^n  be set S and R(x)=6x+4 be set T

ST=4,16,64,256,1024,4096,16384 , (2^(2k+2),k{W})

S-T=2,8,32,128,512,2048,8192,, (2^(2k+1),k{W})

Congruence of R(x) = 6x + 4 and 2^n (The Final Continuum)

Proposition 1

(6x+42^x)=(2^(2n+2),n{Z+})

2^(2n+2)=6(_0^(k=n-1)▒2^(2k+1))+4, where n{Z+}

Proof- Use base two arithmetic

            2^(2n+2)=(0*2^0 )+(0*2^1 )+(0*2^2 )++(1*2^n ),where n is an even integer

Or 1000…000 (base 2)

            (_0^(k=n-1)▒2^(2k+1))= (0*2^0 )+(1*2^1 )+(0*2^2 )++(0*2^(n-1) )+(1*2^n )where n is an odd integer

Or 1010…1010 (base 2)

 

Let x[y] n denote x followed by string y n times to the right.

Let n[x]y denote y followed by string x n times to the left.

Let [x] n denote the string x repeated n times to the right.

Let n[x] denote the string x repeated n times to the left.

 

Then 2^(2n+2)=(1[0]2n+2)base 2 and (_0^(k=n-1)▒2^(2k+1))= (n[10])base 2

(n[10])base 2*(110)base 2=(2n[1]00)base 2

(2n[1]00)base 2+(100)base 2=(1[0]2n+2)base 2

Therefore

2^(2n+2)=6(_0^(k=n-1)▒2^(2k+1))+4, where n{Z+}

QED

This proof unites every power of two to the specific type of R(x) described above because every even power of two is even, thus the division by two is dictated by the second operation of the Collatz conjecture, resulting in an odd power of two.

Proposition 2

Every odd power of two is not congruent to 1 modulo 3 but 2 modulo 3.

Proof- Use base two arithmetic

2^(2n+1)=(1[0]2n+1)base 2

(1[0]2n+1)base 2-(1)base 2=([1]2n+1)base 2

([1]2n+1)base2 / (11)base 2 =([n10])base 2 Remainder 1

Therefore 2^(2n+1)2(mod 3)                                                         

                                                                                                                        QED

The significance of this proof states that no positive odd integer corresponds to an odd power of two in one step of the Collatz conjecture.

Proposition 3

All residue numbers R(x) divided by two are congruent to 2 modulo 3.

Proof

R(x)=6x+4

R(x)/2=3x+2

2^(2n+1)=3(_0^(k=n-1)▒2^(2k+1))+2, where n{Z+}

                                                                                                                        QED

Specific values of residue numbers divided by two are odd powers of two.

As proven above, a residue number R(x) will always lead to a different residue number if it is not equal to an even power of two. Since there is infinitude of residue numbers and powers of two, all residue numbers that are not an even power of two will eventually become one under the operations of the Collatz conjecture.

Summary

1) Every positive even number will become an odd number when it follows the operation so of the Collatz conjecture. Consequently, all positive even integers are unified with the set of positive odd integers.

2)6n+4,n{Z+}  The function of all positive odd numbers of the Collatz conjecture, called the residue numbers. Hence the function binds every odd integers to a residue number.

3) Every residue number will lead to a different residue number following the operations of the Collatz number.

            2^(2n+2)=6(_0^(k=n-1)▒2^(2k+1))+4  and 2^(2n+1)=3(_0^(k=n-1)▒2^(2k+1))+2, where n{Z+}

4) There exists an infinite number of residue numbers and powers of two. According to point number three, every residue number will lead to a different residue number under the Collatz conjecture. Therefore, every residue number will eventually become an even power of two.

5) Every power of two where the power is a positive integer will terminate at the number one and never diverge into the infinite when it follows the operations of the Collatz conjecture.

Conclusion

            Intuitively, according to the summary, every positive integer greater than one will reach the number one under the turbulent operations of the Collatz conjecture before it enters an eternal loop of digits and without diverging into the infinite.

Sources

            Stewart, Ian. Professor Stewart’s Hoard of Mathematical Treasures. London: PROFILE BOOKS LTD. 2009

            Alex Lopez-Ortiz. Collatz Problem. University of Waterloo. 28 Jan 2011 http://www.cs.uwaterloo.ca/~alopez-o/math-faq/node55.html

            Collatz Problem. Wolfram Mathworld. 28 Jan. 2011 http://mathworld.wolfram.com/CollatzProblem.html

 



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Posted 5 Years Ago


I'm amused to see the word "intuitively" used in a discussion of mathematical theory, but this has nothing to do with writing, which is what the lessons are supposed to be about.
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