aa r X i v : . [ m a t h . G M ] M a r A NEW METHOD TO PROVE THE COLLATZ CONJECTURE
DANIAL KARAMI
Abstract.
The Collatz conjecture is an unsolved problem in mathematicswhich introduced by Lothar Collatz in 1937. Although the prize for the proofof this problem is 1 million dollar, nobody has succeeded in proving this con-jecture. However in this article, we will discuss the results of the author’sresearch and come closer to the proof of this conjecture. Introduction
Collatz conjecture which is an unsolved problem in mathematics, explains abouta sequence which can be define as follows: ”Start with any positive integer, if thatwas odd, triple it then add one to the consequence but if that was even divide itby 2. Anyway this conjecture explains that the value of the selected number is notimportant, at the end the sequence will reach one.”This conjecture has been checked by computer test till 5 × so most of thescientists believe that probably this conjecture is true about all of the naturalnumber, and this possibility is very much. But till now, nobody has succeeded tomention a proof which can convince us that this conjecture is correct or not. Lemma 1.1.
Actually the configuration of this unsolved problem in mathematicscan be viewed as: any chosen number several converts −−−−−−−−−−→ Theorem 1.2.
Choose a natural number, then do the below operation on it:If it was even, divide that number by two.If it was odd, triple that number and then add one.To show this conjecture as a function f, It can be define as follows: f ( n ) = ( n/ if n ≡ n + 1 if n ≡ Now by performing this action for many times, implement a sequence. Beginwith any Natural number and take the result of each step as the input for the nextstep. In notation: a i = ( n for i = 0 f ( a i − ) for i > Such that:(Amount of f applied to n i times, is equal with a i ; a i = f i ( n ) .) [1] Example 1.3.
Lets check some examples about it:17 → → → → → → → → → → → → Mathematics Subject Classification.
Primary 11Y16, 97A20; Secondary 33D99, 68W01.
Key words and phrases.
Collatz conjecture, Lothar Collatz, 3n+1, n/2, Function.
Definition 1.4.
Here the modified form of Collatz function discussed as: f ( n ) = ( n/ n ≡ n + 1) / n ≡ . [2]Accordant to the above function, always the consequence of 3n+1 is even suchthat n ∈ N . To realize that why always the result 3n+1 is even in condition that n = 2k+1, let’s see the following paragraph: Proof.
In mathematics, the odd numbers can be shown as 2k+1 or 2k-1. Addition-ally in this conjecture, we have to do 3n+1 just in condition that n = 2k+1. Sowith replacing 2k+1 instead of n , The formulation of the 3n+1 will be define as;3(2k+1)+1 = 6k+4 = 2(3k+2) ⇒ (cid:3) ”About the following function: T ( x ) = ( x/ x ≡ x + 1) / x ≡ Z loop of 2-adic integers is persistentand measure preserving. In fact, its dynamics has recognised to be ergodic. Youcan show the parity vector function Q which perform on Z as: Q ( x ) = ∞ X k =0 (cid:0) T k ( x ) mod 2 (cid:1) k As a result, the above function Q is a 2-adic isometry. So precisely, all of theinfinite and unlimited parity sequence just happens for one 2-adic integer, so that in Z , almost all of the routes are acyclic. Anyway now the equivalent and same formfor the formulation of this problem in mathematics can be viewed as: Q ( Z + ) ⊂ Z .”This section has been widely studied more in [3] Iteration on the real or complex digits : ”As restriction to the numbers ofthe smooth real and complex map, Collatz conjecture map can be seen as: f ( z ) = 12 z. cos (cid:16) π z (cid:17) + (3 z + 1) . sin (cid:16) π z (cid:17) If in standard Collatz map which has been explained above, we improve it byreplacing (3n+1)/2 instead of 3n+1, this map can be shown as; f ( z ) = 12 z. cos (cid:16) π z (cid:17) + (3 z + 1)2 . sin (cid:16) π z (cid:17) Chamberland examined his opinion about the repetition on the real line in 1996.He demonstrated that when there are many infinity fixed points, this conjecturedoesn’t hold for the real numbers. Furthermore, the orbits and cycles escapemonotonously to boundlessness.” For more details study [4].
Definition 1.5. ”The result of 3k+1 is always equal with an even number, unlessthe situation that k = 2˝k (proof is in definition 1.4). So with a ≥ a ´k. From the set I of odd integers into function f such that f ( k ) = ´ k ,syracuse function will appear. Some properties of this function are: • ∀ k ∈ I ; f(4k+1) = f(k). Because 3(4 k + 1) + 1 = 12 k + 4 = 4(3 k + 1) • in more generalization: ∀ p ≥
1, h = 2k+1 ; f p − (2 p h-1) = 2 × p − h-1. (Here f p − is function iteration notation) • ∀ h = 2k+1 ; f(2h-1) ≤ h − The Collatz conjecture is commensurate to express that, for every k ∈ I , thereis an integer n ≥
1, s.t. f n (k) = 1.” the accurate information is in [5] NEW METHOD TO PROVE THE COLLATZ CONJECTURE 3
Anyway, the author of this article tends to study this conjecture conversely.Here in the main result of this paper, some particular patterns have been mentionedwhich every numbers would pass them in their approaching to 1. In fact, in contrastthat each number has its own algorithm, somewhere they have to follow some sameparticular edict in their approaching to 1.2.
Main result
Theorem 2.1.
To find a step before , lets lead 1 to other numbers by doing theformulation of the conjecture backward. The author of this paper concluded that astage before , certainly, there are n s such that n ∈ N .Proof. n n ÷ −−−−→ n − n − − − ... (for n times) −−−−−−−−−−−−−−−→ n − n = 2 = 1Dividing 2 n by 2 will continue until the number of − n , so after that we have 2 n − n = 2 = 1. (cid:3) Definition 2.2.
This section tends to study about a step before 2 n s. In thisconjecture, the numbers before 1, are in two categories.The first group can be define as: { n | n ∈ N } , on the other hand the secondone can be define as: { x ∈ N | x =2 n , x =1 } .In contrast to the second group, the first group would arrive to 1 directly justby dividing them by 2 (proof in theorem 2.1). As a result, it is necessary to relateother numbers in the second group with 2 n s because a step before 1 there are 2 n s.Now lets lead 2 n s to other numbers. because 2 × i = 2 i +1 = 2 n , we wont getany result except 2 n by multiplying 2, i times by 2 and due to the theorem 2.1,2 n s have been mentioned as one stage before 1. So the only way to lead 2 n s toother numbers, is subtracting 1 from 2 n s, and then divide them by 3, in fact Thebackward form of 3n+1. (except 2 because − = 1 and 1 is our destination nota step before 2 n s). Proposition
To identify a step before 2 n s, if we divide 2 n − Proof.
Proof by contradiction : Instead of considering that (2k+1)/3 = 2´k+1,consider that (2k+1)/3 = 2´k+1 so (2k+1)/3 = 2˝k. Now we have: k +13 = 2˝k ⇒ ⇒ ⇒ k ′′′ Exactly the last step has a contradiction. It shows that ¬ ( k +13 = 2´k+1) is falsebecause 2k+1 = 2 k ′′′ , so the phrase of k +13 = 2´k+1 is true. (cid:3) Remark . Regarding to definition 2.2, there is an important question. Does every2 n -1 divisible by 3? First of all, to answer to this question, lets see the followingalgorithm for better understanding: { − = 3 a , 2 − b , 2 − = 3 c , 2 − d , 2 − = 3 e , ... } As you can see in the above set of numbers, just 2 k − m −
1, isdivisible by 3 not 2 k +1 − Proposition k − m -1 is divisible by 3, for any integer n ∈ N . Proof.
Let S m = 4 m -1 = 3 r such that r ∈ N . In condition that k = m , we wouldhave S k = 4 k − r and we must show that for the next turn which is S k +1 ,4 k +1 − t such that t ∈ N . The reason is that, here m is the representor of thenatural numbers. D.KARAMI
Anyway lets multiply S k by 4 and then add 3 to the consequence:4 × (4 k -1)+3 = 4 × (3 r )+3 ⇒ k +1 -4+3 = 12 r +3 → k +1 -1 = 12 r +3And by factoring out a 3, the result can be viewed as:4 k +1 − r +1) ⇒ t = 4 r +1Here 4 k +1 − r +1) which it is divisible by 3, and the proof is over. (cid:3) Proposition k +1 -1 is not divisible by 3 for any integer k ∈ N . Proof.
According to the previous proof, 2 k -1 = 3 r and it is divisible by 3 such that r ∈ N . Anyway lets multiply this equality by 2:2 × (2 k −
1) = 2 × (3 r ) ⇒ k +1 -2 = 6 r and now lets add 1 to the both sides of the equality:2 k +1 -2+1 = 6 r +1 ⇒ k +1 -1 = 6 r +1As a result, 2 k +1 -1 is not divisible by 3 because r +13 = j such that j ∈ N . Sothe proof comes to an end. (cid:3) Due to the proofs which were written in remark 2.3, just before 2 k = 4 m , couldbe there an odd number. So whenever 3 × (2k+1)+1 = 4 m , that odd number (2k+1),is a step before 2 n s. But here there is a exception and that is 1, the reason is that,when you arrive to 1, as conjecture said, you should’t do any math operation on it;also as it wrote at the end of the definition 2.2, 1 is the destination of this conjecturenot a step before 2 n s . Definition 2.4.
Till now, this article shows that a stage before 2 n s , there existthe odd numbers. So any chosen number except 2 n and 1, would arrive to one ofthem in a step before 2 n , and then by doing 3 n +1 on it, it will reach 2 n s ; afterwardregarding to the theorem 2.1, it will arrive to 1 clearly. So the form of the Collatzconjecture till now can be viewed as:... → × −−−−→ m = 2 n ( ÷
2) for n time −−−−−−−−−−→ n +1 = 2˝k+1, n must be an even numberand it is unacceptable. In this conjecture there is no permission to do 3 n +1 whenwe have an even number. As a result, before any odd number there is an evennumber. So lets improve the above configuration by adding (2 k ) before 2´k+1:... → k ( ÷ −−−−−−−−−−−−−→ × −−−−→ m = 2 n ( ÷
2) for n time −−−−−−−−−−→ n , there is an odd number.Also in definition 2.4 we mention that before any 2´k+1 always there exist a 2 k .So the thing which needs to be proven is that why always the even numbers willarrive to the odd numbers, after they being divided many times by 2. the proofhas written in theorem 2.5. Theorem 2.5.
Always the even numbers would arrive to the odd numbers.Proof.
This part is going to prove that how the even numbers will reach the oddnumbers. At first, choose a natural number which is even and afterward divide itby 2 for one time, then consider the following operation on it:If 2k ÷ ÷ NEW METHOD TO PROVE THE COLLATZ CONJECTURE 5 again by 2 until you reach an odd number. Actually, as a function f in mathematicslanguage, it can be viewed as follows: f ( n ) = ( n/ n = 2 k n if n = 2 k + 1 . Now make a sequence by performing this operation for many times, Anyway, it’slogical that it will arrive to an odd number at the end because:All of the even numbers are divisible by 2, so in condition that x = 2´k+1, theeven numbers can be define as 2k = 2 n × x, so we deduce 2k ÷ n = x . Here (n) isdescribing that how many times 2k needs to be divide by 2 till reaching x which isan odd number. For instance, 48 = 2 ×
3, here 48 will reach 3, after we divide itfour times by 2. (cid:3)
Remark . In this section, up to 2 stages before 2 n s, have specified. By doing n − on any 4 n , (the backward form of 3n+1), you will reach some numbers whichthey are a step before 4 n s except 4 (reason at the end of the definition 2.2). Thesenumbers in the following sequence of numbers, are a step before 2 n s:Set of A = {
5, 21, 85, 341, 1365, 5461, ... } .In your calculation, whenever 2k+1 × −−−−→ n , verily that odd number (2k+1)is a member of the above category ( A ).Anyway, lets study about the second stage before 2 n s. As it proved in defini-tion 2.4, in this conjecture before any odd number, there exist an even number.So by multiplying the numbers among the above set of numbers by 2 (backwardform of the n/2), we can form a sequence which is a step before the set of A andsequentially it can be define as:Set of B = {
10, 42, 170, 682, 2730, 10922, ... } In your calculation, whenever 2´k ÷ −−→ B ). Theorem 2.7.
Actually each passel in remark 2.6, is following its own rule to getthe next digit, anyway the algorithm of the set of A can be define as: A = { × −−−−→ × −−−−→ × −−−−→ × −−−−→ ... }⇒ t n = n +1 − And about the set of B , it can be define as: B = { × −−−−→ × −−−−→ × −−−−→ × −−−−→ ... }⇒ t n = n +1 − Corollary 2.8.
Until now, in theorem 2.5 this essay clarified that all of the evennumbers will reach the odd numbers and this proof is appropriate to prove that howthe numbers among the set of B would arrive to the numbers among the set of A too.Anyway in remark 2.6 we reached the first set ( A ) by doing the backward form ofthe 3n+1 on every m , so it is logical that these numbers in this set of numbers, willarrive to m s = n s by doing 3n+1 on them. Afterward regarding to theorem 2.1it is clear that how m s = n s arrive to 1.From the results which have obtained in this essay till now, the configuration ofthis problem received to here:... → { x | x ∈ B } → { y | y ∈ A } → n → According to the results of this paper, the author of this essay have clarified andsummarized these results below:
D.KARAMI
Corollary 2.9.
The starting integer can be an even number or an odd number.In the situation that the starting integer was even, for sure it will reach to an oddnumber (proof were written in theorem 2.5). So it’s better to consider that theconsequence of the primary steps is an odd number. If the obtained odd numberwas 1, you have arrived to the destination also if it was among the set of A , itwill arrive to m = n in the next step (due to the remark 2.6), and after that, itwill reach 1 clearly (proof in theorem 2.1). But in condition that it was neither ofthem, as conjecture said, certainly, with tripling it by 3 and then by adding 1 tothe consequence, it will reach an even number (proof in definition 1.4). Afterwardagain with dividing that even number by 2 until it’s possible, it will reach anotherodd number (proof in theorem 2.5).This loop between odd and even numbers before the set of A or B , will repeatagain and again. It won’t stop until it reaches one of the numbers which is amongthe set of A or B . So chance is helping to each number for many times to it canconvert to a special number which is among the set of A or B .So by considering the results which have achieved in this essay, the general andfinal form of the Collatz conjecture can be viewed as bellow:2k 2´k+1 { x | x ∈ B } { y | y ∈ A } m = n { x | x ∈ B } { y | y ∈ A } m = n Corollary 2.10.
Because in this conjecture each number has its own way to 1, wecan not write an specific algorithm for every numbers. But it’s logical that alter-nation between odd numbers and even numbers, is making chance for the selectednumber to it can convert to a number which is among the set of A or B , and afterthat every thing is clear that how it would arrive to 1. From the results of this article, we conclude that because the result of the primarysteps is odd, if somebody prove that how odd integers would arrive to the numbersin set of A or B , this conjecture will be prove. In fact now, the thing which needsto be prove is that, how the loop between odd and even numbers which introducedin corollary 2.9, would reach the numbers among the set of A or B . And it was thelast approach to prove the Collatz conjecture. Acknowledgements.
I am really grateful to all of those who taught me to learnand i was honored to meet them.
NEW METHOD TO PROVE THE COLLATZ CONJECTURE 7