aa r X i v : . [ m a t h . G M ] F e b Complete Proof of Collatz’s Conjectures
Farzali Izadi ∗ Mathematics Department, Urmia University
Abstract
The
Collatz’s conjecture is an unsolved problem in mathematics. It is named after LotharCollatz in 1973. The conjecture also known as Syrucuse conjecture or problem.Take any positive integer n . If n is even then divide it by 2, else do ”triple plus one” andget 3 n + 1. The conjecture is that for all numbers, this process converges to one.In the modular arithmetic notation, define a function f as follows: f ( x ) = ( n if n ≡ n + 1 if n ≡ . In this paper, we present the proof of the Collatz conjecture for many types of sets definedby the remainder theorem of arithmetic. These sets are defined in mods6 , , , , , , , , , m, for positive integers m . Keywords: Collatz Conjecture, Syracuse Conjecture, 3 x + 1 Conjecture, Ulam conjecture,Hailstone sequence or Hailstone numbers, or as Wondrous numbers. AMS Classification:Primary: 11D09 ; Secondary: 11E16, 14H52. Introduction
The Collatz conjecture is an unsolved problem in mathematics as it has been for more than60 years. It is named after Lothar Collatz in 1973. The conjecture also known as Syrucuseconjecture or problem. [1, 2, 3]. Although the problem on which the conjecture is built isremarkably simple to explain and understand, the nature of the conjecture makes provingor disproving the conjecture exceedingly difficult. As many authors have previously stated,the prolific Paul Erdos once said, mathematics is not ready for such problems.” Thus far allevidence indicates he was correct (see [4]). This paper is not a bibliography of previous works,instead its is an original paper which I analysed the Collatz Conjecture and provided my own ∗ [email protected] , , , , , , , , ,
108 andwe took only odd positive remainders to work with. It is not difficult to prove that the sameresults are true for any mod 12 m, for positive integers m . Syrucuse Conjecture If k is an odd positive integer, then 3 k + 1 is an even integer. So we can write 3 k + 1 = 2 a k ′ ,where k ′ is an odd positive integer and a ≥
1. Furthermore if k is an even positive integer,then k = 2 a k ′ , where k ′ is an odd positive integer. Then dividing by 2 a , we can get anodd positive integer. Thus it is enough to work with only odd positive integers. We define afunction f from a set I of odd integers into itself, called Syrucuse function by taking f ( k ) = k. The Syrucuse conjecture is that for all k in I , there exists an n ≥ f n ( k ) = 1.Equivalently, let E be the set of odd integers k for which there exists an integer n ≥ f n ( k ) = 1, then the problem is to show that I = E [6]. The following is the beginningof an attempt to prove the Syrucuse conjecture by induction:1 , , , E . Let k be an odd integer greater than 9. Suppose thatthe odd positive integers up to and including k − E . Let us try to prove that k is in E . Remark 1. If n is obtained by a finite number of Collatz’s functions from k , and n is aCollatz’s number, then k is a Collatz’s number - here we need at most three consecutiveCollatz functions. First of all, by the remainder theorem of arithmetic, the remainders of any positive number by6 are 0 , , , , , , . Moreover if the number is odd, we are left with just the three numbers1 , , . I) If k = 6 m + 1, and k ′ = 2 m , then f ( k ′ ) = k where k ′ ≤ k −
2. Since k ′ ∈ E then k ∈ E . Proof.
We have 3 k ′ + 1 = 6 m + 1 = k , then f ( k ′ ) = k , and k ′ ≤ k −
2. By inductionhypothesis, it follows that k ′ ∈ E , namely ∃ n ∈ N such that f n ( k ′ ) = 1. As f ( k ′ ) = k ⇒ n ≥
2, so f n − ( f ( k ′ )) = 1 ⇒ f n − ( k ) = 1 which implies that k ∈ E .II) If k = 6 m + 5, and k ′ = 4 m + 3, then f ( k ′ ) = k where k ′ ≤ k −
2. Since k ′ ∈ E then k ∈ E . 2 roof. We have 3 k ′ + 1 = 12 m + 10 = 2(6 m + 5), then f ( k ′ ) = k , and k ′ ≤ k −
2. Byinduction hypothesis, it follows that k ′ ∈ E , namely ∃ n ∈ N such that f n ( k ′ ) = 1. As f ( k ′ ) = k ⇒ n ≥
2, so f n − ( f ( k ′ )) = 1 ⇒ f n − ( k ) = 1 which implies that k ∈ E .III) If k = 6 m + 3, then k ∈ E. Proof.
For this part we consider two cases: (i) m = 2 h −
1, and (ii) m = 2 h for h ≥ . (i) We have k = 12 h −
3, hence 3 k + 1 = 36 h −
9+ 1 = 36 h − h − . It follows that f ( k ) = 9 h −
2, and 9 h − ≤ h − k − . Then by induction hypothesis f ( k ) ∈ E, namely ∃ n ∈ N such that f n ( f ( k )) = 1 = f n +1 ( k ) = 1 which implies that k ∈ E .(ii) We have k = 12 h + 3 , hence 3 k + 1 = 36 h + 10 = 2(9 h + 5) . It follows that f ( k ) = 9 h + 5 = 6(3 h ) + 5 = 6 m + 5 , where m = 3 h for h ≥ . Then by part II we have f ( k ) ∈ E, namely ∃ n ∈ N such that f n ( f ( k )) = 1 = f n +1 ( k ) = 1 which implies that k ∈ E . Remark 2.
Note that the induction process are intertwined in the three different parts.To clarify this, let us see some examples. First of all, suppose that , , , and are in E . Let us prove that k = 17 is in E . Then we need to take k ′ = 11 , where k = 6 × and
11 = 6 × are both in part II while for k = 23 = 6 × we need to take k ′ = 15 , where k is in part II while is in part III. So we first need to prove and are in E . For the second example, let us have a closer look at part I where k = 6 m + 1 . For these numbers we took k ′ = 2 m = 2 a h, where a ≥ and h is an odd positive integer.We note that k ′ not being an odd positive integer is not in E , but f ( k ′ ) = h in E can beproved easily from which one can get an integer n ≥ such that f n ( k ′ ) = 1 and k ′ ∈ E by abuse of notation. The other important point is that sometimes we need to applySyrucuse function couple of times to get the appropriate results. For an example, let k = 27 . Then k + 1 = 82 = 2 × , hence f ( k ) = 41 = l . Next l + 1 = 124 = 4 × which gives rise to f ( l ) = 31 . Finally to get we need to take k ′ = 2 = 10 and get
31 = 3 k ′ + 1 . It follows that is in E so do l and k . Note also that is in part Iand
41 = 6 × is in part II. Remark 3.
It is clear that the density of the three parts together are the same asthe density of the positive odd integers. This proves the Syrucuse Conjecture which isequivalent to the Collatz Conjecture. As k is an odd integer, k − k + 1 are both even so we can write: k ± p h where h is odd and p ≥ . Then k = 2 p h ±
1. 3 emark 4.
For all odd h , f (2 h − ≤ h − .Proof. Let k = 2 h − ⇒ k + 1 = 3(2 h −
1) + 1 = 6 h − . Then we have 3 k + 1 = 2(3 h − k + 1 = 2 h ′ where h ′ = h − , since 3 h − h ′ = 2 a h ′′ where h ′′ is not divisible by 2, and a ≥
0. It follows that f (2 h −
1) = f (2 a +2 h ′′ ) = h ′′ ≤ h ′ = 3 h − . Now we have:I) If p = 1 then k = 2 h − h odd. We have 3 k + 1 = 6 h − h −
1) and3 h − h ′ , since h is odd. Let h ′ = 2 a h ′′ where h ′′ is not divisible by 2, and a isnon-negative integer, then 3 h − a +1 h ′′ ⇒ f ( k ) = h ′′ ≤ h ′ = h − ≤ h −
1, since h − ≤ h − ⇔ h − < h − ⇔ ≤ h . Then f ( k ) < k . On the other hand, f ( k ) = h ′′ is odd and k is odd then f ( k ) ≤ k − ⇒ f ( k ) ∈ E (by induction hypothesis),and so ∃ n ∈ N , f n ( f ( k )) = 1, namely f n +1 ( k ) = 1 ⇒ k ∈ E . The same argument showsthat k = 2 h + 1 ∈ E. II) If p ≥
2, and h is multiple of 3 , k = 2 p · h −
1. Let k ′ = 2 p +1 h −
1. We have3 k ′ + 1 = 2 p +1 · h − p · h − f ( k ′ ) = 2 p · h − k and k ′ = 2 p +1 h − < p · h − k . Since both k and k ′ are odd , one gets k ′ ≤ k −
2. Itfollows that k ′ ∈ E , namely ∃ n ∈ N such that f n ( k ′ ) = 1. As f ( k ′ ) = k ⇒ n ≥
2, so f n − ( f ( k ′ )) = 1 ⇒ f n − ( k ) = 1 which implies that k ∈ E . It is easy to show that for k = 2 p · h + 1, we can choose k ′ = 2 p h. Remark 5.
For all n , n − and n − + 1 are both divisible by . Remark 6. If h is not a multiple of , then h ≡ , . III ) (i) If p = 2 n , n ≥ h = 3 m + 1 and k = 2 n (3 m + 1) + 1, k ′ = 2 n +1 m + [ n +1 +13 ] then f ( k ′ ) = k where k ′ ≤ k −
2. Since k ′ ∈ E then k ∈ E . Proof.
We have 3 k ′ + 1 = 2 n +1 (3 m ) + 2 n +1 + 1 + 1= 2 n +1 (3 m + 1) + 2= 2[2 n (3 m + 1) + 1]4hen f ( k ′ ) = 2 n (3 m + 1) + 1 = k . Furthermore, k ′ < k. Since both k and k ′ are odd,then k ′ < k ⇒ k ′ ≤ k −
2. It follows that k ′ ∈ E (by hypothesis induction ) ∃ n ∈ N suchthat f n ( k ′ ) = 1. Since f ( k ′ ) = k , then n ≥
2. Thus f n − ( f ( k ′ )) = 1 or equivalently f n − ( k ) = 1 ⇒ k ∈ E .(ii) If p = 2 n − n ≥ h = 3 m +1 and k = 2 n − (3 m +1) − k ′ = 2 n − m +2[ n − − ]then f ( k ′ ) = k where k ′ ≤ k . Since k ′ ∈ E then k ∈ E . Remark 7.
In particular for n = 1 , we get m + 1 . Proof.
We have 3 k ′ + 1 = 2 n − (3 m ) + 2 n − − n − (3 m + 1) − f ( k ′ ) = 2 n − (3 m + 1) − k . Furthermore, by direct verification k ′ ≤ k. Itfollows that k ′ ∈ E (by hypothesis induction ) ∃ n ∈ N such that f n ( k ′ ) = 1. Since f ( k ′ ) = k , then n ≥
2. Thus f n − ( f ( k ′ )) = 1 or equivalently f n − ( k ) = 1 ⇒ k ∈ E . IV ) (i) If p = 2 n − n ≥ h = 3 m + 2 and k = 2 n − (3 m + 2) + 1, k ′ = 2 n m + 2[ n − ] + 1.Then f ( k ′ ) = k where k ′ ≤ k . Since k ′ ∈ E then k ∈ E . Remark 8.
In particular for n = 1 , we get m + 5 . Proof.
We have 3 k ′ + 1 = 2 n (3 m ) + 2 · n − n − (3 m + 2) + 1]then f ( k ′ ) = 2 n − (3 m + 2) + 1 = k and k ′ ≤ k as k ′ < k and both are odd. As in theprevious cases, it follows that k ∈ E .(ii) If p = 2 n , n ≥ h = 3 m + 2 and k = 2 n (3 m + 1) − k ′ = 2 n m + 2[ n − ]. Then f ( k ′ ) = k where k ′ ≤ k . Since k ′ ∈ E then k ∈ E . Proof.
We have 3 k ′ + 1 = 2 n (3 m ) + 2 · n − n (3 m + 2) − f ( k ′ ) = 2 n (3 m + 2) − k . Furthermore k ′ ≤ k − k ∈ E . V ) If k = 2 n − (3 m + 1) + 1 , then for n = 1, k ∈ E .5 roof. For n = 1 , we have k = 6 m + 3 . Then We have already proved this case in theFirst set of assertions.We see that these numbers clearly contain the set of odd positive integers which againprove the Syrucuse Conjecture and consequently the Collatz Conjecture.
After reading the paper by Jan Kleinnijenhuis, Alissa M. Kleinnijenhuis and Mustafa G.Aydogan [7] where they have done arithmetic in mod 6 and 18 and obtained congruenceclasses with mod 96 with the idea of the Hilbert hotel and Collatz tree, I started to examinemore closely my 2nd proof and 3rd proof to see what difference does it make to work withmods 6, 12, 15, 18, 24, 36, 48, 60, and 12m in general and I found out that mod 15 and mod18 don’t give rise to complete solutions with my approach but any mod 12m does. We haveseen that in the 1st and 2nd proofs, the numbers 4 and 3 or arithmetic with mod 6 playthe crucial roles. One can easily figured out that working with numbers mod 12 m, i.e., thenumbers in the set S = { , , , , , , , , , · · · } give rise the same conclusions and in contrast, working with the odd or twice times of theodd modules will run into obstacles. The main idea comes from the fact that for each number k ∈ I, either we have to find a number k ′ such that 3 k ′ + 1 = k and k ′ < k, or (3 k + 1) / ∈ E. If we do the modular arithmetic in mod 15 , then for each i = 0 , , , · · · , , the results ofthe Collatz function 3 x + 1 to the number 15 m + i never going to be divisible by 2, andconsequently get bigger and bigger. If we do the modular arithmetic in mod 18, then for each i = 0 , , , · · · , , the results of the Collatz function 3 x + 1 to the number k = 18 m + i wouldbe 2 . m +3 i +1 which can be at most divisible by 2 where the resulting number is still biggerthan k. On the other hand, for each number s in the set S , the first term in k = 3( sm ) + 3 i + 1is divisible by 4 and if the term 3 i + 1 also divisible by 4, then the dividing by 4 gives rise to anumber such as k ′ less than k and f ( k ) = k ′ . If this approach doesn’t work for any number k ,then we can try to use the second trick to find a number k ′ such that 3 k + 1 = k ′ and k ′ < k. In this case f ( k ) = k ′ ∈ E. Let us try mod s = 12 , , , , . We will see the remaindersbehave the same when we are working with the different modules. The remainders of s aredivided into four different types based on their arithmetic characteristics with a few numberscan fall in two types, 13 and 33 for instance.(1) Firstly, let us take s = 12 . In this case the remainders are: U = { , } ,V = { , } ,W = { } = { } . (2) Secondly, we take s = 24 . Then the remainders are the followings: U = { , , , } ,V = { , , } ,W = { , , } ,X = { , } . (3) Thirdly, we take s = 36 . Then the remainders are: U = { , , , , , } ,V = { , , , } ,W = { , , , , } ,X = { , , } . (4) Fourthly, for the s = 60 , we have U = { , , , , , , , , , } ,V = { , , , } ,W = { , , , , , , , , } ,X = { , , , , , , } . Finally, we take s = 108 and prove that all numbers belong to the set E. The proofs forall other s ∈ S are all the same. U = { i + 1 | i = 0 , , , , , · · · , } ,V = { , , , , , , , , , } ,W = { , , , , , , , , , , , , , , , , } = { i + 2 | i = 3 , , , , , , , , , , , , , , , , } ,X = { , , , , , , , , } = { i + 3 | i = 0 , , , , , , , , } . For s = 108, and for any odd positive number k , we can write k = 108 + j, where j = 1 , , , , · · · , . The remainders of k, are 54 odd positive numbers which fall into fourdifferent types due to their characteristics. First of all, for the remainders in the set U , wehave k = 108 m + (6 i + 1) . For these numbers we use the first trick to find k ′ = 36 m + 2 i. Thenwe get 3 k ′ + 1 = 3(36 m + 2 i ) + 1 = 108 m + 6 i + 1 = k. Clearly k ′ < k and by the definitionof Syrucuse function we have f ( k ) = k ′ . The remaining arguments are clear from the firstproof. Secondly, the remainders of the second set V are divisible by 4 via the function 3 x + 1,hence if k = 108 m + j, for j ∈ V , we can write k = 108 m + j. In this case we need to usethe second trick i.e., 3 k + 1 = 3(108 m + j ) + 1 = 4(81 m ) + 3 j + 1 = 4(81 m ) + 4((3 j + 1) / j , the number 3 j + 1 is divisible by 4 . It follows that k = 4(81 m + (3 j + 1) / f ( k ) = 81 m + (3 j + 1) / f ( k ) ≤ k − . Theresult follows. Thirdly, for the remainders of the set W we may take k ′ = 2(36 m + i ) + 1 . Then 3 k ′ + 1 = 2(108 m + 3 i ) + 3 + 1 = 2(108 m + 3 i + 2) , hence f ( k ′ ) = 108 m + 3 i + 2 = k .Here we see again that f ( k ′ ) ≤ k − . Thus f ( k ′ ) ∈ E and consequently k ∈ E. Finally, for k = 108 m +(12 i +3) = 6(18 m +2 i )+3 , we have 3 k +1 = 6(3 · m +6 i )+10 = 2[6(27 m +3 i )+5] . f ( k ) = 6(27 m + 3 i ) + 5 = 6 h + 5 , where h = 27 m + 3 i. BY the part III of the firstproof, it follows that f ( k ) ∈ E and consequently k ∈ E. We clearly see that the density ofthe numbers in this proof also equals 1 , and the proof is complete. References [1] J. C. Lagarias,
The x + 1 Problem: An annotated bibliography , (196-199).[2] J. C. Lagrarias,
The x + 1 Problem: An annotated bibliography , II (2000-).[3] J. C. Lagrarias,
The x + 1 Problem and its generalizations , American MathematicalMonthly, Volume 92, 1985, pp.3-23.[4] Chamberland, M.,
An update on the 3x+1 Problem (Catalan, trans- lated by ToniGuillamon i Grabolosa). Butlleti de la Societat Catalana de Matematiques, 18:19-45.[5] Chamberland, M.,
A continuous extension of the 3x+1 problem to the real line . Dynamicsof Continuous, Discrete and Impulsive Systems, 2:495-509, 1996.[6] http://en.wikipedia.org/wiki/Collatz_conjecturehttp://en.wikipedia.org/wiki/Collatz_conjecture