aa r X i v : . [ m a t h . G M ] J a n Proof of Collatz Theorem
Benyamin Khanzadeh H. ∗† February 5, 2021
Abstract
In this article, we will show that Collatz is theorem and we proof it bymethod that we made in section 2 and 3. In section 1, first we introductionCollatz problem and idea of mathematician about this problem then wechange the function of this problem and we make a new definition ofCollatz set and generalize Collatz problem in the set theory. In section 2we decrease all of natural numbers to Z and make a model with lemmathat we said. Then in section 3 we say 3 properties of numbers that arein our models and then we make a new definition of coloring of graphto complete our model and make a new model to explain Collatz systemwith 3 numbers. Finally in section 4 we begin proof some part of firstmodel and we use properties that we proved in section 3, to proof ourmodel completely. Keywords:
Collatz, Number theory, conjecture, Proof of Collatz, Graphtheory.
The Collatz conjecture is a conjecture in mathematics that concerns a sequencedefined as follows: start with any positive integer n. Then each term is obtainedfrom the previous term as follows: if the previous term is even, the next termis one half of the previous term. If the previous term is odd, the next term is 3times the previous term plus 1. The conjecture is that no matter what value ofn, the sequence will always reach 1.[1]The conjecture is named after Lothar Collatz, who introduced the idea in1937, two years after receiving his doctorate.[2] It is also known as the 3 n + 1problem, the 3 n + 1 conjecture, the Ulam conjecture (after Stanis law Ulam),Kakutani’s problem (after Shizuo Kakutani), the Thwaites conjecture (afterSir Bryan Thwaites), Hasse’s algorithm (after Helmut Hasse), or the Syracuseproblem.[3][4]Paul Erd˝os said about the Collatz conjecture: ”Mathematics may not beready for such problems.”[6] Jeffrey Lagarias stated in 2010 that the Collatzconjecture ”is an extraordinarily difficult problem, completely out of reach ofpresent day mathematics.”[7] ∗ Benyamin Khanzadeh Holasou,
E-mail address: [email protected] † Thanks to professor Farzali Izadi that helps me with editing this article and helps me tounderstand how can I write a mathematics form article
1o Collatz conjecture is a problem that no one has been able to solve com-pletely for the past 80 years. This problem states that:If, in a set of natural numbers, we multiply each odd number by three andadd it to a sum and divide each even number by two, it finally comes to oneafter a certain number of steps. In other words: ∀ x ∈ N , f n ( x ) = 1 ⇐⇒ f ( x ) = ( x + 1 ⇐⇒ x ∈ O x ⇐⇒ x ∈ E In other words we can say it as: if f ( x ) = ( x + 1 ⇐⇒ x ∈ O x ⇐⇒ x ∈ E rule beestablished, will any natural number reach 1 or we have a different value thatfor it, function can’t reach 1?By trying Collatz on several numbers, we find that: the number one has theinfinite period in the Collatz conjecture because: 1 → → → → → · · · and this period of rotation never ends.The problem with this periodicity is that any number that reaches one ex-periences an infinite periodicity.To eliminate the periodicity, we extend the properties of Collatz as follows: g ( x ) = x + 1 ⇐⇒ x ∈ ( O − { } ) x ⇐⇒ x ∈ E ⇐⇒ x = 1If we want to generalize the definition of Collatz we can say: Definition 1.2.
If a number such as x reach 1 with g ( x ) function then we saythat it’s in set of truths or in other words: x : reach 1 with g ( x ) function ⇒ x ∈ T Or we can define T as: T = { x | x reach 1 with g ( x ) function } And now set of false is a set numbers which are not in T , are in it, or inother word: F = T ′ = { x | x / ∈ T } To proof Collatz with this definition, we must show that any natural numberis in T set.In Section 2, we create a graph that reduces the infinite number of naturalnumbers to 10 by several lemma and theorems. We must prove that no numberhas an infinite period, which is possible in Section 4. (In Section 3, we reduce10 to 3 to make proofs more straightforward and easier.) In order to draw a graph for the Collatz system, we have to use lemma anddraw this graph through them.If we put an odd number in f , the result will be even: x = 2 k + 1 ⇒ x + 1 = 6 k + 4 = 2 K But if the number is even, the result can not be said, because: . N = { , , , , ... } = 2 k ⇒ x = k Because it’s not possible to say k we will give two lemma to say what will happento k . Lemma 2.1.
Suppose that x = P ni =1 a i i − is given number. As a i ∈ { x ∈ W , ≤ x ≤ } and x is odd then f ( x ) will become even if and just if f ( a )become even, or in other words: x = P ni =1 a i i − , x = 2 k + 1 ⇒ f ( x ) = 2 K ⇐⇒ f ( a ) = 2 A P roof.
To prove the given lemma, it is sufficient to place the rule in the Collatzsystem, since x is an odd, then a will be an odd, so: a = 2 q + 1 , ≤ q ≤ ⇒ f ( P ni =1 a i i − ) = 3 P ni =1 a i i − + 1Now with extensive writing we will have to:3 P ni =1 a i i − + 1 = 3 P ni =2 a i i − + 3 a + 1 =3 P ni =2 a i i − + 3(2 q + 1) + 1 = 3 P ni =2 a i i − + 6 q + 4So 6 q + 4 = 2 A ≡ A (mod10) is new unit that it’s even so lemma is true. (cid:4) But we can not say about a number that is even, because we can not saythat k is even or odd. Following lemma provide something that tell what willhappen to k . Lemma 2.2.
Suppose that x = P ni =1 a i i − is given number. As a i ∈ { x ∈ W | ≤ x ≤ } then in f ( x ), new a become in the form of a + 5 if and just ifold a is odd and new a become in the form of a if and just if old a is even,or in other words: x = P ni =1 b i i − ⇒ ( b = a ⇐⇒ a = 2 kb = a + 5 ⇐⇒ a = 2 k + 1 P roof.
In the first state we choose even a : x = P ni =1 a i i − = P ni =3 a i i − + 10 a + a = P ni =3 a i i − + 20 A + a x = ( P ni =3 a i i − + 20 A + a ) / P ni =3 a i i − ) / A + a ∴ b = a So the first state is true. Now let choose odd a : x = P ni =1 a i i − = P ni =3 a i i − + 10 a + a = P ni =3 a i i − + 20 A + 10 + a x = ( P ni =3 a i i − + 20 A + 10 + a ) / P ni =3 a i i − ) / A + 5 + a ∴ b = a + 5So given lemma is true. (cid:4) If we want to make a graph to describe the Collatz theorem, we must notethat the field of Z is a complete field for describing the Collatz system, becausein lemma 2.1 and 2.2 we saw that only by knowing the unit can the future ofCollatz be predicted and thus proved.(We do not need to know tens because weknow what the next numbers will be during lemma 2.2 .)Now suppose the graph G is the same as the assumed graph, now we defineit as follows: . At all of this article a i is in 0 ≤ ∀ a i ≤ , ∀ a i ∈ W . . W = { , , , , . . . } . Finished proof of Lemma 2.1 . Finished proof of Lemma 2.2 : ( V ( G ) = { v i | i ∈ W , ≤ i ≤ } , v i = iE ( G ) = {−−→ v i vj | i = x, j = y } The reason for choosing the vertices is that we are looking at one, not theother. (Because it depends on what happens to the unit [we proved this inlemma 2.1 and lemma 2.2 .]).Now we express the x and y in the edges. Since each edge is defined accordingto the Collatz instruction, then v j is defined according to the Collatz rule, whichcan be accurately defined by lemma 2.1 and 2.2. Definition 2.1.
Degree is the same definition as for a graph, but with a numberof differences, the output degree being denoted by deg + and the output degreedenoted by deg − . For example : A BC
In the graph G : deg + ( A ) = 1 means that the outputdegree A is equal to 1.In the graph G : deg + ( B ) =deg − ( B ) = 1, the input degree (def − ) and thedegree degree (deg + ) are equal.Now, considering this issue, we state two theorems to create the given graph: Theorem 2.1.
If it is i ∈ O and we assume v i as the vertex then:deg( v i ) ± = 1 P roof.
It is clear from one that an odd number will only go to an even number.So according to lemma 2.1, the output degree is completely correct, and now weprove the input degree of the given graph, we know from lemma 2.2 and lamme2.1 that only even numbers make up an odd number now we show in lemma 2.3that these numbers are unique and we finish our proof with respect to the inputvertex.
Lemma 2.3.
For any even x there is only and just only one odd x in Z . P roof.
Suppose this is not the case, then x and x will reach x , so we measurethem relative to each other according to lemma 2.2 If we assume x = x , theresult is correct (because it gives x = x when we multiply the sides by 2.)So suppose the second case is x/ x / x = x + 5Multiply the sides by 2. And we have: x = x + 10That in a given field: 4 = x So given lemma is true and thus the theorem is proved, because only a singlevertex reaches the hypothetical vertex and the hypothetical vertex reaches aunique vertex, the hypothetical is completely wrong. (cid:4) (cid:4) Theorem 2.2.
If it is i ∈ E and we assume v i as the vertex then:deg( v i ) ± = 2 P roof.
It is clear from lemma 2.2 that the output degree is correct. So, con-sidering lemma 2.2, half of the theorem is absolutely correct. Now to prove thedegree of input, we will deal with the fact that each odd number will go to aneven number, so we only have to prove one thing that any even number willreach an even number, it is unique that when we put an even instead of an oddin lemma 2.3, the problem is completely proven.(This means that according toLemma 2.2, one of a / a / + ( v i ) =deg − ( v i ) = 2 , i ∈ E iscorrect (because according to Lemma 2.1 a person reaches an even number Andaccording to Lemma 2.2 a pair reaches a pair.)) (cid:4) Now according to the said theorems any v j is f ( v i ) that they are in Z field.For example v with −−→ v v edges will reach to v . So E ( G ) can be defined asfollows: E ( G ) = {−−→ v i v j | f ( v i ) ≡ v j (mod 10) } Now, according to the lemmas and theorems, the graph G can be drawn asfollows: v v v v v v v v v v ff f f ffff f f ff fff Graph 2.1: Graph of Colltaz system in Z . Finished proof of Lemma 2.3 . Finished proof of Theorem 2.1 . Finished proof of Theorem 2.2 v i = i so we can have this graph too:
03 6589 42 71 ff f f ffff f f ff fff
Graph 2.2: Graph of Colltaz system in Z Now here we will reduce 10 numbers ( that we reduced them to 10 in section 2) to 3 numbers that have the same properties.In sub-section 3.1, we will talk about numbers properties and we will provetheir propertirs.Then in sub-section 3.2, we will make a new graph with a new definition ofcoloring that we made there.
Theorem 3.1.1.
According to lemma 2.1, we can make a set that it makes ofodd numbers. (Because according to Lem 2.1, all odd numbers have the sameproperty): S = { , , , , } P roof.
According to lemma 2.1, these numbers determine what will happen tothe numbers of the natural person of our choice. ( ∀ x ∈ O ⇒ ∀ f ( x ) ∈ E ⇐⇒∀ a ∈ O ⇒ ∀ f ( a ) ∈ E )Since each odd number reaches another even number, and this number is nec-essarily different, the theorem is clear and correct. (cid:4) Theorem 3.1.2.
According to lemma 2.2, we can create two sets, each withdistinct properties: S = { , , } S = { , } . Finished proof of Theorem 3.1.1 roof. To prove the theorem, we examine and prove the properties one by onein two distinct theorems.
Theorem 3.1.3. S has the following properties: x = P ni =2 a i i − + a , a ∈ S ⇒ ( (i) f ( a ) ∈ E ⇐⇒ a ∈ E (ii) f ( a ) ∈ O ⇐⇒ a ∈ O P roof. (i) According to lemma 2.2: f ( a ) = a / Definition 3.1.1. S can be defined as S = { x | x = 4 k, ≤ k ≤ , ∀ k ∈ W } .Since a ∈ S then it can be concluded that: a ∈ S ⇒ a = 4 kf ( a ) = f (4 k ) = k = 2 k ⇒ a , k ∈ E According to the final result, the first part of the proposition was proved.(ii) According to lemma 2.2: f ( a ) = a + 5According to definition 3.1.1, we will have: a ∈ S ⇒ a = 4 kf ( a ) = f (4 k ) = k + 5 = 2 k + 5 = 2( k + 2) + 1 = 2 k ′ + 1 ⇒ a , k ′ + 1 ∈ O According to the final result, the proposition was proved and by proving it, thetheorem was also proved. (cid:4) Theorem 3.1.4. S has the following properties: x = P ni =2 a i i − + a , a ∈ S ⇒ ( (i) f ( a ) ∈ O ⇐⇒ a ∈ E (ii) f ( a ) ∈ E ⇐⇒ a ∈ O P roof. (i) According to lemma 2.2: f ( a ) = a / Definition 3.1.2. S can be defined as S = { x | x = 4 k +2 , ≤ k ≤ , ∀ k ∈ W } .Since a ∈ S then it can be concluded that: a ∈ S ⇒ a = 4 k + 2 f ( a ) = f (4 k + 2) = k +22 = 2 k + 1 ∴ a ∈ E , f ( a ) = 2 k + 1 ∈ O (ii) According to lemma 2.2: f ( a ) = a + 5According to definition 3.1.2, we will have: a ∈ S ⇒ a = 4 k + 2 f ( a ) = f (4 k + 2) = k +22 + 5 = 2 k + 6 = 2( k + 3) = 2 k ′ ∴ a ∈ O , f ( a ) = 2 k ′ ∈ E According to the results, Theorem 3.1.4 and Theorem 3.1.2 are proved and areabsolutely correct. (cid:4) (cid:4) . Finished proof of Theorem 3.1.3 . Finished proof of Theorem 3.1.4 . Finished proof of Theorem 3.1.2 .2 Binary arithmetic (binary system) and coloring It is very interesting that the number of sets ( S , S , S ) is exactly equal to themaximum number of coloring G ( χ ( G )).We can only coloring G with three colors. For this reason, we can notcalculate and consider all coloring schemes.Now we place each graph number in its binary state. We will see that G Binary is: ff f f ffff f f ff fff
Graph 3.2.1: Graph of Colltaz system in binary arithmetic.We create an easy rule with a binary arithmetic for coloring graphs and thatrule is: c i , i ∈ E ⇒ j ∈ G Binary , j ∈ O c i , i ∈ O ⇒ j ∈ G Binary , j ∈ E According to the law, a graph can be colored similar to the graph below:8 c c c c c c c c c ff f f ffff f f ff fff Graph 3.2.2: Coloring system of Collatz graphOf course, this is one of the forms of coloring according to those two rules.With a little attention to the graph and its comparison with the theoremsthat proved the properties of even numbers, we will have:Due to Graph 3.2.2 ⇒ (i) v = c , v = c Due to subsection 3.1 ⇒ (ii) v and v have common propertiesDue to our rule ⇒ (iii) c and c heve not common properties((i) < (ii), (iii)) ⇒ ∴ The assumed graph coloring is incorrectSo we have to coloring the graph in a different way and according to the rule wesaid, but since −−→ v v are next to each other, the type of coloring will always bedifferent, and this creates a paradox that can not be used to continue coloring.Because of these events, violations, and problems, we need to create a newdefinition of coloring.We know that our definition of coloring is as follows: Definition 3.2.1. (Normal type coloring ( C )). We say that C is the coloringof the graph G if and just if the two vertices that are connected to each otherin G by an edges have different colors.The above definition is the simplest definition of graph coloring.Now we create a new type of coloring by breaking some of the rules ofcoloring. Definition 3.2.2. (Collatz type coloring ( C Collatz )). We say that the graph of G is coloring as C Collatz if and only if C Collatz is coloring based on the propertiesof G and must be coloring with the minimum ( χ ( G )) allowable amount of G coloring, and each member G define a directional graph, and in its coloring, twovertices connected by an edge do not necessarily have to have distinct colors.By generalizing definition 2, we will have: Definition 3.2.3. (Beta type coloring ( C β ). If C Collatz is for graphs that haveno direction, C β is defined. 9ne of the properties that we find according to the definition of C Collatz and C β is that: it is not necessary to coloring the two vertices that are next to eachother, and this coloring should only be based on the properties and sets and nomore than χ ( G )If we want to coloring the graph based on C Collatz or C β , we will have: c c c c c c c c c c ff f f ffff f f ff fff Graph 3.2.3: C Collatz ( C β ) Coloring system of Collatz graphAnd the result of this graph is as follows: ∀ c ∈ S , ∀ c ∈ S , ∀ c ∈ S In this section, we first prove part of the Collatz conjecture by the graph wecreated in section 2, and then we prove the Collatz by the graph we created insection 3 and the question variations we created in section 1 in the definition ofCollatz conjecture.
Theorem 4.1.1. If G is defined as G : ( V ( G ) = { v } E ( G ) = E ( G [ V ( G )]) then G is true in Collatz (it is in the set of truths), or in other words: G : f ⇒ G ⊂ T roof. For proof given theorem assume the number n is in v . During the defini-tions of the assumed graph in 2, n is defined in the form of n = P ki =2 a i i − +10.Or the definition of n can be generalized to n = 2 a · b · (2 k + 1). (The thirdnumber that is multiplied must be odd because if it is not odd then it can bemultiplied by increasing the power of 2.) In Collatz function, n , after a steps,finally will reach to: 5 b (2 k + 1). Now we put new number in our field ( Z field)and: 5 b (2 k + 1) ≡ · b − k + 5 b ≡ b ≡ G ⊂ T . (cid:4) Theorem 4.1.2. If G is defined as G : ( V ( G ) = { v , v , v } E ( G ) = E ( G [ V ( G )]) then G is true in Collatz (it is in the set of truths), or in other words: v v v f f ⇒ G ⊂ TG : f P roof.
First assume that G is in F for a distinct value.( ∃ x ∈ G ⇒ x ∈ F )Now we will show that this distinct value will not be except 1 and negativenumbers and we will err in the proposition and thus we will conclude that thetheorem is true.We begin with a number that it is in x = σ + 1 = P ni =2 a i i − + 1 form. After3 steps with calculating, it will become in σ + 1 form: f ( P ni =2 a i i − + 1) → f (3 P ni =2 a i i − + 4) → f ( P ni =2 a i i − + 2) → P ni =2 a i i − + 1 = σ + 1We know that ∀ x ∈ O : x +14 ≤ x , x +14 = x ⇒ x = 1: x +14 = x ⇒ x + 1 = 4 x ⇒ x − x = x And if x +14 > x ⇒ x < x +14 > x ⇒ x + 1 > x ⇒ > x − x = x And as we told before, 1 x (because it makes an infinity ring). And negativeintegers can’t too.(because our numbers is in N .) So equal and bigger part itnot true: . Finished proof of Theorem 4.1.1 x + < x, ∀ x ∈ O \ { } This inequality shows that: numbers will become lesser than, untill it becomesequal 1 (because we can say if it is false so it becomes infinity ring) and dueto g(x) function, we will see that 1 will reach 1 so this graph is true for any N number or we can say: x ∈ N , x ∈ G ⇒ x ∈ T . So G ⊂ T . (cid:4) In theorem 4.1.2 and theorem 4.1.1, we proved that G , G ⊂ T and due to logicof propositions, we will have: ∀ x ∈ G , G ⇒ x ∈ T or: −−→ c c , −−→ c c , −−→ c c , −−→ c c , c , c , c ∈ T Because G , G is based of E ( G ) , E ( G ) , V ( G ) , V ( G ), we can say all of their( E, V ) members are in T . ( G = E ( G ) ∪ V ( G ) , G = E ( G ) ∪ V ( G ) ∴ G , G ⊂ T ⇐⇒ E ( G ) , E ( G ) , V ( G ) , V ( G ) ⊂ T )Now according this, we can say any sub-graph is true because edges of sub-graphs must have c or c or c and because of that, we proved all of them.For making sure, we will make a theorem and then we will proof it. Theorem 4.2.1.
Any given sub-graph of graph G is true in Collatz. P roof . Hypothesis that theorem is not true and we have an unique sub-graphthat is not in truths set and we call it G k .G k is defined as G k : ( V ( G k ) = { v i | i have some properties } E ( G k = E ( G [ V ( G k )])And if G k is in F so we will see that: E ( G k ) , V ( G k ) ⊂ F And due to we proved that any c , c , c is in T , second part ( V ( G k ) ⊂ F ) iswrong and if we see graph we will in any sub-graph edges, we have −−→ v v or −−→ v v or −−→ v v or −−→ v v and we proved that these edges are in T , so first part ( V ( G k )) iswrong too.So we can understand that this hypothesis is not true, so we will see that: E ( G k ) , V ( G k ) ⊂ F ≡ F ∼ G k * F ∼ G k ⊂ F ′ = T So our theorem is true. (cid:4) Due to this theorem we will see: S ki =1 G i = G ⊂ T Now we propose the last theorem:
Theorem 4.2.2. (Collatz theorem). Any N number will reach 1 with g ( x ) = x + 1 ⇐⇒ x ∈ O \ { } x ⇐⇒ x ∈ E ⇐⇒ x = 1 function after some steps. P roof.
Hypothesis that the theorem is incorrect and does not reach one for adistinct value of x , and that one enters an infinite periodicity in which one isnot. . Finished proof of Theorem 4.1.2 . Finished proof of Theorem 4.2.1 x must be part of natural numbers, so its unit must be variable from 0to 9, and since we have already proved that any natural number whose unit isbetween 0 and 9 does not form an infinite loop and is true. So x must necessarilybe true and not create an infinite loop, thus proving. (cid:4) References [1] https://en.wikipedia.org/wiki/Collatz conjecture[2] O’Connor, J.J.; Robertson, E.F. (2006). ”Lothar Collatz”. St Andrews Uni-versity School of Mathematics and Statistics, Scotland.[3] Maddux, Cleborne D.; Johnson, D. Lamont (1997). Logo: A Retrospective.New York: Haworth Press. p. 160. ISBN 0-7890-0374-0. The problem is alsoknown by several other names, including: ”Ulam’s conjecture, the Hailstoneproblem, the Syracuse problem, Kakutani’s problem, Hasse’s algorithm, andthe Collatz problem.”[4] According to Lagarias (1985),[5] p. 4, the name ”Syracuse problem” wasproposed by Hasse in the 1950s, during a visit to Syracuse University.[5] Lagarias, Jeffrey C. (1985). ”The 3x + 1 problem and its gen-eralizations”. The American Mathematical Monthly. 92 (1): 3–23.doi:10.1080/00029890.1985.11971528. JSTOR 2322189.[6] Guy, Richard K. (2004). ””E17: Permutation Sequences””. Unsolved prob-lems in number theory (3rd ed.). Springer-Verlag. pp. 336–7. ISBN 0-387-20860-7. Zbl 1058.11001.[7] Lagarias, Jeffrey C., ed. (2010). The ultimate challenge: the 3x + 1 prob-lem. Providence, R.I.: American Mathematical Society. p. 4. ISBN 978-0821849408. . Finished proof of Theorem 4.2.2. Finished proof of Theorem 4.2.2