Proof to the Collatz conjecture

Discussion in 'Alternative Theories' started by Sean Gilligan, Jul 13, 2022.

  1. Sean Gilligan Registered Member

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    In the proof to the Collatz conjecture we use 3 types of odd numbers
    A number divisible by 3 (3n) 9,15,2, 27 etc
    A 3n+2 11,17, 23, 29...etc
    A 3n-2 13, 19, 25, 31...etc
    When we do this we spot a pattern.
    1. Every division by 2^odd number (2,8,32,128 etc) leads to a 2+1
    2. Every division by 2^even number leads to a 2-1.
    3. There are 2 functions which lead to the same stop-start value of x only 1234 can satisfy both functions simultaneously.
    3x+1÷2^n
    3x+1/3÷2^n
    Some 2 separate numbers can together satisfy both functions to arrive at the same stop-start number but they cannot be in the same sequence.
    Because
    For a loop there must be 2 ways in and one way out. This is not possible.
    4. A number can have 2 ways in and one way out for example 47 can be reached by 31×3+1÷2 or by 125×3+1÷8 but the 2 ways in cannot happen in the same sequence.
    3a. Because both come from an initial number that must be part of the sequence, for example 27 is the starting x which leads to the number 31×3+1÷2=47 but 27 is divisible by 3 so cannot be the arrived at by a descent from 3x+1÷2. 97 is a 3n-2 which cannot be descended into from above because.
    4b. A starting x that is a 3n-2 cannot be descended into from above because in order for a 3n-2 to be arrived at by the function 3x+1÷2 it must be arrived at from below because any (3n-2)×3+1÷2 can only happen where that 3n-2 is a lower value than the initial x because in order to arrive at the same slope between x and 2x it must also satisfy the slope between x and 4x (3x+1/3) in a direct descent so therefore any 3n-2 must arrive at most at 2x. So therefore such stop start number would be a 3n+2. If it is a 3n+2 then 2x must at most be 1/2 of 1/4 of 8x which has the same slope as between x and 2x relative to 0 but this is impossible because then the 8x would be 16x relative to the stop start number which is impossible because all divisions by 2^even number (base 4) end in a 3n-2. 
    So there can be no descent from higher up it must come from below. There cannot be any number within infinity that can satisfy this requirement because the 1st number to violate the descent to 1234 must theoretically come from above (which I have just shown is impossible) not below because all lower numbers would have already decended to 1234.
    4c. No 2 different 3n+2 can descend at the same slope to arrive at the initial value of X see illustration attached.
    No 2 3n+2 numbers can both satisfy the same end number after 3x+1÷2^n because any 2 numbers satisfying the function descend at a different slope relative to a direct line to 0 so cannot both arrive at the same slope that is between x and 2x in a direct descent. So any initial x must have been arrived at from only one possible route, a sequence which must begin at the initial value of x (or lower which as already established is impossible because the theorised 1st number to violate the drop to 1234 cannot be arrived at from below). Also because for any theoretical slope to arrive at the stop-start x which is a 3n+2 it must satisfy 3x+1÷2 from (a) a lower value of x which must have been arrived at from operations of the function beginning at the initial x which means a descent from a higher value of 3x+1 also is impossible because then it cannot also satisfy the function 3x+1/3 to descend into the same slope between 2x and x. If it did arrive from higher it cannot also arrive from lower which is obvious but may not be obvious to anyone under stress.
    So no loop is possible arising from any number within infinity for these simple reasons. The Collatz conjecture absolutely definitely cannot have a loop.
    The reason we cannot have an ascent into infinity is because of the hidden sequence of an descending value of ÷2^n. With every incremental rise of 1.504.....1.503.... etc in 3 or more consecutive operations of 3x+1÷2^1 the value of each next 3x+1 is divisible by half the last 2^n. When 2x+1 reaches a number that is not divisible by 2^n it must and does jump to a higher value of 2^n to continue. This value constantly rises so the magnitude of the descent is always more than the magnitude of the ascent over any given range. It can fluctuate but any fluctuation overall leads to a drop in value because it cannot avoid arriving at a value of 2^n that brings the value below the initial value in the sequence. This is because any incremental rise always leads to a subsequence that begins with divisions leading to either a prime or numbers divisible by a prime so the next sequence of divisions is inevitalby going to be divisible by a number with a higher power of n in 2^n.
    Eg: in the sequence 27 to 1 we arrive at 479 which +1 is
    480÷32=15 479×3+1 is
    720÷16=45 719×3+1 is
    1080÷8=135 1079×3+1 is
    1620÷4=405 1619×3+1 is
    2430÷2=1215 2430÷2 is the lowest possible division in this sub-sequence so the next number must be divisible by a higher value of n in 2^n. As one can see every 1.50..... increment rise in the main sequence +1 divisible by 2^n leads to a next value in the sub-sequence which is divisible by 2^n where n is one less than the last. Simultaneously the final number triples in value to the last end number arrived at by 3x+1 which in this case is always divisible by 5 until it can no longer comtinue the sequence. Therefore it cannot rise in an increment of 1.5.
    It therefore must begin a new sequence with a higher value of n in 2^n.
    In this example it jumps to a division by 8
    2429×3+1=7288÷8=911 which begins a new sub-sequence. 911(*note below)×3+1=
    1368÷8=171 1367×3=2051
    2052÷4=513 2051×3=3077
    3078÷2=1539 3077×3=9231 this is the last possible division at 2^1 so a new sequence begins and we get a new sub-sequence where the main sequence is in divisions by 4
    577+1=
    588÷4=147
    434÷2=217 this is at 2^1 in this sub sequence which then jumps to a higher value of division in this case 16.
    So the fact this subsequence is always divisions that decrease it must jump to a new number which is divisible by a higher value of base 2 the further we go into the sequence it forces the value down over a few interations. Sometimes a generous few but always a few which then go down and down because the number derived at by 3x+1 can not avoid being divided by a 2^n into descent which leads to a number below the original x in the sequence therefore back to 1234 loop.
    So no rise to infinity is possible.
    Combined with the fact no loop is possible means the Collatz conjecture is absolutely, definitely....true.
    From the same person who solved the double slit experiment 3 years ago which unlocked a full grand unified field theory, Theory of Everything but whom the big bang, particle model lobby have been trying to keep silent.
    The proof is presented in video format in English, math and international sign language in this short video.

    And with an image of the slopes of descent examples plotted on a graph on the main webpage at doubleslitsolution dot weebly dot com/collatzconjectureproof dot https://doubleslitsolution.weebly.com/collatzconjectureproof.htmlhtml
    PS the people claiming to offer money for these prizes have been breaking the law, advertising their names and businesses for free using intellectual property that does not belong to them, they are aware of this but don't care for the law or the property of others. The terms of these so called "prizes" amount to a scam and anyone dealing with these criminals do so at their own risk. Let them be the last to find out when this goes viral please.
     
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  3. DaveC426913 Valued Senior Member

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    17,124
    Why have you not published this in a peer-reviewed paper?

    Why must this be so? Setting aside the logic that follows; I see the above as a premise. But I don't see the logic that shows why it must be granted.

    A loop is simply a sequence of numbers that start at n and, after a few rises and falls, returns to n. By definition, it has zero ways in and zero ways out, unless you are using the terms in a way unfamiliar to me.

    I see no illustration attached.
     
    Last edited: Jul 13, 2022
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  5. Sean Gilligan Registered Member

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    I explain why there cannot be a loop linked to the starting number. Any such number must come from below x which is impossible because every previous x to the 1st x theorised to escape the descent to 1234 is known to lead to 1234
    I tried to upload an image but it wouldn't work. It is in the video explanation on my youtube channel Sean g 137.
    The reasons this is not for peer review because it is as watertight as Euclids proof to the primes. It needs no review it is not an opinion or a theory it is mathematical fact. It belongs to me not anyone else, no one has the rigbt to delay, profit from or jeopardize this reaching the public, everyone has a right to the truth freely on equal terms regardless of economic means or education, it is the law as enshrined in the UDHR, other reasons are explained on the Collatz Conjecture proof on my website doubleslitsolution dot weebly dot com/collatzconjectureproof dot html
     
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  7. DaveC426913 Valued Senior Member

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    17,124
    No, it is not. It behooves you to look up what a 'fact' is.

    It is not truth; it is a claim, yet to be vetted.

    I would discuss the nature of the loops further but I think it's important to help you get over your own ego first.
     
  8. Sean Gilligan Registered Member

    Messages:
    3
    Still not understanding r ye?!! Here i explain the Proof to the Collatz Conjecture using primary school math and simple common sense, explained in the first 5 minutes. EVERYONE should be able to inderstand this. With what it leads to explained thereafter. It is done using a graph and plotting each step of the function as a function of time so best you watch to understand.
     
  9. Michael 345 New year. PRESENT is 72 years oldl Valued Senior Member

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    12,679
    What am I missing?

    I would think
    • divide even number by 2 you will get down to 2 and finish with 1
    • if you divide number by 2 and get odd number adding 1 to make it even to keep going you will get to 2 and finish with 1
    What am I missing please?

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  10. exchemist Valued Senior Member

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    11,391
    Ah, you must be yet another nutter. It's perfectly obvious, to anyone knowing the first thing about intellectual property, that it breaks no law to offer a prize for solving a problem.
    Furthermore there is no IP in a mathematical theorem or conjecture.

    And you have not proved the Collatz Conjecture. If you had, you would be lauded to the skies by the professors of mathematics. Have you even submitted your "proof" to the mathematics department of a university?
     
  11. DaveC426913 Valued Senior Member

    Messages:
    17,124
    I see that you have posted this elsewhere, with unsatisfactory responses.
    https://www.thenakedscientists.com/forum/index.php?topic=85167.0
    Perhaps you'll find SciFo more responsive. I hope you'll have plenty of opportunity to answer questions.

    Since it is simple to explain, then you can simply explain it here in text. It is generally considered bad form on a science discussion forum to just flog your YouTube video in lieu of an actual discussion.

    I asked you a question:
    I assume this is intended as a response:
    although it doesn't seem to actually address my comment directly. Allow me to elaborate.

    1. Can you explain what is special about the number 1234? Citing a reference in published works that examines it would be fine.
    2. How does this address the issue of a starting number x that is larger than any starting numbers already analyzed? For example: how can you be sure of the behavior if a starting number larger than 2.95×10^20? Are you claiming that any such number must reduce to 1234? If so, defend that claim, or cite a reference that proves it.
     
    Last edited: Jul 15, 2022
  12. DaveC426913 Valued Senior Member

    Messages:
    17,124
    Read up on the Collatz Conjecture.
    The algorithm is:
    If starting number is even, divide by 2,
    If starting number is odd, multiply by 3 and add 1.
    The conjecture is that all whole numbers larger than 1 eventually reduce to 1.

    Two unproven possibilities exist:
    1. a starting number that evolves upward with no bound
    2. a loop - a starting number that evolves (after some number of iterations) back to itself - i.e. a closed loop - and therefore is isolated from the primary tree, and never reaches 1.
     
  13. Michael 345 New year. PRESENT is 72 years oldl Valued Senior Member

    Messages:
    12,679
    Ya thanks

    I did read up a bit more. What was missing I didn't realise when you arrive at 1 you had to keep going ie multiple 1 by 3 and you are in a loop

    As for the proof, people with far more powerful computers than you and I have access to have computed numbers with more than 18 digits and have not solved it

    Also why post here when there is a Million dollars on offer for a solution?

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  14. DaveC426913 Valued Senior Member

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    Brute force solving cannot prove the conjecture true; it can only prove it false. So far, it has not done so.

    It would not matter if they solved it to 40 decimal places, it's not proven till there's a formal proof that covers the entire domain (i.e. all positive integers).
     
  15. Michael 345 New year. PRESENT is 72 years oldl Valued Senior Member

    Messages:
    12,679
    Yes I understand but If someone turns up with a 16 digit number and claims this is what you are looking for they already know it's not

    As you say
    • the conjecture is that all whole numbers larger than 1 eventually reduce to 1. ((you cannot do ALL numbers)
    • a starting number that evolves upward with no bound (you cannot teach no boundary)
    • closed loop - and therefore is isolated from the primary tree, and never reaches 1. (Is there a possibility one of these could be found?)
    Has any PATTERN been established wth iterationns of large numbers?

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  16. BdS Registered Senior Member

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  17. DaveC426913 Valued Senior Member

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    17,124
    Sure. This looks pretty patterny:

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    but a pattern is not a proof.
     
  18. origin Heading towards oblivion Valued Senior Member

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    I have a love/hate relationship with math, this thread is the part of math that I hate. I read about 2 lines of the OP and my eyes glaze over and I start to nod off...
     
  19. DaveC426913 Valued Senior Member

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    17,124
    Sure, so did mine. The OP is exploring his(her) own method of analysis, which takes time to absorb, but I know enough about the Collatz Conjecture and about mathematical logic to spot when the OP seems to be making unwarranted leaps to conclusions - such as this fixation with the number 1234. That's new to me; I haven't encountered it before and a quick internet search hasn;t relaved anything (except the OP's post on another forum) so I've asked the OP to point to some research that supports it. So far, nada.
     
  20. Michael 345 New year. PRESENT is 72 years oldl Valued Senior Member

    Messages:
    12,679
    Was thinking along the lines of a pattern leading to a formula with which you could plug in the number you were thinking of testing and getting some sort of output giving either how many iterations it would produce before reaching 4 2 1 or showing it does not

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  21. DaveC426913 Valued Senior Member

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    Yes. That's called brute forcing it. You have to pick a start number and process it. One number, one solution. But learning the fate of one number doesn't help you with another.

    If we had a pattern that could shortcut that process, then yes that would be (at least part of) a proof.
     
  22. James R Just this guy, you know? Staff Member

    Messages:
    37,359
    Moderator note: This thread has been moved out of Physics & Math for now.

    If it is established that the purported proof is valid, then I will move the thread back to our Science sections.

    This seems unlikely, because it would make sense to first publish any such proof in a peer-reviewed journal, rather than on an internet forum like this one.

    ---

    I haven't read through the wall of text that is the opening post. A skim read gives me the impression of poor presentation, which makes me reluctant to devote time and effort to finding the flaws that are likely to be there.

    Nevertheless, it seems that this has attracted some interest, so there's no reason it can't be discussed further by others.
     
  23. BdS Registered Senior Member

    Messages:
    497
    Made a quick app in Python.

    The output:

    Enter a seed number X: 9663
    28990
    14495
    ...
    3389303
    10167910
    5083955
    15251866
    7625933
    22877800
    11438900
    ...
    160
    80
    40
    20
    10
    5
    16
    8
    4
    2
    1
    Seed Number X = 9663
    Steps = 184
    X Highest = 27114424



    The line "if (int(X) % 2) == 0:" checks if X is currently odd or even and calculates it accordingly.
     

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