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  4. Proof to the Collatz Conjecture
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Proof to the Collatz Conjecture

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Offline Sean g (OP)

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Proof to the Collatz Conjecture
« on: 13/07/2022 04:26:18 »
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 many in 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 the latest short video on my <link removed>

And on the main webpage at < link removed>

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 for everyone's safety.

* IMG_20220712_152641.jpg (1876.2 kB, 2304x3249 - viewed 267 times.)
« Last Edit: 13/07/2022 06:35:17 by Colin2B »
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Offline Colin2B

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Re: Proof to the Collatz Conjecture
« Reply #1 on: 13/07/2022 06:43:58 »
Hi there and welcome to the forum.
In the main physics and general science sections we ask that all posts are titled with a question and ask a science question in the first post. For that reason your post has been moved here. We also limit the posting of links for new members and ask that you do not try to circumvent that.
One of the rules of the forum, that you agreed to when registering, is that you do not post material that has already been published elsewhere and that includes YouTube, web sites and blogs.
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and the misguided shall lead the gullible,
the feebleminded have inherited the earth.
 

Offline Sean g (OP)

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Re: Proof to the Collatz Conjecture
« Reply #2 on: 13/07/2022 09:43:19 »
Ok if you know anything about maths this is one of the simplest problems and now solutions to understand yet considered THE hardest problem in math. It is inevitable that it will go viral, somewhere, if you decide to keep it in an obscure place, that is your choice, it may turn out to be the biggest lost opportunity your site has ever had to really raise your profile.
If you don't see that, I can ask any amount of questions regarding physics, the field is full of issues, that can be done. However I can't provide the info without some reference to where it came from,  the implications of my work is a matter of life and death for many. Which would be unfortunate if you decide to be a stickler for the rules.
In reference to your quote, Id say be careful not to fall into the trap of thinking quoting a quote means you are immune to the negativity it points out and risk becoming one of the feeble minded ruling the world.
I hope and pray you reinstate it. Every rule has an exception. I suggest this is that exception.
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Offline Bored chemist

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Re: Proof to the Collatz Conjecture
« Reply #3 on: 13/07/2022 10:55:45 »
Quote from: Sean g on 13/07/2022 09:43:19
the implications of my work is a matter of life and death for many.
Who gets killed by the truth of the conjecture?

All we need to do is observe them and see if they die.
If they do then we know that the conjecture is true.
If not, we have shown it to be false.

Either that, or you are wrong to say it's a matter of life and death.


Quote from: Sean g on 13/07/2022 09:43:19
Every rule has an exception. I suggest this is that exception.
Everyone who "proves" something difficult considers that they should be "exceptional".
If you are right, you should be talking to someone like this.
https://www.prnewswire.com/news-releases/bakuage-offers-prize-of-120-million-jpy-to-whoever-solves-collatz-conjecture-math-problem-unsolved-for-84-years-301326629.html
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