User talk:83.137.6.165

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Hello, I'm Yoshi24517. I wanted to let you know that I reverted one of your recent contributions—specifically this edit to Potter Payper—because it did not appear constructive. If you would like to experiment, please use the sandbox. If you have any questions, you can ask for assistance at the Help desk. Thanks. Yoshi24517^Chat _Online 23:42, 28 January 2022 (UTC)[reply]

Absolute proof a loop other than 1421 is impossible in the Collatz conjecture has been posted on the Collatz page on Reddit. It should not need peer review because it is an inequality which proves it, which is in essence just a little more sophisticated than 5+3≠7. The mathematician is homeless, has no links to academia and no bank account so states he has been unable to get peer review. It is now in the public domain. Here is a copy of the proof from Reddit

A HIGHER LOOP IS IMPOSSIBLE IN THE 3X+1 PROBLEM, THE COLLATZ CONJECTURE IN A SIMILAR STYLE TO EUCLIDS SIMPLE PROOF TO THE PRIMES For there to be a higher loop than 13421 in the 3x+1 problem (Collatz conjecture) we must be able to cancel out all falls from all rises in all numbers between the first and last value of X (capital X for the theorised number which the loop would hinge on) in the loop to leave a value of zero.

This is proven impossible in 5 steps of deduction.
In this solution I refer to 3x+1 as y for convenience.
1. Between the first X and the first y-1 we have 2x which we can cancel out with the final y to the final x. This leaves a value of y-3x between the final y to the final x(also the first x).
y is always even, x is always odd, 3x is always odd so an odd number from an even number leaves an odd number so y-3x is always odd.
2. So we now know we must get an odd number as a net rise for all values between all y's to x's and all x's to y-1's between the 1st y and final y-1 to cancel out with the final y-3x and leave 0.  
3a)  In the case where the value of x is higher than the previous x there is always 1x between y to x and 2x from that x to the next y-1. So we can cancel out the 1x between y and x with 1x between x and the next y-1. This leaves a value of 0 in the descent between y and x and a value of 1x  in the rise from x to the next y-1. This always gives us a net rise of 1x an odd number between y and the next y-1 where the value of x is greater than the previous x.
3 b) In the case where the value of x is lower than the previous x there is always an odd number of x's in the descent between y to x and 2x from that x to the next y-1. So we can cancel out the 2x in the rise between x and y-1 with the lowest 2 x's in the descent between y and x. This leaves a value of 0 in the rise between x and y-1 and a value of y-3x  in the descent from y to x. This always gives us a net descent of y-3x (always an odd number) between y and the next y-1 where the value of x is lower than the previous x.
4.  This means we need either
a) an odd number of net descents subtracted from an even number of net rises to leave an overall net rise of an odd number between the first y and final y-1 Or
b) an even number of net descents subtracted from an odd number of net rises to leave an overall net rise of an odd number between the first y and final y-1
Because we must have either a) or b) it means we must have an odd number of x's between the first and final x.
5. Next we look at all the values of x's (blue in image) left between the first and final X after all the values between x's and y's have been cancelled (yellow in image) Because we need an odd number of x's between the first and final X to satisfy condition 2 it means that when we cancel out all values of x's one from the other we will be left with an odd number which cannot be zero.
In order to be left with 0 we would need to have an even number of x's to cancel one from the other to be left with a net value of 0 but as I have proven this is impossible because we need an odd number of x's between the first and final x to cancel out and get a net value of 0 in order to satisfy condition 2.
So this inequality proves that a loop other than the 13421 loop is impossible in the 3x+1 problem regardless of the value of any x in any sequence within infinity.

NOTE: In this proof I had the second last x a higher value than the final X(and firstX). However suppose the final X were a higher value than the 2nd last x in a hypothetical loop the inequality still stands because in that case we would have 1X (an odd number) as a net rise in the first deduction (condition 1) so we would still need an odd number of x's between X and X to get the values between x's and y's to be a net fall an odd number to cancel with 1X and get to 0 but if we do then we cannot get the individual values of x's to cancel to 0. So such a loop cannot exist. 6. In case anyone is thinking it should only be the difference between the original X and each individual x between the 1st and final X (coloured red in the image) which should be considered for condition 5 and these are even. It makes no difference because we still have an odd number of even numbers where each value between the first X and each individual x is different and one is greater than all others we still will be left with a hypothetical net value which is greater or less than the first X. I say or less because even if some of the values were a lower value than the original X we would still have c) an odd number of values greater than X and an even number of values lower than X or d) an even number of values greater than X and an odd number of values less than X we still can't get them to cancel to the original X because if we cancel all values above X to a single number and all below to a single number one is greater than X and one is lower or one is equal to x and the other is greater or lower, so the difference between them cannot be zero to cancel to a final net rise or drop equal to the first X. 7. To look at that another way, If all the values between y's and x's and the 1st and last 3X cancel to 0 then because this would leave a number other than 0 for the total net value between the 1st and last X and each individual x then that non zero number must be either added or subtracted from the overall value of 0 between the y's x's and the 1st and last value of 3X to leave a non zero number then the remainder between X and X cannot be zero either. So a loop other than 13421 is impossible in the 3x+1 problem (Collatz conjecture) regardless of any value of any x within infinity. Proven by

Sean A Gilligan
April 2023 refined to its current simplest form November 2023

It is available in a short video demo using 6 colour images to demonstrate how such a hailstone sequence cannot ever exist here youtu dot be forward slash cbzF7C0l23A?si=399lqQ1liiaJPf5c 83.137.6.165 (talk) 19:41, 21 November 2023 (UTC)[reply]

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