Collatz conjecture 2nd loop proven impossible
The proof a 2nd loop is impossible in the Collatz Conjecture is proven using a simple subtle inequality that holds for every possible sequence within infinity. Made visually easy to follow using 6 colour images of a hypothetical hailstones sequence, also with voiceover. Having no home or bank account nor contacts in academic institutions, peer review has been impossible to get.
If sharing please use the link, CC attribution video is only 4 minutes 50 seconds 0-3:45 proof plus info, credits and references. https://youtu.be/88Ow6wogUTw?si=B_P8obXM13p3ZhW0
Sean A Gilligan
April*2023*refined and revised most recently on 22 May*2024
Abstract
Prove the function x×3+1/2^n when repeated will always go to 1.
The origin of the function is attributed to being proposed by German mathematician Lothar Collatz in*1937*[1] Take any odd number multiply it by 3 and add one then divide by 2 until one arrives at the next odd number, repeat as many times as possible, so far every number goes to one and loops between*1421. The task is to prove it always will, 2 possible exceptions have been hypothesised, one where a sequence returns to the same value of x and loops forever or where it rises eternally higher towards infinity. In this paper I prove such a hypothesised loop is impossible.
Introduction
In order to have a loop in the 3x+1 problem the value of all rises VR must equal the value of all falls VF between the 1st and final* X(capital X) so that VR-VF=0
We can deduce whether that is possible by working backwards and using only full values of x's in 5 simple steps of logical deduction.*
Where any 3x+1=y y is always even.
1. the 1st rise we cancel 2X between X and y-1 (which is 3X leaving X from 0 to X and 0 from X to 3X) with 2X in the final fall between the final y(fy) and X.* This leaves fy-3X in the final fall between fy and X.* y is always even x is always odd so fy-3x is always odd
2. So we know we must get a net rise (between all y to x to y-1's) between the 1st y and the final y-1(fy-1) to cancel to 0
3)a) Between X and X, where any x is greater than the previous x then y=2x so we cancel 1x in the descent from y to x with 1x in the rise from x to 3x this leaves a net rise (NR) of 1x from x to 3x (or y-1) (and one value of x between 0 and x).* 1x is always odd.
3. b) Where x is less than the previous x y=x×2^n(n greater than 1) in the fall we can cancel 2x in the rise from x to 2x with 2x in the fall from y to x leave a net fall (NF)=y-3x always odd (plus 1x between 0 and x).
Add up all NR to a total net rise TNR of all values between x to y's and y to x's , add up all NF(not including fy-3X) to a total net fall TNF. Subtract one from the other to leave an overall net ascent ONA (from all y to x to y-1's). If ONA=fy-3X we could cancel one from the other to leave a final value of 0 between all x's and y's.
4. This would need an odd number of x's between the starting X and yf because we need an odd TNR minus an even TNF or an even TNR minus an odd TNF to leave an odd number equal to fy-3X. So when one is cancelled from the other we are left with 0 for all values between x and y's between Xand X.
5. However here arises an inescapable inequality. For a loop to happen every value of x must be linked with no breaks in the chain so calculating from the lowest odd value of x in the loop as the starting X (Xn) we know the next odd value of x must be higher so now measuring from the final y to (Xn+1)×3 instead of from 3Xn we are still left with an odd number in the descent from the final y. However now we have an even number of odd x's between the 2nd y ((Xn+1)×3)+1 and the final y-1 so we can't get an odd number remainder as a net rise between the 2nd y ((Xn+1)×3)+1 and the final y-1 (fy-1) to cancel with the final y-((Xn+1)×3) to leave a net rise of 0
So a 2nd loop is impossible in the 3x+1 problem, regardless of any value for X in any sequence within infinity.. This part of the conjecture is now ruled out as being impossible.
Please share with anyone interested in mathematics or science in general
With no regular income except from busking and no bank account or contacts in academic institutions I have been unable to get peer review so far. So sharing would be most welcome and appreciated.
Other acknowledgements,
catonmat.net*for the Collatz sequence generator by Peter Krumins, which may prove invaluable for proving the second part of the conjecture, which Peter says is Alien technobody from the future, thanks Alien techies, great app. also
Youtube videos by Veritasium and Numberphile.
Lists of primes used during research
naturalnumbers.org
wolframalpha.com
Thanks for reading.
If sharing please use the link, CC attribution video is only 4 minutes 50 seconds 0-3:45 proof plus info, credits and references. https://youtu.be/88Ow6wogUTw?si=B_P8obXM13p3ZhW0
Sean A Gilligan
April*2023*refined and revised most recently on 22 May*2024
Abstract
Prove the function x×3+1/2^n when repeated will always go to 1.
The origin of the function is attributed to being proposed by German mathematician Lothar Collatz in*1937*[1] Take any odd number multiply it by 3 and add one then divide by 2 until one arrives at the next odd number, repeat as many times as possible, so far every number goes to one and loops between*1421. The task is to prove it always will, 2 possible exceptions have been hypothesised, one where a sequence returns to the same value of x and loops forever or where it rises eternally higher towards infinity. In this paper I prove such a hypothesised loop is impossible.
Introduction
In order to have a loop in the 3x+1 problem the value of all rises VR must equal the value of all falls VF between the 1st and final* X(capital X) so that VR-VF=0
We can deduce whether that is possible by working backwards and using only full values of x's in 5 simple steps of logical deduction.*
Where any 3x+1=y y is always even.
1. the 1st rise we cancel 2X between X and y-1 (which is 3X leaving X from 0 to X and 0 from X to 3X) with 2X in the final fall between the final y(fy) and X.* This leaves fy-3X in the final fall between fy and X.* y is always even x is always odd so fy-3x is always odd
2. So we know we must get a net rise (between all y to x to y-1's) between the 1st y and the final y-1(fy-1) to cancel to 0
3)a) Between X and X, where any x is greater than the previous x then y=2x so we cancel 1x in the descent from y to x with 1x in the rise from x to 3x this leaves a net rise (NR) of 1x from x to 3x (or y-1) (and one value of x between 0 and x).* 1x is always odd.
3. b) Where x is less than the previous x y=x×2^n(n greater than 1) in the fall we can cancel 2x in the rise from x to 2x with 2x in the fall from y to x leave a net fall (NF)=y-3x always odd (plus 1x between 0 and x).
Add up all NR to a total net rise TNR of all values between x to y's and y to x's , add up all NF(not including fy-3X) to a total net fall TNF. Subtract one from the other to leave an overall net ascent ONA (from all y to x to y-1's). If ONA=fy-3X we could cancel one from the other to leave a final value of 0 between all x's and y's.
4. This would need an odd number of x's between the starting X and yf because we need an odd TNR minus an even TNF or an even TNR minus an odd TNF to leave an odd number equal to fy-3X. So when one is cancelled from the other we are left with 0 for all values between x and y's between Xand X.
5. However here arises an inescapable inequality. For a loop to happen every value of x must be linked with no breaks in the chain so calculating from the lowest odd value of x in the loop as the starting X (Xn) we know the next odd value of x must be higher so now measuring from the final y to (Xn+1)×3 instead of from 3Xn we are still left with an odd number in the descent from the final y. However now we have an even number of odd x's between the 2nd y ((Xn+1)×3)+1 and the final y-1 so we can't get an odd number remainder as a net rise between the 2nd y ((Xn+1)×3)+1 and the final y-1 (fy-1) to cancel with the final y-((Xn+1)×3) to leave a net rise of 0
So a 2nd loop is impossible in the 3x+1 problem, regardless of any value for X in any sequence within infinity.. This part of the conjecture is now ruled out as being impossible.
Please share with anyone interested in mathematics or science in general
With no regular income except from busking and no bank account or contacts in academic institutions I have been unable to get peer review so far. So sharing would be most welcome and appreciated.
Other acknowledgements,
catonmat.net*for the Collatz sequence generator by Peter Krumins, which may prove invaluable for proving the second part of the conjecture, which Peter says is Alien technobody from the future, thanks Alien techies, great app. also
Youtube videos by Veritasium and Numberphile.
Lists of primes used during research
naturalnumbers.org
wolframalpha.com
Thanks for reading.
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