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.