48 hours - the most isolated camp on earth

Maybe you're not such a faithless generation after all?  Here is the 1st part of the Proof to the Collatz conjecture which absolutely proves there is only one possible way in in any one sequence so no loop is possible as tight as and explained as simply as Euclids proof to the primes though it does have more factors. The other half is on the collatzconjectureproof page on my website which I'm sure you know by now.

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.


4a. 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.