I jokingly suggested for someone to prove the Collatz Conjecture, and they came up with their own proof. I have no idea how to disprove proofs, so can anyone tell me either what is wrong with this proof or how I could have determined that myself?
"So, if odd, apply (6n+2)/2, else apply n/2. If 2 is not a factor of x, then multiplying x by any number not divisible by 2 would produce a number also not divisible by two. Three is not evenly divisible by two. By definition, neither is the odd variable. By definition, an odd number mod 2 will give you 1. Adding one to one will give you two. Two mod two is 0. The sum of two values is the modulus of the sum of the moduli of the values. Therefore, 3n+1 makes an odd n even. However, using induction, for the postulate to be true, the number must converge to 1 faster than it diverges. Thus, 3n+1 must produce a number that is divisible by 4 at least a quarter of the time that it produces values, as |2n| < |3n|. And (4a+2)/3=1 shows that a=25%. This is where the proof becomes a little bit more challenging. Under what conditions does (3n+1)%4=0? For 3n+1 to be divisible by 4, 4m+1=n for all integers (According to WolframAlpha, at least). This approaches a quarter of all values as infinity, and the original problem requires a quarter of the values to be divisible by 4 as the number approaches infinity. By the squeeze theorem, QED?"