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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?

Proof:

"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?"

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closed as off-topic by Did, kingW3, Lord Shark the Unknown, Especially Lime, Leucippus Jun 2 '17 at 0:10

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    $\begingroup$ To be honest, the "proof" is barely readable. For example, what is x? $\endgroup$ – Andrew D Jan 8 '13 at 3:44
  • $\begingroup$ I think he mixed up his 'x' and 'n' variables, where x = n $\endgroup$ – Waffle Jan 8 '13 at 3:47
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    $\begingroup$ Then he managed to get confused proving the trivial statement that "if n is odd, 3n+1 is even" (which somehow took five lines to prove). I mean, I can excuse mixing up variables, but spending too long proving trivial matters and not nearly enough time justifying his assertions isn't good proof writing (and the latter is enough to question a proof aswell). $\endgroup$ – Andrew D Jan 8 '13 at 3:56
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    $\begingroup$ Adding one to one will give you two. Can't argue with that. $\endgroup$ – mjqxxxx Jan 8 '13 at 4:11
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    $\begingroup$ The first 9 sentences only makes sense as a (sloppy) description of the iteration.After that ("Using induction.." it is gibberish, and cannot be critiqued. It can be flushed. $\endgroup$ – DanielWainfleet Sep 25 '15 at 9:23
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The best way to disprove a purported proof is to use the technique to prove something manifestly false.

For instance, as N goes to infinity, the percentage of all numbers less than N that are divisible by 4 goes to 25. But so does the percentage of all numbers less than N that are 3 mod 4. Therefore, all numbers that are divisible by 4 are also 3 mod 4 by the squeeze theorem.

When you do this, your friend will insist that you are using the squeeze theorem wrong. Ask how your friend's usage of the squeeze theorem differs from this.

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  • $\begingroup$ (to clarify, maybe your friend is NOT making this fallacy - it's unclear what he's using the squeeze theorem for at all. the point is that the question forces him to clarify.) $\endgroup$ – user29743 Jan 8 '13 at 5:01
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Apply to the Quollatz function (which I just made up): Q (n) = C (n) except that Q (137) = 274.

The "proof" applies to the Quollatz function just as well, but is obviously wrong.

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The final piece of punctuation is a good sign that this is not a proof. The author is indicating here that he or she does not have a firm conclusive argument. The onus is on the writer of the proof to fill in this glaring gap, not on you to disprove it.

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"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%"

Not true, there are no restrictions on the expectancy of terms in the collatz sequence, other then that the number of terms is finite. Anything he extrapolated from this is therefore not based on facts, making his whole proof invalid.

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