Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?"

share|improve this question
2  
To be honest, the "proof" is barely readable. For example, what is x? –  Andrew D Jan 8 '13 at 3:44
    
I think he mixed up his 'x' and 'n' variables, where x = n –  Waffle Jan 8 '13 at 3:47
    
how he can uses the squeeze theorem in the proof? i don´t find the way... –  dwarandae Jan 8 '13 at 3:50
1  
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). –  Andrew D Jan 8 '13 at 3:56
    
Or 'x' could have been an entirely new variable in a separate statement (n would have been fine as well) –  Waffle Jan 8 '13 at 3:58
show 3 more comments

3 Answers 3

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.

share|improve this answer
    
(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.) –  user29743 Jan 8 '13 at 5:01
add comment

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.

share|improve this answer
add comment

"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 false, making his whole proof false.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.