Re: Another Reason Why Collatz is Unprovable
- From: "Craig Feinstein" <cafeinst@xxxxxxx>
- Date: 1 Jun 2006 09:18:54 -0700
My proof on arXiv.org rules out a proof by contradiction because it
rules out any proof. It is a completely logical proof. No one on usenet
has found any flaws in the proof. The only things people have been
doing are claiming that my definition of "random" is not rigorous
(because it does not specify a formal language, which is really
irrelevant in the context of my proof) and giving straw-man arguments
which prove false statements and claiming that my proof uses the same
type of arguments. If you have faith in logic, then you should have
faith in my proof.
Another poster gave another argument against my proof which I neglected
to mention. That is, "We already knew one cannot prove a "for all n" by
treating every integer individually, that's why mathematicians have
learned to make general arguments."
This is the most bizarre argument I've heard, and I never really
responded to it. But I might as well respond now as it appears that
more than one person believes it and I may be able to help some people
improve their critical thinking skills by changing their minds:
My proof never claims that you have to treat every integer
individually. My proof says: let's pretend that we have a proof of
Collatz with L bits. It then shows that there is a specific n for which
any proof that Collatz halts at one with input n requires at least L+1
bits. From this, I conclude that any proof of Collatz must have at
least L+1 bits, so we have a contradiction, as our pretend proof only
has L bits. Therefore, Collatz is unprovable.
So we see that there is absolutely no merit to arguments that it is
possible to make some sort of general argument which doesn't talk about
any specific integers and is clever enough to get around my proof. A
general argument must cover all specific cases. If the shortest proof
for any specific case is at least L+1, then it logically follows that
the shortest proof for any general case is at least L+1.
Craig
.
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