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riddles >> general problem-solving / chatting / whatever >> The 3n+1 Conjecture is Unprovable?!
(Message started by: Barukh on Nov 3rd, 2004, 4:33am)

Title: The 3n+1 Conjecture is Unprovable?!
Post by Barukh on Nov 3rd, 2004, 4:33am
Read this article (http://arxiv.org/PS_cache/math/pdf/0312/0312309.pdf).

:-/

Title: Re: The 3n+1 Conjecture is Unprovable?!
Post by THUDandBLUNDER on Nov 3rd, 2004, 5:25am
Yet he admits the possibility of the conjecture being false, either provably or not.

Actually, this paper has been rubbished on sci.math (http://mathforum.org/discuss/sci.math/t/179676?hi_n=611231,611233,611313,611328,611566,610003,610095,610170), particularly by mathematicians Robin Chapman and Torkel Franzen. Expert opinion seems to indicate that his approach is fundamentally flawed. For example:

Try reading the Feinstein paper. "Theorem 2" states
that "For any m, n in N, if each T^{(k)}(n) is distinct
(for k = 0, 1, ..., n-1) then it is impossible to determine
the value of T^{(m)}(n) without performing at least m computations.

Leaving aside the private language (what does Feinstein mean by
"m computations"? He doesn't tell us) note the universal
quantifiers on the m and n ("For any"). The family of counterexamples
m arbitrary and n = 2^m immediately come to mind. Each of these
satsifies the hypothesis, and it takes no "computations" to prove
that T^m(n) = 1 for these m and n.

(Chapman)

AND

> Feinstein proved in theorem 2 that it takes m steps to prove that
> T^m(n)=1 for some m.

 By all means let us stipulate that he did. The idea is that "since
the value of F(m_0) is determined in showing that there exists an m
such that F(m)=1", where m_0 is the smallest m such that F(m)=1,
proving that there exists an m such that F(m)=1 must involve at least
as many steps as proving that F(m_0)=1. This idea is mistaken, or
perhaps just confused. Proving the existence of an m such that F(m)=1
does not in general involve any "determination" of F(m_0) in the sense
of actually computing F(m_0).

(Franzen)





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