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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). :-/ |
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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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