Why can no one in sci.math understand my simple point?
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From: Virgil on 20 Jun 2010 22:18 In article <4c1eb4dc$0$1027$afc38c87(a)news.optusnet.com.au>, "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Virgil" <Virgil(a)home.esc> wrote in message > news:Virgil-22B7E6.14010020062010(a)bignews.usenetmonster.com... > > In article > > <ccb88bb1-d8a6-4f04-bc6a-ca3346f7c8ad(a)c10g2000yqi.googlegroups.com>, > > WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > > >> On 20 Jun., 02:04, "Mike Terry" > >> > > >> > No, you're misunderstanding the meaning of computable. > >> > > >> > Hopefully you will be OK with the following definition: > >> > > >> > A real number r is computable if there is a TM (Turing machine) > >> > T which given n as input, will produce as output > >> > the n'th digit of r. > >> > >> Whatever might be the true meaning: The Turing machine need a finite > >> definition. Therefore the computable number has a finite definition. > >> > >> There are only countable many finite definitions. And every diagonal > >> of a defined Cantor list has also a finite definition. > > > > Cantor's argument does not require that any member of a list of reals be > > computable beyond a finite number of decimal places, so enough of each > > can be finitely defined for the proof to work. > > Exactly the same as mine. > > Cantor calculates the anti-diagonal to n places (for all n) using only the > first n digits of the first n items. So do I. Technically speaking,CANTOR'S anti-diagonal argument involves only binary sequences, not real numbers at all, though the real number form of the argument is almost always attributed to him.
From: Tim Little on 20 Jun 2010 22:29 On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > No, I only need a finite algorithm which will calculate it to n > decimal places with finite input. You have previously stated that you can do this with "about three lines of Java code". Here's your Java method signature: int antidiagonalDigit(BigInteger n). Here are the first twenty digits of the first twenty members of my list: 93525532854532512838 32127313472944276266 70595916184994935423 20733652719572401688 30472031767118774150 47190325821263633948 74236814853458351851 58521903865615844550 91701104659863267390 39510921669610680229 19656091025330568974 49591084533660072011 81683520683673830124 93720622611168810054 50284245443806050152 11702670934156383526 58534679962278312978 68478827933494896765 83579375050000862329 50241229144880453593 Now let's see your Java code. If I then give you the first 10^100 list entries to 10^10 digits, would the same code suffice? Would it have to be enlarged? How large would it be if I required any digit up to the 10^100'th? How large would it have to be if I asked for an algorihm that produces *any* digit on request? - Tim
From: Tim Little on 20 Jun 2010 22:40 On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Of course it is an algorithm. > > It accepts the first n digits of the first n terms on the list That's where you can stop. An algorithm for a computable real accepts as input *only* the desired precision. Nothing else. Since you mentioned Java previously in the thread (and may understand that better), you could think of it as follows: public interface ComputableReal { public int getDecimalDigit(BigInteger n); } If you can implement that interface with a finite self-contained algorithm, it's a computable real. If you can't, it isn't. Note that the interface has no signature for a method that takes "the first n digits of the first n terms on the list". If you want list data, you'll have to embed it within the implementation somehow. - Tim
From: Tim Little on 20 Jun 2010 22:47 On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message > I do understand that. But I don't actually ask for a "finite algorithm" in > my proof, I only require that I am provided a purported list of all > comptuable Reals. Later in the proof, you claim that the antidiagonal real is computable. In other words, that there exists a finite algorithm for computing it. > Yes, in practice, this means that each item is specified by a finite > algorithm, but so what? I ask for a purported list of all computable Reals > and show there is at least one missing. At least one *what* missing? A real? Fine - Cantor's proof shows that there is a missing real. What's that? You want the antidiagonal to be a *computable* real? Well, then - you are required to show that there always exists a finite algorithm to compute it, and that is not present in Cantor's proof. So your modified "proof" is either invalid, or you must include significant amounts of material not present in Cantor's proof. As is well-known to competent mathematicians and computer scientists, no amount of material will make it valid, as the modified "theorem" is simply false. It is *not true* that any list of computable reals has a computable antidiagonal. - Tim
From: Tim Little on 20 Jun 2010 22:52
On 2010-06-21, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > Actually, he shows that any purported list of all Reals must miss at last > one Real. > > *not* the same as being uncountable Yes it is. Look up the definitions. > witness the same argument applied to computable Reals. Applying the same argument shows that any purported list of all computable reals must miss at least one real. So? >> Actually Cantor's original "diagonal" proof was based on the set of >> all functions from the naturals to a two element set, > > No. Yes, actually. There are many more modern reforumlations of Cantor's diagonal construction, and you are likely confusing those with the original. The more modern ones are more streamlined with the benefit of hindsight. That is why you were asked by one poster *exactly which* version of Cantor's diagonal argument you are talking about. So far it seems to be a strawman proof you invented yourself and are ascribing to Cantor. - Tim |