Another AC anomaly?
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From: Ross A. Finlayson on 22 Dec 2009 20:27 On Dec 19, 5:20 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <500d09c4-6322-417e-a4d9-f9716b19e...(a)c34g2000yqn.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > > > > > On Dec 19, 12:14 pm, Virgil <Vir...(a)home.esc> wrote: > > > In article > > > <2fac8bb1-4c90-4421-b559-1ea7f0301...(a)e27g2000yqd.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > Aha, you are clearly a mindreader. Well, as far as I know > > > > > mindreading > > > > > > > is not part of mathematics. Anyhow, I can think of numbers larger > > > > > than > > > > > > > that path. > > > > > > > But that is completely irrelevant. I am able to think about a set > > > > > that > > > > > > > contains all natural numbers, you apparently are not. > > > > > > > > > > > > How do you know that without confirming it by thinking the last too? > > > > > > Why need I to think about a last one (which there isn't) to be able to > > > > > think > > > > > about a set that contains all natural numbers? Apparently you have > > > > > some > > > > > knowledge about how my mind works that I do not have. > > > > > Yes. A very convincing and often required proof of completenes of a > > > > linear set is to know the last element. > > > > Knowing the allegedly last element in a set does not work unless one has > > > imposed a linear ordering on the set, which is in no wise necessary for > > > sethood. Which is the "last" point of the set of points on a circle? > > > > > One of the rebuttals has meanwhile been changed. Peter Webb > > > > recognized: It is true that you cannot show pi as a finite decimal, > > > > but you can't show 1/3 as a finite decimal either. > > > > > Just what I said. > > > > > > > > > The tree contains all paths that can be constructed by nodes, > > > > > using > > > > > > > > the axiom of infinity. Which one would be missing? > > > > > > > > > > > > > > The infinite paths because you stated a priori that your tree did > > > > > not > > > > > > > contain infinite paths. So it is impossible to construct in your > > > > > tree > > > > > > > infinite paths by the axiom of infinity. > > > > > > > > > > > > The axiom of infinity establishes the set N from finite numbers. > > > > > > It establishes the *existence* of a set N of finite numbers. > > > > > What else should be established? > > > > > > > It establishes the infinite paths as well in my tree from finite > > > > > > paths. > > > > > > No. That is impossible because you stated that the paths were finite. > > > > > What it *does* establish is the extistence of a set P of finite paths. > > > > > It is rather silly to argue about the uncountability of the set of > > > > paths. > > > > Then stop doing it. There is a perfectly adequate proof that one cannot > > > have a list of paths that contains all paths, or , equivalently, one > > > cannot have a list of all subset of N. > > > Since being capable of being listed is a necessary requirement for > > > countability, that proof eliminates countability for the set of all > > > paths s well as for the power set of any infinite set. > > > > > Only minds completely disformed by set theory could try to > > > > defend the obviously false position that there were uncountably many > > > > paths. > > > > Minds thus "deformed" by set theory are often capable of creating valid > > > proofs that WM not only cannot create but also cannot even follow. > > > > > But "10 Questions" will give you the answer why there are not > > > > uncountably many paths. There are no infinite decimal expansions of > > > > real numbers. There are not due paths in the tree. > > > > 1 = 1.000... is a real number with an infinite decimal expansion, and in > > > ZF there are infinitely many more of them. > > > > I know not what a "due path" is, but in my infinite binary tree, whose > > > nodes are the members of N, every path is an infinite subset of N, and > > > there are more of them than even WM can count. > > > > > It is as impossible to express any real number by an infinite decimal > > > > expansion as it is impossible to express 0 by a unit fraction. The > > > > rest will be explained in "10 Questions". Therefore I will stop with > > > > this topic here. > > > > Then in WM's world, 1.000... is not a real number. > > > > Odd of him, but he has always been odd. > > > There you assume the consistency of the statement "Dedekind/Cauchy is > > complete" > > Not at all! > > I merely note that every rational integer is included among the reals, > in both the Dedekind construction and the Cauchy construction, and 1 is > nicely both rational and integral. Oh, so 1 = 1.000... now, eh. Good luck with that. Regards, Ross F.
From: Marshall on 22 Dec 2009 20:38 On Dec 22, 5:27 pm, "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > On Dec 19, 5:20 pm, Virgil <Vir...(a)home.esc> wrote: > > > Not at all! > > > I merely note that every rational integer is included among the reals, > > in both the Dedekind construction and the Cauchy construction, and 1 is > > nicely both rational and integral. > > Oh, so 1 = 1.000... now, eh. > > Good luck with that. He won't need it. 1 the ratio is the same number as 1 the natural, is the same number as 1 the member of any other set. Unless you confuse numbers and their representations. Marshall
From: master1729 on 23 Dec 2009 06:26 jesse hughes wrote : > WM <mueckenh(a)rz.fh-augsburg.de> writes: > > >> But you are not a mathematician. > > > > Do you think so? I studied mathematics and a > university council > > appointed me to teach mathematical lessons. Have > you better insights > > than they? Or have you only a different definition > of mathematics? Do > > you mean that I am not a matheologian? Then you are > right! > > That you teach mathematics is a damned shame. I'm > not sure why you > draw attention to this sad mistake and I don't know > why the > administration hasn't the sense to fix their error. > > -- > Jesse F. Hughes > "[I]f gravel cannot make itself into an animal in a > year, how could it > do it in a million years? The animal would be dead > before it got > alive." --The Creation Evolution Encyclopedia typical sci.math. nothing can convince them. nothing is a proof of anything. a degree in math or teaching math does not mean a thing. one wonders why one asks for credentials anyways or why one has to study anyway , given that whatever you reach , it will never ' count ' , ' prove ' or ' convince ' anyone or anything. like if you have a phd , but you use gmail , you must still be an idiot , and similar nonsensical reasoning !! i do not intend to judge or defend neither hughes or WM , this is just a typical sci.math response which in general does not make sense as explained above. tommy1729
From: Jesse F. Hughes on 23 Dec 2009 17:12 master1729 <tommy1729(a)gmail.com> writes: > typical sci.math. > > nothing can convince them. > > nothing is a proof of anything. > > a degree in math or teaching math does not mean a thing. > > one wonders why one asks for credentials anyways or why one has to > study anyway , given that whatever you reach , it will never ' count > ' , ' prove ' or ' convince ' anyone or anything. > > like if you have a phd , but you use gmail , you must still be an > idiot , and similar nonsensical reasoning !! > > i do not intend to judge or defend neither hughes or WM , this is > just a typical sci.math response which in general does not make > sense as explained above. I am convinced that WM is incompetent because I have read his mathematical arguments. *Of course* credentials cannot convince me otherwise. His ignorance is plain as day, and the fact that his administration overlooks it is a damned shame, not evidence of his competence. -- "I told her that I loved her. She said she loved me too. Neither one was lying, Yet it wasn't true." -- Del McCoury Band
From: WM on 27 Dec 2009 12:57
On 22 Dez., 15:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Even a matheologian should understand that: If there is no digit at a > > finite place up to that the sequence 0.333... identifies the number > > 1/3, then there is no digit at a finite place up to that the number > > 1/3 can be identified. > > Right, there is no digit at a finite place up to that the number 1/3 can be > identified. And as there are no digits at infinite places that appears to > you to be a paradox. It is not. There is *no* finite sequence of digits > that identifies 1/3. But there is an *infinite* sequence of digits that > does so. Why not pronounce it a bit clearer? We know that every digit is the last digit of a finite initial segment of digits. There is no other digit, is it? Therefore your statement can be expressed in the following way: There is no finite initial segment that identifies 1/3. But there is an infinite finite initial segment that identifies 1/3. If you read this sentence, perhaps you get an impression of what matheology is? > > I said, if there is a sequence that identifies > > 1/3, then the identifying digits must be at finite places. > > Right, all identifying digits (there are infinitely many) are at finite > places. Therefore your statement can be expressed in the following way: There is no finite initial segment that identifies 1/3. But there is an infinite finite initial segment that identifies 1/3. > > > The union of finite initial segments cannot ield an infinite initial > > segment? > > Yes. But as you have stated that your tree contained finite paths only, > such an infinite initial segment is not (according to *your* definition) > a path. The axiom of infinity states: There is an infinite set such that with every n the follower of n belongs to the set. This is true for the binary tree. With every finite path 1, ..., n the follower 1, ..., n+1 is in the tree. The union of all those paths of course is finite, as I said, because there is no infinite path, but according to the axiom it is infinite. > > > Does the sequence of 1/3 not consist of a union of all finite > > initial segments? > > It is, but also (according to *your* definition) it is not a path. There can be no question: The union of all finite initial segments is finite. But according to the axiom, it is infinite. So we can both be well pleased. Regards, WM |