Why can no one in sci.math understand my simple point?
in [Logic]
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From: Transfer Principle on 15 Jun 2010 16:34 On Jun 14, 9:13 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > There's no need for bullying (George), it's just a maths theory. > Address the statements and questions and add your own. I want to avoid using five-letter insults in a thread unless another poster in the thread has introduced that word. Here, Herc calls George Greene a "bully." Now whether Greene deserves to be called "bully" is open to a debate, but still, there must be an underlying reason that posters like Herc and galathaea use the word "bully" to describe posters like Greene. I see that WM has joined this thread as well. Normally, I would be grateful to see Cooper and WM work together to stand up to the "bully" Greene -- except I noticed that Greene has yet to post in this thread or respond to Herc at all. Also, in this thread, WM calls Jesse Hughes a "crank," and then Daryl McCullough argues that the word "crank" more accurately describes WM than Hughes. I agree with McCullough to some extent. Based on his posting habits, Hughes is more likely to agree with posters who merit the "bully" label than those who merit the "crank" label. But, as I mentioned earlier, I will not divide posters into group, but let other posters group for themselves. And so if WM believes that "crank" is the grouping label that best describes Hughes, than who am I to interfere with that?
From: Virgil on 15 Jun 2010 16:39 In article <4e980d72-6504-45da-9126-0718bfa712e1(a)z8g2000yqz.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 14:45, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > wrote: > > Computing the diagonal number is actually very easy. > > Therefore the diagonal number is a computable real. Given a list of computable numbers, it is, but one can also have a list containing uncomputable numbers. > Then Cantor's proof works exclusively on a countable set, namely the > set of computable reals. WRONG! All one requires is that for each listed number a different decimal digit position can be computed. So that, for example if the list is such that the nth number can be computed to its nth decimal place, the list need not contain ANY computable numbers at all. > > And it shows that this set is uncountable. Since any such "antidiagonal" is at least computable from the list, all it shows is that any list of reals is incomplete. > This result is wrong. That is what WM claimed years ago, and still has not proved.
From: WM on 15 Jun 2010 16:39 On 15 Jun., 21:41, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > > > >On 15 Jun., 18:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> >We should not use oracles in mathematics. > > >> On the contrary! Many real numbers in physics are not computable > >> to infinite precision (for example, the fine structure constant). > > >Numbers are computable. The fine structure constant is a name. It has > >soem 20 letters. > > I can see that you don't understand the distinction between > use and mention. "Daryl" is a name having five letters. Daryl > is a human being. If you think that the fine structure constant is a number, then you should try to study a bit of physics in order to understand the difference between facts and names. Daryl would exist even it had no name. A name that cannot be named cannot exist. A number that cannot be defined does not exist. Your incompetence does probably not allow you to understand that. But even you should be able to understand the following: Whatever language may be chosen: The number of finite definitions in all languages over finite alphabets is a countable set. Therefore Cantor's "proof" proves that a countable set is uncountable. A fine result. Regards, WM
From: WM on 15 Jun 2010 16:42 On 15 Jun., 21:46, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > |-|ercules says... > > > > >"Daryl McCullough" <stevendaryl3...(a)yahoo.com> wrote... > >> That's *all* that matters, for Cantor's theorem. The claim > >> is that for every list of reals, there is another real > >> that does not appear on the list. > > >Yes but HOW does Cantor show that? > > You've been told many times. He shows that for every > list L of reals, there is another real antidiag(L) that > is defined in such a way that > > forall n, antidiag(L) differs from the nth real in L at > the nth decimal place. > > From this, it follows: > > forall n, antidiag(L) is not equal to the nth real. > > From this, it follows: > > antidiag(L) is not on the list L. No, that is the conclusion of a crank. It follows for all n: The antidiagonal is not in the first n lines. That is true. But it cannot be extrapolated to the whole set N. Regards, WM
From: Virgil on 15 Jun 2010 16:45
In article <f78b53d6-24d1-42e2-86bd-1dd0893b81a9(a)q12g2000yqj.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 16:06, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > > > > > >On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > > > >> (B) There exists a real number r, > > >> Forall computable reals r', > > >> there exists a natural number n > > >> such that r' and r disagree at the nth decimal place. > > > > >In what form does r exist, unless it is computable too? > > > > r is computable *relative* to the list L of all computable reals. > > That is, there is an algorithm which, given an enumeration of computable > > reals, returns a real that is not on that list. > > > > In the theory of Turing machines, one can formalize the notion > > of computability relative to an "oracle", where the oracle is an > > infinite tape representing a possibly noncomputable function of > > the naturals. > > We should not use oracles in mathematics. WM would prohibit others from doing precisely what he does himself so often? > A real is computable or not. My list contains all computable numbers: > > 0 > 1 > 00 > ... > > This list can be enumerated and then contains all computable reals. If that list is .0, .1, .00, ..., then it contains no naturals greater than 1. If that list is 0., 1., 00., ..., then it contains no proper fractions. In either case it is massively incomplete. |