Why can no one in sci.math understand my simple point?
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From: Sylvia Else on 21 Jun 2010 09:11 On 21/06/2010 1:39 PM, Newberry wrote: > > Not sure why you think you had to tell us how the anti-diagonal is > defined. You claimed you could CONSTRUCT it. Please go ahead and do > so. I'm sure he will - right after you provide the list of reals. Sylvia.
From: Jesse F. Hughes on 21 Jun 2010 09:18 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> writes: >> Yes, it was. Polite correction of your errors wasn't getting >> anywhere, and you did not appear to be aware of how bad it looks to be >> declaiming "Cantor didn't prove that the reals were uncountable!" >> while obviously not grasping some of the most basic concepts in set >> theory, proof, and computability. >> > > Again, rude. But accurate and apparent to all but one person following this thread. -- "You are beneath contempt because you betray mathematics itself, and spit upon the truth, spit upon decency, and spit upon the intelligence of the world. You betrayed the world, and now it's time for the world to notice." -- James S. Harris awaits Justice for crimes against Math.
From: Inverse 19 mathematics on 21 Jun 2010 10:17 On Jun 15, 9:06 am, 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. > > -- > Daryl McCullough > Ithaca, NY WHY THIS OBSESSION with infinite sequence of PI number illusion, which after all is derived as an approximation of tape measurements. Why the Xenophobia for a new way to look at this via the -1 tangent mathematics. The Pi is more related to 180/19 / 3.016666666667 == 3.14044780459 or close , depending on the mathematics. 180/3.14044780459= true radian of around 57.31666. This finally will lead to a correction of the degree value itself , but these values are very close. If any Pi sequence is truly infinite, its value should be infinite and there is no evidence therof Hope research
From: WM on 21 Jun 2010 11:15 On 21 Jun., 15:05, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 21/06/2010 7:51 PM, WM wrote: > > > > > > > On 20 Jun., 22:18, Virgil<Vir...(a)home.esc> wrote: > >> In article > >> <f92c169d-ee85-40c2-aa82-c8bdf06f7...(a)j4g2000yqh.googlegroups.com>, > > >> WM<mueck...(a)rz.fh-augsburg.de> wrote: > >>> On 20 Jun., 17:51, "Jesse F. Hughes"<je...(a)phiwumbda.org> wrote: > > >>>> Cantor was this utterly insane freak who chose not to accept Newberry's > >>>> word for it, and instead *prove* that there was no list of all real > >>>> numbers. Obviously, his proof is nonsense, because, after all, Newberry > >>>> said there was no list. > > >>> His proof is nonsense because it proves that a countable set, namely > >>> the set of all reals of a Cantor-list and all diagonal numbers to be > >>> constructed from it by a given rule an to be added to this list, > >>> cannot be listed, hence, that this indisputably countable set is > >>> uncountable. > > >> That is a deliberate misrepresentation of the so called "diagonal proof". > > > But this proof can be applied to this countable set and shows its > > uncountability. > > Let's see. How might that work... > > OK, by way of example, take the set of rationals. It's countable, so we > can list them. > > Now construct an anti-diagonal. It's clearly not in the list, so... > > ... um, well what, exactly? It is an irrational number. Add it at first position to the former list L0, obtain L1. Now construct the diagonal of L1 (according to a fixed scheme). It is certainly an irrational number. Add ist to the list L1 at first position, obtain list L2. Now construct the diagonal, ... and continue. I this way you get a countable set consisting of all rationals and of all diagonal numbers of these lists. This set is certainly not countable, because you can prove that there is always a diagonal number not in the list. On the other hand, the set is countable by construction. What now? Regards, WM
From: Mike Terry on 21 Jun 2010 15:35
"WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:0e1c7c2c-32e9-4f67-b79f-502b81c21aa4(a)y11g2000yqm.googlegroups.com... > On 21 Jun., 15:05, Sylvia Else <syl...(a)not.here.invalid> wrote: > > On 21/06/2010 7:51 PM, WM wrote: > > > > > > > > > > > > > On 20 Jun., 22:18, Virgil<Vir...(a)home.esc> wrote: > > >> In article > > >> <f92c169d-ee85-40c2-aa82-c8bdf06f7...(a)j4g2000yqh.googlegroups.com>, > > > > >> WM<mueck...(a)rz.fh-augsburg.de> wrote: > > >>> On 20 Jun., 17:51, "Jesse F. Hughes"<je...(a)phiwumbda.org> wrote: > > > > >>>> Cantor was this utterly insane freak who chose not to accept Newberry's > > >>>> word for it, and instead *prove* that there was no list of all real > > >>>> numbers. Obviously, his proof is nonsense, because, after all, Newberry > > >>>> said there was no list. > > > > >>> His proof is nonsense because it proves that a countable set, namely > > >>> the set of all reals of a Cantor-list and all diagonal numbers to be > > >>> constructed from it by a given rule an to be added to this list, > > >>> cannot be listed, hence, that this indisputably countable set is > > >>> uncountable. > > > > >> That is a deliberate misrepresentation of the so called "diagonal proof". > > > > > But this proof can be applied to this countable set and shows its > > > uncountability. > > > > Let's see. How might that work... > > > > OK, by way of example, take the set of rationals. It's countable, so we > > can list them. > > > > Now construct an anti-diagonal. It's clearly not in the list, so... > > > > ... um, well what, exactly? > > It is an irrational number. Add it at first position to the former > list L0, obtain L1. Now construct the diagonal of L1 (according to a > fixed scheme). It is certainly an irrational number. Add ist to the > list L1 at first position, obtain list L2. Now construct the > diagonal, ... and continue. I this way you get a countable set > consisting of all rationals and of all diagonal numbers of these > lists. No... What you get is an infinite sequence of lists (L0, L1, L2, ...) Each Ln in the sequnce contains all rationals, and Ln contains the antidiagonals for L0,...L(n-1). Ln doesn't contain its own antidiagonal, and doesn't contain any antidiagonals for Lm with m>n. > This set is certainly not countable, because you can prove that > there is always a diagonal number not in the list. What set? It seems you're thinking there is some "limit list" for the sequence of lists (L0, L1, L2, ...), or do you mean a specific list Ln for some n? If you mean some kind of limit list, please define exactly what this is. > On the other hand, > the set is countable by construction. What now? That depends on how you answer my previous question! Regards, Mike. |