Why can no one in sci.math understand my simple point?
in [Logic]
|
From: K_h on 15 Jun 2010 18:13 "fishfry" <BLOCKSPAMfishfry(a)your-mailbox.com> wrote in message news:BLOCKSPAMfishfry-9EA4D6.23584214062010(a)news.giganews.com... > In article <87ocucFrn3U1(a)mid.individual.net>, > "|-|ercules" <radgray123(a)yahoo.com> wrote: > >> Consider the list of increasing lengths of finite >> prefixes of pi >> >> 3 >> 31 >> 314 >> 3141 >> .... >> >> Everyone agrees that: >> this list contains every digit of pi (1) >> > > No, I don't agree, so "Everyone agrees that ..." is false. > > The list consists of a collection of integers. Item n on > the list are > the first n digits of pi, starting from 3 and ignoring the > decimal > point. So the 1000th item on the list is 31... pi to 1000 > places. > > There is no one element of the list that contains pi in > its entirety. > And the reason is because each 'n' represents a FINITE > NUMBER. Like 6, > or 100043, or a zillion eleven. And on that line we find a > zillion > eleven digits of pi. But no more! > > No one item on the list contains pi in its entirety. True, there is no entry for pi, in its entirety, on the list but all of the digits of pi are there along the diagonal. _
From: George Greene on 15 Jun 2010 18:24 On Jun 15, 4:34 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > 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? Well, that depends on why you are here. If you are here because your being here might somehow be constructive (in your opinion) FOR ANY human being other than your personal solipsistic self, then you should be attacking people for doing things that are harmful. You should be categorizing bad actors as bad actors and attempting deterrence as best you can conceive it. Otherwise, by not being part of the solution, you remain part of the problem.
From: Virgil on 15 Jun 2010 18:25 In article <b6b07da2-13e2-4172-91ab-d2cf297c1522(a)y11g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Jun., 21:03, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > WM says... > > > > >On 15 Jun., 18:46, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > > >> I'm not sure what you are saying. The fact is, we can prove > > >> that for every real r_n on the list, d is not equal to r_n. > > > > >Of course. Every real r_n belongs to a finite initial segment of the > > >list. > > >That does not yield any result about the whole list > > > > On the contrary, the definition of "d is on the list" > > is that "there exists a natural number n such that r_n = d". > > We proved "forall n, r_n is not equal to d". So that > > means "there does not exist a natural number n such that r_n = d", > > so that means "d is not on the list". > > > > We have thus proved something about the whole list. > > > > >> That means that d is not on the list. There is no extrapolation > > >> involved. > > > > >Look here: We can prove for any finite segment > > >{2, 4, 6, ..., 2n} > > >of the ordered set of all positive even numbers that its cardinal > > >number is surpassed by some elements of the set. > > > > >Nevertheless this appears not be a proof that the cardinal number of > > >the whole set is less than some elements of the set. > > > > So there we have an example of an illegitimate extrapolation. > > If you prove Phi(n) for an arbitrary natural number n, then you are allowed > > to conclude: > > > > forall natural numbers n, Phi(n). > > > > So you can conclude: > > > > forall natural numbers n > 0, the set of all even numbers less than or > > equal to 2n has a cardinality less than 2n. > > > > That's true. That's a legitimate proof. On the other hand, it is not > > legitimate to conclude: > > > > The set of all even numbers has a cardinality that is less than > > some even number. > > > > That's an unwarranted extrapolation. > > > > So there are legitimate proofs, and there are bogus proofs. > > And Cantor's proof shows something for every natural number: Every > line numerated with a natural number from 1 to n does not contain the > diagonal. Not just from 1 TO n, but from 1 PAST every n. > But to conclude that the whole set, i.e., the infinite list, does not > conclude the diagonal is a bogus conclusion. Is that "conclude" supposed to b "contain"? I will presume so. So that WM regards as sometimes valid a statement that what is true for each member of a set need not be true for every member of that set? That is even more outre that WM usually gets.
From: George Greene on 15 Jun 2010 18:26 On Jun 15, 5:02 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > So WM calls me a crank. Big deal. Exactly. WM is himself a crank and a bully at best, and possibly (at worst) A TROLL. We should all consume the contempt that we receive from the contemptible as a vitamin.
From: Virgil on 15 Jun 2010 18:29
In article <87sk4ohwbt.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > Virgil <Virgil(a)home.esc> writes: > > > Note that it is possible to have an uncomputable number whose decimal > > expansion has infinitely many known places, so long as it has at least > > one unknown place. > > You need infinitely many unknown places. If the value of some decimal digit of a number depends on the truth of an undecidable proposition, can such a number be computable? |