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From: cbrown on 24 Dec 2006 02:55 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > > > > > > > I can calculate 1 + 1/2. > > > > > > > > Good for you! > > > > > > Bad for you, if you can't. > > Can't you? > > > > I know that 2/3 < 1 is a convenient way to express 2 < 3. > > > > For heaven's sake. "2/3 < 1" is not just "a convenient way" of > > expressing "2 < 3". > > It is *also*, among others, a convenient way of expressing "2 < 3 > Let G be the additive group of rational numbers. Let H be the additive group of dyadic rationals (i.e., rationals of the form x/2^y). Then H < G. Do you think it then makes sense to say "1 < G/H"? Are you aware that it is more usual to say "G/H is isomporphic to the additive group of rationals with odd denominator? > > > > E(n)/P(n) -> 2 is a feature only of finite n. |E(n)| < |P(n)| is a > > feature only of infinite n. /Many/ things that can be said in the > > finite case are not true in the infinite case; > > But 1 + 1/2 + 1/4 + ... < 1 is *not* among those. > You say this because you "feel" it, not because you have a proof of it. What exactly do you mean by "... < 1"? > > for example E(n) is > > finite for natural n, and not finite for infinite n. > > For example {1,2,3,...n} <--> {2,4,6,...2n} is a bijection for finite > sets. It is obviously false for infinite sets. Huh? There /is/ a bijection between the set of all naturals and the set of even naturals. It is the function f(x) = 2*x. > Never thought about that topic? Presumably not, because you had to learn > that stuff from authorities. Here you see how harmful belief in > authorities can be. > What "authorities" do you claim I am quoting when I state "the function f(x) = 2*x is a bijection between the set of all naturals, and the set of all even naturals"? I am simply applying specific definitions of the terms "function", "bijection", "the set of all naturals" and "the set of all even naturals", using the usual logic. > It is when you try to argue "lim n-> oo E(n)/P(n) = 2 contradicts > > E(n)/P(n) < 1 in the infinite case" that your confusing notation > > "looks" like a contradiction (to you; to me it looks like nonsense). > > Then you really do accept 1 + 1/2 + 1/4 + ... < 1 ? Of course not. How do you infer that from my statements? Cheers - Chas
From: Franziska Neugebauer on 24 Dec 2006 03:47 ,----[ <458b675c$0$97264$892e7fe2(a)authen.yellow.readfreenews.net> ] | mueckenh(a)rz.fh-augsburg.de wrote: | | > Franziska Neugebauer schrieb: | | ,----[<4589b948$0$97268$892e7fe2(a)authen.yellow.readfreenews.net> ] | | >> ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ] | | >> >> | >> >> > If an unbounded diagonal exists, then obviously an | | >> >> | >> >> > unbounded line must exist. [(*)] | | [...] | | >> >> That does not prove (*) since the whole diagonal is not | | >> >> bounded | | >> >> by any such n. | | >> > | | >> > If you doubt my proof, | | >> | | >> Which proof? You simply stated: | | >> | | >> (1) forAll elem in diag Exists line (m is in line) | | >> (2) forAll elem in diag (elem is finite natural number) | | >> (3) forAll natural n (forAll elem in diag <= n | | >> is in line n) | | >> | | >> So please show us step-by-step how (*), which states | | >> | | >> (*) notExists natural n (forAll m in diag (m <= n)) | | >> -> Exists natural n' (notExists natural n(forAll m | | >> in line n' (m | | >> <=n)), | | >> | | >> follows from (1), (2) and (3). | | > | | > See my recent reply to William Hughes. | | | | Please copy and paste. I am not able to guess what you have in mind. | `---- | | [...] | > We talked about the potentially infinite set N. | >> Which means that for any n we can find a single line L(n) that | >> works | >> for this n. It does not mean that that we can find a single | >> line L_D, that works for all n. | > | > That is impossible, beause there is no "all n". | | If you can't proof (*) you should write so. | | > If "all n" were somewhere, then we had no problem to put them in one | > single line. | | We are not disputing "all n are somewhere" but the validity of (*). | Until now (*) is not valid. | | > Due to the finiteness of every n, every position of the | > line was finite. And due to the fact that a number with only finite | > indexes is a natural number, the whole number was a natural number. | | I don't know hat "whole number" you are writing about but it seems to | me that you compulsively try to apply your well-known unproved AND | invalid convictions ("N cannot be infinite because A n e N (n is | finite)"). `---- mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> > Recently we had the following claim (by Virgil) >> > >> > The number of equations >> > >> > 1+1+1+...+1 = n >> > >> > with finitely many 1 is infinite. Considering the fact that there >> > are exactly as many 1 as different numbers, it is impossible that >> > the number of numbers can be infinite while the sme number of 1 is >> > finite. >> >> The number of sums (left to the "=") and the number of n's (right to >> the "=") is not bounded. What precisely is your problem? > > The infinitely many sumands in > > 1 > 1+1 > 1+1+1 > ... 1. Every line contains finitely many summands. 2. Every n e N is finite. Do you agree? > yield an infinitude of sums but not an infinitude of summands. 3. Every single sum contains only finitely many summands. 4. You may revisit my posting of November 21: ,----[ <4562fc49$0$97261$892e7fe2(a)authen.yellow.readfreenews.net> ] | Condider the following scenario: | | state 1: Given a parking space with *two* cars having their | wheels mounted. (each car has 4 wheels) | | state 2: Unmount all the wheels of both cars and put them on one | stack of wheels. (stack of wheels has 8 wheels) | | Proposition 1: "There are cars in state 1 which have as many wheels | as the stack of wheels in state 2 has." | | Proposition 2: "There is *a* car in state 1 which has as many | wheels as the stack of wheels in state 2 has." | | Now my interpretation: Claim 1 states that there are at least two (due | to the plural form "cars" and due to the "are" instead of "is") cars | each of which have the property of having as many wheels as the stack | of wheels has (8). That is wrong in my view. | | Claim 2 states that there is (at least) one car which has as many | wheels as the stack of wheels has (8). Which is wrong, too. | | Question 1: Do you agree with my interpretation? `---- > How can finitely many summands in this linear order yield an > infinitude of sums? That is your problem? F. N. -- xyz
From: Franziska Neugebauer on 24 Dec 2006 03:58 context: > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > ,----[<4589b948$0$97268$892e7fe2(a)authen.yellow.readfreenews.net> ] > > | >> ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ] > > | >> >> | >> >> > If an unbounded diagonal exists, then obviously an > > | >> >> | >> >> > unbounded line must exist. [(*)] > > | [...] > > | >> >> That does not prove (*) since the whole diagonal is not > > | >> >> bounded > > | >> >> by any such n. > > | >> > > > | >> > If you doubt my proof, > > | >> > > | >> Which proof? You simply stated: > > | >> > > | >> (1) forAll elem in diag Exists line (m is in line) > > | >> (2) forAll elem in diag (elem is finite natural number) > > | >> (3) forAll natural n (forAll elem in diag <= n > > | >> is in line n) > > | >> > > | >> So please show us step-by-step how (*), which states > > | >> > > | >> (*) notExists natural n (forAll m in diag (m <= n)) > > | >> -> Exists natural n' (notExists natural n(forAll m > > | >> in line n' (m > > | >> <=n)), > > | >> > > | >> follows from (1), (2) and (3). > > | > > > | > See my recent reply to William Hughes. > > | > > | Please copy and paste. I am not able to guess what you have in > > | mind. > > `---- > > > > [...] > > > We talked about the potentially infinite set N. > > >> Which means that for any n we can find a single line L(n) that > > >> works > > >> for this n. It does not mean that that we can find a single > > >> line L_D, that works for all n. > > > > > > That is impossible, beause there is no "all n". > > > > If you can't proof (*) you should write so. > > > > > If "all n" were somewhere, then we had no problem to put them in > > > one > > > single line. > > > > We are not disputing "all n are somewhere" but the validity of (*). > > Until now (*) is not valid. > > > > > Due to the finiteness of every n, every position of the > > > line was finite. And due to the fact that a number with only finite > > > indexes is a natural number, the whole number was a natural number. > > > > I don't know hat "whole number" you are writing about but it seems to me > > that you compulsively try to apply your well-known unproved AND invalid > > convictions ("N cannot be infinite because A n e N (n is finite)"). > > > > > Recently we had the following claim (by Virgil) > > > > > > The number of equations > > > > > > 1+1+1+...+1 = n > > > > > > with finitely many 1 is infinite. Considering the fact that there are > > > exactly as many 1 as different numbers, it is impossible that the > > > number of numbers can be infinite while the sme number of 1 is finite. > > > > The number of sums (left to the "=") and the number of n's (right to the > > "=") is not bounded. What precisely is your problem? mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> > In short: If any natural number n does exist, then there exists a >> > line in which this and any smaller number does exist. >> > This holds for every number n. >> >> Every natural n. >> >> > You know that the logical symbol for every is forall. >> >> I don't see how this could support your "famous" quantifier reversal. >> Hence that does not prove the existence of a _single_ line (*). > > Show me a number which is not in a single line. I do not claim that there is a _natural_ number which is not in a single line. Hence I do not have to show what you have asked for. > Or show me two lines, each of which lacks a number the other one does > not lack. I do not claim that there are two lines each of which mutually lacks a number the other one does not lack. Hence I do not have to show what you have asked for. I still wait to read your proof of (*) which is derived from (1), (2) and (3). F. N. -- xyz
From: Franziska Neugebauer on 24 Dec 2006 04:06 mueckenh(a)rz.fh-augsburg.de wrote: > no irrational number can be approximated better than to about 10^-100. Linguistic Question: Is it meaningfull to speak of an "approximation" if one denies the existence of the thing which is approximated (the irrational number)? F. N. -- xyz
From: mueckenh on 24 Dec 2006 04:30
stephen(a)nomail.com schrieb: > Newberry <newberry(a)ureach.com> wrote: > > > What is supposed to follow from all this? > > That what is true "in the limit" is not necessarily true for the infinite > case. For instance the diagonal number of Cantor's proof differs from any list entry of a finite list. But it certainly does not differ from any list entry of an infinite list. (Compare the case 1.000...1 = 0.999...) For instance the bijection {1,2,3,...,n} <--> {2,4,6,...,2n} holds for every finite sequence, but it does not hold for an infinite sequence (obviously, because there are more natural numbers than even atural numbers). It is a pity that mathematicians have not known your theorem earlier. It would have spared us much confusion. Regards, WM |