Cantor Confusion
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From: mueckenh on 25 Dec 2006 16:45 Virgil schrieb: > In article <1166909252.274423.224430(a)i12g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Are you saying that x^2 = 2 does not have a solution? > > > > Yes. But that is another topic. (In short: There are less than 10^100 > > bits in the universe. Therefore, no irrational number can be > > approximated better than to about 10^-100. > > Who says one has to be satisfied with an approximation? Everybody knows it - or could know it. There is no n-ary representation of an irrational with more than 10^100 bits. But, alas, for his diagonal argument, Cantor needs an n-ary representation with more than 10^100 bits. ======================================== >> You like to snip in such a way that a wrong meaning of my words becomes >> manifest? > Since your words are full of wrong meanings, anything that reveals this > truth is to be supported. Yeah. THAT's it. Heil Virgil! Have you meanwhile changed your opinion on: > 2^omega, representing the set of functions from omega to {0,1}, > is demonstrably not countable. 2^omega is demostrably a countable ordinal. You are in error (same as with the set of paths). Have you meanwhile changed your opinion on: >>You cannot have an actually infinite finite number. > WM may not be able to, but his inadequacies are personal, not universal. Would you consider to change also your wrong opinion on: > If WM were the least bit amenable to reason, he would see > his logic requires that every infinite path in a infinite binary > tree must have a terminal (leaf) node at its "end". Because then you could see that every separated path (or bunch of paths) must have an initial node where it starts as a separated path (or bunch of paths). Regards, WM
From: mueckenh on 25 Dec 2006 16:52 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > [...] the problem of the tree [...] > > Today there is only one "finite" tree I care for. A green one. Fröhliche Weihnachten! > > >> That is your problem? > > > > Not really. I know that it is not a problem but only the relegious > > confession of some matheologists. > > I thought you want to discuss _your_ problems. I am so happy to have none. I wish to help you. > > >> Linguistic Question: Is it meaningfull to speak of an "approximation" > >> if one denies the existence of the thing which is approximated (the > >> irrational number)? > > > > Not as long as not all deny its existence. > > To me this reads > > As long as all deny the existence of the entity to be approximated > it is not meaningful to speak of an "approximation". > > Is this correct? > > I do not deny its existence. Do you? What meaning does have an > "approximation" to you if the approximated entity (irrational number) > is supposed to not exist? Ther number does not exist, but the idea does exist. One can approximate an idea by a number. Regards, WM
From: mueckenh on 25 Dec 2006 16:58 Newberry schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Newberry schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Newberry schrieb: > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > > You need to construct a surjection. > > > > > > > > No, I don't. > > > > > > > > > > > > > > Yes you do, if you want to show that there exists a surjection > > > > > > > from edges to paths. > > > > > > > > > > > > You cannot construct a surjection because irrational numbers do not > > > > > > exist. > > > > > > > > > > Are you saying that x^2 = 2 does not have a solution? > > > > > > > > Yes. But that is another topic. (In short: There are less than 10^100 > > > > bits in the universe. Therefore, no irrational number can be > > > > approximated better than to about 10^-100. The Cauchy criterion for > > > > converging series fails.) > > > > > > What cannot be approximated? > > > > You know the Cauchy criterion. For every epsilon > 0 you must be able > > to find an n_0 such that for n > n_0 we have |a_n - pi| < epsilon. > > > > If you have less than 10^100 bits available, then you cannot realise an > > a_n which reproduces more than the first 10^100 bits of a number like > > pi. (Unless pi was a number showing some pattern, like the number > > 0.111.., which would make it possible to express every a_n using less > > than 10^100 bits.) > > > > You cannot find such a number a_n without having pi already. But this > > number shall be used to establish the existence of pi. This is > > impossible. pi does not exist *as a number* (it exists as an idea > > So it does exist after all. Yes, the idea exists. But in order to apply Cantor's diagonal proof to irrational numbers, we cannot use the ideas but need a decimal (or another n-ary) representation. This is impossible. And that is frequently overlooked. > BTW, an algorithm exists for calculating > ALL its digits. The algorithmus by BBP is not sufficient to calculate digits with very large indexes. (See their examples.) Anyhow, no algorithmus allows to store and to know more than 10^100 digits of any irrational number simltaneously. Merry Christmas. Regards, WM.
From: mueckenh on 25 Dec 2006 17:01 David R Tribble schrieb: > muecken h wrote: > >> Not yet a number, unless you can specify it such that one (at least > >> you) can decide whether q < n or q = n or q > n for any natural number > >> n given to you. Only saying that it is not equal to any natural number > >> given is not enough. > > > > David R Tribble schrieb: > >> I did not say that q is not equal to any given natural number. > >> I said that q is not equal to any natural number that has been > >> "brought into existence". Trichotomy is well defined for q. > >> For any natural n given, if n has been "brought into existence", > >> then q > n. > >> > >> Since you believe that only those numbers that have been "brought > >> into existence" exist, you should readily grasp the definition of q. > > > > mueckenh wrote: > > Yes. You mean a number q which exists but is larger than any existing > > number. > > [...] > > It is not my assumption that a non existing number does exist. Could > > well be an axiom of set theory. > > Well, now I'm confused. Could you provide an example of a natural > number that does not exist? Take the first 10^100 digits of pi (if you can - but you cannot). It is impossible to bring this number to existence in the whole universe. Regards, WM
From: Franziska Neugebauer on 25 Dec 2006 19:03
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: [...] >> >> Linguistic Question: Is it meaningfull to speak of an >> >> "approximation" if one denies the existence of the thing which is >> >> approximated (the irrational number)? >> > >> > Not as long as not all deny its existence. >> >> To me this reads >> >> As long as all deny the existence of the entity to be >> approximated it is not meaningful to speak of an >> "approximation". >> >> Is this correct? >> >> I do not deny its existence. Do you? What meaning does have an >> "approximation" to you if the approximated entity (irrational number) >> is supposed to not exist? > > Ther number does not exist, but the idea does exist. 1. So you agree that there _exist_ two entities x1, x2 e R which solve the equation x^2 = 2? 2. Is it true that what "we" call an "irrational number" (x1, x2)) is _identical_ to your "idea"? 3. If so, I cannot see what a meaningful, substantial difference between your "irrational idea" and the common "irrational number" could be. What - besides pure terminology - is that difference? > One can approximate an idea by a number. I do not see that one can in general numerically approximate an idea. F. N. -- xyz |