Cantor Confusion
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From: Poker Joker on 29 Sep 2006 21:43 "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message news:1159578634.539857.299400(a)b28g2000cwb.googlegroups.com... > Poker Joker wrote: >> >> By analogy, what you're saying is: >> >> >> >> For ANY x >> >> there is a procedure to find a y such that x/y = 1. >> > >> > I said no such thing by analogy or otherwise. >> >> I did. > > Right, so when you said "By analogy, What YOU're saying is", [all caps > added], the word 'your'e' does not belong there. > >> >> Because we are using the verbage "ANY", we don't >> >> have to worry about special cases like when x = 0. >> >> That's how mathematicians work? >> > >> > No, you've got it reversed. >> >> No, I don't. You neglected to point out the difference in >> the analogy. > > I did. There's no analogy with any signficant feature of what I said, > and the reason I say that is just as I posted to which you just > responded by saying I don't understand the analogy. > > My points stand. You've not refuted uncountability proofs nor Arturo's > point to which you originally responded. > > If you reject first order logic, then fine. No one says you have to > accept first order logic. Then maybe ou can tell us what system of > logic for mathematics you use instead. I understand everything you understand. I just understand more, and that confuses you.
From: MoeBlee on 29 Sep 2006 21:51 Poker Joker wrote: > No its not. Its like if you give me a proof that zero isn't a real > number: > > For ANY x, there is procedure to construct a y, such that x = 1/y. > When x = 0, there is no y. > Therefore x is not a real number. > > If everybody neglects the fact that the construction isn't valid > for x=0, then the proof is flawless. The point is that we are NOT be able to prove that for any x there is y such that x=1/y. Whatever we argue about x, if there is special case blocking the generalization from x, then then our argument will NOT go through for an arbitrary real number x. If we prove that for any x there is a y such that x=1/y, then would have had to use the assumption that x not= 0. So that would NOT be an argument regarding an arbitrary real number x. Look, here's what we have: Mr. Mathematician says: Let p be an arbitrary prime number. [Insert proof here that p has the property that there is a prime number greater than p.] Therefore, for ALL prime numbers p, we have that p has the property that there is a prime number greater than p. Goober Boober says: But what if p is even? What if p is greater than 100? What if p is a successor of a prime? Indeed, what if p is such that there is NOT a prime greater than p? Hey, Mr. Mathematician, you didn't take into consideration all these special cases, did you? No, you didn't. So your argument is invalid. MoeBlee
From: MoeBlee on 29 Sep 2006 21:59 Poker Joker wrote: > "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message > news:1159579135.514717.296590(a)m7g2000cwm.googlegroups.com... > > > Look, suppose x is an arbitrary even number and I prove that therefore > > x has property P. I then conclude that all even numbers have property > > P. > > The property could be say, that x is prime (assuming you neglect the > special case of 2 and nobody notices like you've done below). So what if x is prime? We proved that as long as x is an even number, then it has property P, and we assumed nothing about x other than that it is an even number. So whether x is prime or not prime, we proved that x has property p, therefore all even numbers have property p. > > Then you say, "What if x is prime? What if x is greater than 100? What > > if x is a successor of a prime? What if x is square? What if any number > > of special cases hold?" > > > > Then I say, as any mathematician would say, "If any of those special > > cases disqualified some even number from having property P, then I > > would not have been able to prove that P holds for an ARBITRARY even > > number." > > > > You REALLY don't understand that? > > Yes, I understand. I also understand other things that you don't. I guess > we'll leave it at that. No, you don't understand, otherwise you would not be talking about x being prime. And you would not object to the uncountability proof on the basis that you are. We prove for an ARBITRARY enumeration into the reals that that arbitrary enumeration has the property of not have the set of real numbers as its range. So we conclude that ALL enumerations into the reals have the property of not having the set of real numbers as its range. And we don't have consider YOUR objections that we may not have considered special cases or that we missed some special case. We don't have to consider such objections because, IF there were a special case that blocks the generalization, then it would have blocked the proof for an ARBITRARY enumeration into the reals. That principle of inferring from the arbitrary to the general is just plain logic, ancient or modern logic, it doesn' t matter. It's plain logic. MoeBlee
From: MoeBlee on 29 Sep 2006 22:04 Poker Joker wrote: > > My points stand. You've not refuted uncountability proofs nor Arturo's > > point to which you originally responded. > > > > If you reject first order logic, then fine. No one says you have to > > accept first order logic. Then maybe ou can tell us what system of > > logic for mathematics you use instead. > > I understand everything you understand. I just understand more, and > that confuses you. Like I said, you've not refuted the uncountability of the reals nor Arturo's point to which you originally responded. MoeBlee
From: Virgil on 29 Sep 2006 22:46
In article <DCiTg.25582$QT.2854(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Virgil" <virgil(a)comcast.net> wrote in message > news:virgil-A8F506.18030329092006(a)comcast.dca.giganews.com... > > In article <UeiTg.25580$QT.16960(a)tornado.rdc-kc.rr.com>, > > "Poker Joker" <Poker(a)wi.rr.com> wrote: > > > >> "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message > >> news:efj3bk$120f$1(a)agate.berkeley.edu... > >> > In article <N_YSg.1208$3E2.403(a)tornado.rdc-kc.rr.com>, > >> > Poker Joker <Poker(a)wi.rr.com> wrote: > >> >> > >> >>"Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message > >> >>news:efgfhd$261u$1(a)agate.berkeley.edu... > >> >>> In article <1159410937.013643.192240(a)h48g2000cwc.googlegroups.com>, > >> >>> <the_wign(a)yahoo.com> wrote: > >> >>>>Cantor's proof is one of the most popular topics on this NG. It > >> >>>>seems that people are confused or uncomfortable with it, so > >> >>>>I've tried to summarize it to the simplest terms: > >> >>>> > >> >>>>1. Assume there is a list containing all the reals. > >> >>>>2. Show that a real can be defined/constructed from that list. > >> >>>>3. Show why the real from step 2 is not on the list. > >> >>>>4. Conclude that the premise is wrong because of the contradiction. > >> >>> > >> >>> This is hardly the simplest terms. Much simpler is to do a ->direct<- > >> >>> proof instead of a proof by contradiction. > >> >>> > >> >>> 1. Take ANY list of real numbers. > >> >>> 2. Show that a real can be defined/constructed from that list. > >> >>> 3. Show that the real from step 2 is not on the list. > >> >>> 4. Conclude that no list can contain all reals. > >> >>> > >> >> > >> >>How can it be simpler if the list can be ANY list instead of a > >> >>particular one. > >> > > >> > Because a direct proof is simpler than a proof by contradiction. > >> > > >> >> ANY list opens up more possiblities than > >> >>a single list. > >> > > >> > Any list does not require you to assume that there is a "single list" > >> > which some some particular property. > >> > > >> > ====================================================================== > >> > "It's not denial. I'm just very selective about > >> > what I accept as reality." > >> > --- Calvin ("Calvin and Hobbes" by Bill Watterson) > >> > ====================================================================== > >> > >> We all noticed you neglected this logic: > >> > >> if its true for ANY list, then it must be > >> true for a specific list. So if considering a single specific list > >> shows a flaw, then looking at ANY (ALL of them) list doesn't > >> help. > > > > But the construction method, which applies equally well to all lists > > without modification, if flawless for any list is equally flawless for > > all lists. > > > > And it is flawless for any list. > > Define a real number r between 0 and 1. Denote > r_n to be the nth digit of r. The digits of r are defined > as follows: > > r_n = 5 if r_n = 4, otherwise r_n = 4. > > That construction applies equally well to all lists > without modification. If it is flawless for any list it is > flawless for all lists. > > And it is flawless for any list. > > That's what you're saying. I see where you are > coming from and I won't bother answering such > "flawless" logic anymore, so don't bother. Since Poker Joker has has no legitimate answer I am not surprised that he chickened out. |