Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: Moebius Newsgroups: sci.logic,sci.math Subject: Re: Replacement of Cardinality Date: Thu, 15 Aug 2024 19:41:22 +0200 Organization: A noiseless patient Spider Lines: 84 Message-ID: References: <6c471296-90b8-4cf7-bc9b-480bd34ef190@att.net> <1f25a3d6-7b0e-476d-aa99-ecb003cf763f@att.net> <75e2ce0e-7df8-4266-968b-9c58e4140b03@att.net> <35d8c0a1-dab3-4c15-8f24-068e8200cb07@att.net> <45ad1007-b1a7-49d0-a650-048f02738226@att.net> Reply-To: invalid@example.invalid MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Thu, 15 Aug 2024 19:41:24 +0200 (CEST) Injection-Info: dont-email.me; posting-host="6f15604c04b8929e2981b2c365f0c048"; logging-data="1078736"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX184Ds26BCC7o+aNA1+4X/DN" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:R41s51kl6SMCAU+2HP0LCHz8CfA= Content-Language: de-DE In-Reply-To: Bytes: 4938 Am 15.08.2024 um 19:34 schrieb Moebius: > Am 15.08.2024 um 19:01 schrieb Moebius: >> Am 13.08.2024 um 19:02 schrieb Jim Burns: >>> On 8/13/2024 10:21 AM, WM wrote: >> >>>> [There is] a real [number] x with NUF(x) = 1. >>> >>> INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋ >>> >>> NUF(INVNUF(1)) > 1 >>> Contradiction. >> >> Well, no, this just isn't a proof, sorry about that, Jim. >> >> (Hint: The term "INVNUF(1)" is not defined. Hence you may not even use >> it in a proof by contradiction. Actually, you didn't state an >> assumption in your "proof".) >> >> Proof by contradiction: >> >> Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR such >> that NUF(x0) = 1. This means that there is exactly one unit fraction u >> such that u < x0. Let's call this unit fraction u0. Then (by >> definition) there is a (actually exactly one) natural number n such >> that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by >> definition) 1/(n0 + 1) is an unit fraction which is smaller than u0 >> and hence smaller than x0. Hence NUF(x0) > 1. Contradiction! >> >> (Of course, it's clear that I'm using the same "proof idea" as you >> used in your attempt of a proof.) > > Ok, I'll give it a try. > > Assume that there is an x e IR such that NUF(x) = 1. > > May we (by using this assumption) define INVNUF in such a way that 1 is > in its domain? > > I mean can we define > >        INVNUF(n) = _first_ x with NUF(x) = n     for n e {1, ...} > > now? (This would allow to use the term INVNUF(1) in our proof.) > > For this we would have to show/prove that there is a _first_ x with > NUF(x) = 1. And for this we would have to show that (a) there is an x e > IR such that NUF(x) = 1 (which is our assumption, so nothing to do here) > and (b) that there is no smaller real number x' such that NUF(x') = 1. > But I DOUBT that we will be able to prove/show that (even with our > assumption). > > So I consider this a dead end. > > --------------------------------------------------------------------- > > Oh, wait! From our assumtion we can derive a contradiction (see my proof > above)! Hence we can even prove (b) now (in the context of this proof by > contradiction)! For that we assume ~(b) and immediately get ~~(b) and > hence (b) from the assumption ~(b) and our original assumption, by > contradiction. (The assumption ~(b) is now "discharged"). > > Now we can define INVNUF such that 1 is in its domain. And hence now we > may use your line of thought: > >>> INVNUF(1) > ⅟ ⌊⅟INVNUF(1) +1⌋ > ⅟ ⌊⅟INVNUF(1) +2⌋ >>> >>> NUF(INVNUF(1)) > 1 >>> Contradiction. > > And hence we finally get: there is no x e IR such that NUF(x) = 1.  qed > > Well... Im mean, in this case we would need (the mayor part of) my original proof [...] Let x0 e IR such that NUF(x0) = 1. This means that there is exactly one unit fraction u such that u < x0. Let's call this unit fraction u0. Then (by definition) there is a (actually exactly one) natural number n such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 + 1) is an unit fraction which is smaller than u0 and hence smaller than x0. Hence NUF(x0) > 1. Contradiction! as a "subproof". You see... (unnecessarily complicated).