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From: Moebius <invalid@example.invalid>
Newsgroups: sci.logic,sci.math
Subject: Re: Replacement of Cardinality
Date: Thu, 15 Aug 2024 19:41:22 +0200
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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).