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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: How many different unit fractions are lessorequal than all unit
 fractions?
Date: Thu, 12 Sep 2024 14:20:09 -0400
Organization: i2pn2 (i2pn.org)
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On 9/12/24 1:54 PM, WM wrote:
> On 12.09.2024 14:35, Richard Damon wrote:
>> On 9/12/24 7:18 AM, WM wrote:
>>> Le 12/09/2024 à 03:00, Richard Damon a écrit :
>>>
>>>> So, you can't "index" an unbounded set of unit fractions from 0, as 
>>>> there isn't a "first" unit fraction from that end.
>>>>
>>>> We can "address" those unit fractions with the value, but we can not 
>>>> "index" them from 0, only from 1/1.
>>>
>>> If you can index all unit fractions, then you can index them from 
>>> every side.
>>> Fact is that NUF(x) increases from 0, but at no point it can increase 
>>> by more than 1 because of
>>> ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
> 
>> Nope, that ASSUMPTION just means you can't actually have an infinite 
>> set, as you can't get to the upper end to let you count down.
> 
> Wrong. Dark numbers prevent counting to the end. Dark numbers establish 
> the existence of a set where no end can be seen. That is the only way to 
> make infinity and completeness compatible.
>>
>> 1/n - 1/(n+1) > 0 thus says that no 1/n is the smallest, as there will 
>> exist a 1/(n+1) that is smaller that that.
> 
> That is not a proof of existence. The formula says: If n and n+1 exist, 
> then they differ.
> 
> Regards, WM

No, it says that if n is in the Natural Numbers, then the value of the 
expression 1/n - 1/(n+1) is greater than 0, and thus must exist, and 
thus n+1 must exist.

If you think there is some n that exist that doesn't have an n+1 that 
exist, then your idea of the Natural Number system is just incorrect, 
and you are admitting that you only have  afinite set of numbers, and 
thus NUF(1) must be a finite number, not Alehp_0, so everything you have 
said about it is just a lie.