Re: How many different unit fractions are lessorequal than all unit fractions?

On 9/12/2024 11:20 AM, Richard Damon wrote:

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 :
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So, you can't "index" an unbounded set of unit fractions from 0, as there isn't a "first" unit fraction from that end.
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We can "address" those unit fractions with the value, but we can not "index" them from 0, only from 1/1.

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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

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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.

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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.

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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.

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That is not a proof of existence. The formula says: If n and n+1 exist, then they differ.
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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.

WM seems to like to mix and match systems... WM says is 1.6969 in the naturals?, no therefore its dark. Well... It not so dark because its written down. It tries again. lol.

 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.