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

On 10/28/24 3:52 PM, WM wrote:

On 28.10.2024 12:21, Richard Damon wrote:

On 10/28/24 6:36 AM, WM wrote:

 

NUF increases by 1 or more, but more would violate mathematics.

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No, NUF(x) jumps from 0 to Aleph_0 in the domain of finite numbers, because there is no finite x where it has the value of 0.

 It has the value 0 for all x =< 0. And it cannot jump by more than 1 at any point.

Of course it can. just not at any finite value. Since the Unit Fractions have an accumulation point at 0+, and the slope of NUF is based on the density of Unit fractions in the area, it can jump infinitely between 0 and the positive reals.

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It is just a "undefined" function.

 No, but the first steps happen at undefinable x.

No, it happens at an x that isn't a finite number.
If you allow your NUF to accept infintesimal numbers, you can define a set of number that "act" like Unit Fractions for it to count, but by the time you get to the finites, it has reached Aleph_0

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Hint: For each and every x e IR, x <= 0: NUF(x) = 0
and for each and every x e IR, x > 0: NUF(x) = aleph_0.

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That is blatantly wrong because it would require that ℵo unit fractions exist between 0 and each and every x > 0, i.e., the open interval (0, 1].

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Right, like is what happens.
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It may seem strange to a person stuck in finite logic, but is true when you understand how infinity works.

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This infinity between 0 and (0, 1] is not what I can accept.

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Note, it isn't an "infinity between" it is that the "bottom" of (0, 1] doesn't exist as a definable point.

 That is true. The bottom is dark.

No, the bottom is outside the set.
Your "Darkness" is just your attempt to hide the problems with your logic.

 Regards, WM