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

On 10/31/24 3:32 PM, Chris M. Thomasson wrote:

On 10/31/2024 4:36 AM, Richard Damon wrote:

On 10/30/24 10:04 AM, WM wrote:

On 30.10.2024 13:57, FromTheRafters wrote:

WM explained :

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NUF(0) = 0, NUF(1) = ℵo. Therefore NUF must grow but cannot grow by more than 1 at any point x of the real axis.

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The number of unit fractions less than x is always aleph_zero for positive x.

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Believe what you like without foundation.
If ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 is true, the NUF(x) grows in steps of not more than 1.
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Regards, WM
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No, what that expression shows is that for every n, the unit fraction 1/ n has another unit fraction 1/(n+1) that is smaller than it, so there is no "first" unit fraction in that sense for NUF(x) to get to 1.
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You are just proving you don't undetstand even your own mathematics.

 Does WM think that there is a unit fraction that is "closest" to 0? If so, he is moronic yet again... Wow.

Yep, that is his claim, that there MUST be a smallest unit fraction (and any smaller are somehow "dark").
Which of course just gets him admitting he is wrong, but trying to hide it by defining those extra ones are hidden under that carpet and we can't use them.