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

On 10/29/24 4:36 AM, WM wrote:

On 29.10.2024 00:58, Richard Damon wrote:

On 10/28/24 3:42 PM, WM wrote:

 

I mean that there are unit fractions.
None is below zero.
Mathematics proves that never more than one is at any point.
>

Which doesn't mean there must be a first, as they aproach an accumulation point where the density becomes infinite.

 Their density is bounded by uncountably many points between every pair of consecutive unit fractions:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
The density is one point over uncountably many points, that is rather precisely 0.

>
Something which can't happen your world of finite logic, but does when the logic can handle infinities.

 Where does the density surpass 1/10? Can you find this point? If not it is another proof of dark numbers.
 Regards, WM
 

Remember, density is a measure in a REGION, not at a point.
Passing 1/10 points per unit is easy, just take the interval [1, 11].
To get densities that low, you need to include a lot of the flat space after 1,
Your interval you like to talk about [1/(n+1), 1/n] will have 2 unit fractions in a distance of 1/(n(n+1)) or a total density of 2n(n+1) points per unit. (make the interval (1/(n+1), 1/n] or [1/(n+1), 1/n) and you drop it to just n*(n+1) points per unit which is probably a better estimate so we don't double count the endpoints.
This shows that as our region reaches towards zero, the slope of NUF increases (except that it is infinity there already, so doesn't change.
If we try to move the lower end from being at a unit fraction, to being at zero, the slope just goes up to its limit of infinity, thus showing that NUF(x) CAN jump from 0 to ALeph_0 when moving from 0 t a finite positive value.