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

On 9/3/2024 6:22 AM, WM wrote:

On 03.09.2024 06:25, Jim Burns wrote:

On 9/2/2024 4:37 PM, WM wrote:

On 02.09.2024 19:19, Richard Damon wrote:

as any unit fraction you might claim to be
that one has a unit fraction smaller than itself,
so it wasn't the smallest.

>
Your argument stems from visible unit fractions
but becomes invalid in the dark domain.

>
The darkᵂᴹ domain
  between 0 and visibleᵂᴹ unit.fractions
is empty.

>
Then you could see the smallest unit fraction.

If the smallest unit.fraction existed,
you could see it positive and undercut by
a visibleᵂᴹ smaller.than.smallest unit.fraction.
But you can't see that.
The smallest unit fraction doesn't exist.
If the darkᵂᴹ domain held a positive lower.bound of
visibleᵂᴹ unit fractions,
you couldn't see it , but it would be undercut by
lower.than.lower.bound visibleᵂᴹ unit fractions.
But that can't be.
The darkᵂᴹ domain is empty.

Remember that they are fixed points with
non-empty gaps on the real line.

Around each visibleᵂᴹ ⅟k, mark a gap of size g/2ᵏ
gap size g/2ᵏ decreases exponentially faster than
gap separation ⅟k-⅟(k+1)
If gaps do not overlap initially,
they do not overlap ever.
Gₖ = g/2+g/4+g/8+...+g/2ᵏ
½⋅(g+Gₖ)  =
  ½⋅(g+g/2+g/4+g/8+...+g/2ᵏ)  =
  g/2+g/4+g/8+...+g/2ᵏ+g/2ᵏ⁺¹  =
  Gₖ+g/2ᵏ⁺¹
½⋅(g+Gₖ)  =  Gₖ+g/2ᵏ⁺¹
Gₖ  =  g-g/2ᵏ  <  g
For visibleᵂᴹ unit.fraction ⅟j, let g = ½⋅⅟j²
All the visibleᵂᴹ unit.fractions ⅟ℕᴰᴱꟳ and their gaps
fit between 0 and ⅟j

Hence there is a first one.

For each visibleᵂᴹ unit fraction ⅟k
⅟(k+1) disproves by counter.example that ⅟k is first.
A darkᵂᴹ unit.fraction is between
0 and the visibleᵂᴹ unit fractions.
A darkᵂᴹ unit.fraction implies that
½⋅glb.⅟ℕᴰᴱꟳ both is and is not undercut by
visibleᵂᴹ unit.fractions.
Darkᵂᴹ unit fractions do not exist.
A first unit fraction does not exist.

Each positive point is undercut by
visibleᵂᴹ unit.fractions,

>
No.
Only each visible positive point is undercut by
visible unit.fractions.

Each darkᵂᴹ positive point is positive.
Each darkᵂᴹ positive point is
a positive lower.bound of visibleᵂᴹ unit.fractions.
Each (hypothetical) darkᵂᴹ positive point undercuts
the (hypothetical) positive greatest.lower.bound β of
visibleᵂᴹ unit.fractions.
2⋅β > β (hypothetically)
2⋅β is undercut by visibleᵂᴹ ⅟k
½⋅β is undercut by visibleᵂᴹ ¼⋅⅟k
However,
½⋅β < β (hypothetically)
½⋅β is NOT undercut.
A darkᵂᴹ positive point forces contradiction.
Therefore,
Only visibleᵂᴹ positive points exist.