Re: how

On 5/22/2024 6:43 AM, WM wrote:

Le 22/05/2024 à 01:27, Jim Burns a écrit :

ℝ is ℚ and points between non.∅ splits of ℚ

for any x > 0

>
that you can determine

For any x > 0 in ℚ or between a non.∅ split of ℚ
more.than.any.k<ℵ₀ unit.fractions
sit before x
among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋
0 < rₓ/sₓ < x

for any x > 0
more.than.any.k<ℵ₀ unit.fractions
sit before x
among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋
>

Ax_def > 0: NUF(x_def) = ℵo is right.

>
∀x ∈ ℝ: x > 0 ⇒ NUF(x) = ℵ₀

>
If
there is no unit fraction smaller than all x > 0,
then
there is an x > 0 preventing this.

There is no unit.fraction ⅟k smaller than ⅟k⁺¹
¬∃¹ᐟᴺ ⅟k: ⅟k ≤ ⅟k⁺¹
There is no unit.fraction ⅟k smaller than all x > 0
¬∃¹ᐟᴺ ⅟k: ∀ᴿx > 0: ⅟k ≤ x
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
| Assume otherwise.
| Assume x¹ᐟᴺ > 0:  ∀¹ᐟᴺ ⅟k: x¹ᐟᴺ ≤ ⅟k
|
| b¹ᐟᴺ is the greatest lower bound of unit fractions.
| 0 < x¹ᐟᴺ ≤ b¹ᐟᴺ
|
| 0 < ½⋅b¹ᐟᴺ < b¹ᐟᴺ < 2⋅b¹ᐟᴺ
| No unit.fraction < ½⋅b¹ᐟᴺ
| Unit fraction ⅟k < 2⋅b¹ᐟᴺ
|
| However,
| ⅟k < 2⋅b¹ᐟᴺ
| (⅟k)/4 < (2⋅b¹ᐟᴺ)/4
| ⅟(4⋅k) < ½⋅b¹ᐟᴺ
| Unit.fraction ⅟(4⋅k) < ½⋅b¹ᐟᴺ
| Contradiction.
Therefore,
there is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k

That is merciless logic.

∀y:∃x≠y:x<y  ⟺
¬∃y:¬∃x≠y:x<y  ⟺
¬∃y:∀x≠y:¬(x<y)  ⟺
¬∃y:∀x≠y:y<x  ⟺
¬∃x:∀y≠x:x<y