Re: how

Le 22/05/2024 à 17:48, Jim Burns a écrit :

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

Between each pair of unit fractions, there is a finite distance (with many points x > 0).
Hence not even two unit fractions can satisfy the condition to sit before any x > 0.
Therefore Ax > 0: NUF(x) = ℵo is wrong.

 

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

>
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k

There is an x >= 0 smaller than all unit fractions.

 | Assume otherwise.
| Assume x¹ᐟᴺ > 0:  ∀¹ᐟᴺ ⅟k: x¹ᐟᴺ ≤ ⅟k

That does not destroy this condition:
Between each pair of unit fractions,
there is a finite distance.
Hence not even two unit fractions can satisfy the condition to sit before any x > 0.
Therefore Ax > 0: NUF(x) = ℵo is wrong.
Why do you never address this fact?
Of course if all numbers were visisble, we had a contradiction. That does not change by your repeated proofs of this fact. Try to refute this fact: Between each pair of unit fractions, there is a finite distance. Therefore of many unit fractions, all but at most one are not smaller than every x > 0. Hence
Ax > 0: NUF(x) = ℵo is wrong.
Regards, WM