Re: how

Liste des GroupesRevenir à s math 
Sujet : Re: how
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 23. May 2024, 20:52:39
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <6b388f9a-9309-41a1-b445-39bbb73b020d@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 5/23/2024 8:10 AM, WM wrote:
Le 22/05/2024 à 22:48, Jim Burns a écrit :
On 5/22/2024 3:13 PM, WM wrote:

Disprove this:
Between ℵo unit fractions
there are at least ℵo real numbers x.
For them NUF(x) = ℵo is wrong.
>
For any x > 0
there are more.than.any.k<ℵ₀ unit.fractions < x
>
If you are right,
Arithmetic says
x  >  ⅟⌊(1+⅟x)⌋  >  ⅟⌊(k+1+⅟x)⌋  >  0

If you are right,
then there is a contradiction,
since I am right with absolute certainty.
Hence we have to find a way
to satisfy both statements:

JB:
For any x > 0
there are ℵ₀ smaller unit fractions.
For any x > 0
there are more.than.any.k<ℵ₀ unit.fractions < x
among which are  ⅟⌊(1+⅟x)⌋  ...  ⅟⌊(k+1+⅟x)⌋

WM:
Between two unit fractions
there are ℵo real numbers x.
Between any two ⅟m < ⅟n
there are more.than.any.k<ℵ₀ real.points
among which are (1/n+k⁺¹/m)/k⁺²  (2/n+k/m)/k⁺²
(3/n+k⁻¹/m)/k⁺²  ...   (k⁺¹/n+1/m)/k⁺²

I have shown the way: Dark numbers.
Darkᵂᴹ numbers in ℚ and between splits of ℚ
which are between 0 and all unit fractions
do not exist, neither darklyᵂᴹ nor visiblyᵂᴹ
We can know that they don't exist by starting with
that description and then making not.first.false
claims until we get to a contradiction.
If they exist, the contradiction must be true.
It can't be true, so they don't exist.

In accordance with:
There is no unit fraction smaller than all x > 0,
Also true:
There is no x > 0 smaller than all unit fractions.

and even in accordance with
For any unit fraction there are ℵ₀ smaller real x > 0.
Also true:
For any x > 0 there are ℵ₀ smaller unit fractions.

Note that
points on the real axis are fixed and
not subject to quantifier nonsense.
Also true:
The opposite of a quantifier shift:
∀x:∃y≠x:x<y  ⇒  ¬∃y:∀x≠y:x<y

Date Sujet#  Auteur
2 Apr 25 o 

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