Sujet : Re: How many different unit fractions are lessorequal than all unit fractions?
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 07. Sep 2024, 20:01:09
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <44111bb0-2d71-4714-8c0c-0fdd0a6d1355@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 9/7/2024 7:02 AM, WM wrote:
On 07.09.2024 01:51, Jim Burns wrote:
On 9/6/2024 4:52 PM, WM wrote:
Between 0 and your defined x or epsilon,
not between 0 and every possible x.
>
If x > 0 then there is ⅟⌊1+⅟x⌋ between 0 and x
...strictly between.
x > ⅟⌊1+⅟x⌋ > 0
Then identical unit fractions differ.
Identical unit fractions are identical.
Different unit fractions are different.
In other news,
triangles have three sides.
They are identical because
NUF(x) counts them at the same x,
Counting.at.x a unit.fraction.in.⅟ℕᵈᵉᶠ∩(0,x)
does not mean the unit fraction is at x
No unit fraction in ⅟ℕᵈᵉᶠ∩(0,x) is at x.
Different unit fractions are different.
but they differ because
NUF(x) counts more than 1
...unit fractions at different points, none x...
at this x.
⅟⌊1+⅟x⌋ is not between 0 and every possible x′ > 0
but ⅟⌊1+⅟x⌋ doesn't need to be.
The claim is "between 0 and x".
You (WM) think ⅟⌊1+⅟x⌋ needs every possible
because
you think a quantifier shift is reliable.
A quantifier shift is not reliable.
A quantifier shift is not reliable.
>
That is no quantifier shift but simplest logic.
| ∀x ∈ R⁺:
| ∃u ∈ ⅟ℕᵈᵉᶠ:
| u < x
❀❀❀❀❀❀
❀ SHIFT ❀
❀❀❀❀❀❀
🛇 ∃u ∈ ⅟ℕᵈᵉᶠ:
🛇 ∀x ∈ R⁺:
🛇 u < x
Between 0 and x
there are more.than.any.k<ℵ₀ unit fractions.
>
Between 0 and every epsilon you can define.
>
Between 0 and any epsilon satisfying my description.
⎛ Each ε > 0 is least.upper.bound of
⎜ a foresplit of all positive rationals.
⎜
⎜ Each positive rational is
⎜ the ratio of two positive naturals.
⎜
⎜ Each positive rational can be
⎝ counted.to from 1
I haven't made a claim for other epsilons.
>
Fine.
There I agree.
Every of your epsilons has
ℵo smaller unit fractions in (0, eps).
Each one of my ε > 0 has, in (0,ε),
more.than.any.k<ℵ₀ unit.fractions ⅟n
such that n can be counted.to from 1
⎛ ε = lub.S
⎜ ℚ⁺ ⊇ S ≠ {}
⎜ S ∋ p/q ≤ ε
⎜ p,q ∈ ℕᵈᵉᶠ
⎜ let n = (q+p)÷p [÷ int.div]
⎜ ℕᵈᵉᶠ ∋ n > ⅟ε
⎜ ⅟ℕᵈᵉᶠ ∋ ⅟n < ε
⎜
⎜ ∀j ∈ ℕᵈᵉᶠ: ⅟(n+j) < ε
⎝ |ℕᵈᵉᶠ| = ℵ₀
This proves dark numbers.
A quantifier shift 'proves' darkᵂᴹ.numbers,
or it would prove if it weren't unreliable.
⎛ Quantifier shifts are unreliable
⎜ _even if_
⎝ someone lies and says they aren't using one.
Darkᵂᴹ δ
0 < δ <ᵉᵃᶜʰ ⅟ℕᵈᵉᶠ
proves that,
0 < δ ≤ lub.⅟ℕᵈᵉᶠ
⎛ ½⋅lub.⅟ℕᵈᵉᶠ <ᵉᵃᶜʰ ⅟ℕᵈᵉᶠ
⎝ ¬(½⋅lub.⅟ℕᵈᵉᶠ <ᵉᵃᶜʰ ⅟ℕᵈᵉᶠ)
An impossible consequence proves no.darkᵂᴹ.numbers.