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 : 23. Sep 2024, 18:58:36
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <519db81b-4a4d-417d-8cd2-7fef5a342efd@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
User-Agent : Mozilla Thunderbird
On 9/23/2024 8:57 AM, WM wrote:
On 22.09.2024 22:22, Jim Burns wrote:
On 9/22/2024 2:37 PM, WM wrote:
On 22.09.2024 19:44, Jim Burns wrote:
There is no point next to 0.
>
There is a smallest unit fraction
because
there are no
unit fractions without a first one
when counting from zero.
>
Each visibleᵂᴹ unit.fraction ⅟k has
a counter.example ¼⋅⅟k to its being smallest.
>
Yes.
Consider the point β between
the visibleᵂᴹ unit.fractions and
lower.bounds of the visibleᵂᴹ unit.fractions.
VUnitFractionBounds ᵉᵃᶜʰ≤ β ≤ᵉᵃᶜʰ VUnitFractions
⎛ Imagine β being positive.
⎜
⎜ ½.β is a visibleᵂᴹ.unit.fraction lower.bound
⎜ 2.β is among the visibleᵂᴹ unit.fractions.
⎜ 0 < ½.β < β < 2.β
⎜
⎜ 2.β is among the visibleᵂᴹ unit.fractions.
⎜ For example, for some visibleᵂᴹ ⅟k
⎜ 0 < ½.β < β ≤ ⅟k < 2.β < ⅟1
⎜
⎜ But ¼⋅⅟k is also a visibleᵂᴹ unit.fraction,
⎜ and ¼⋅⅟k is where a visibleᵂᴹ unit.fraction
⎜ cannot be: below the bound.
⎜ 0 < ¼⋅⅟k < ½.β < β ≤ ⅟k < 2.β < ⅟1
⎜
⎜ ½.β is a not.bounding bound,
⎝ which is gibberish.
A positive greatest.lower.bound of
visibleᵂᴹ unit.fractions
is gibberish.
Also,
any positive lower.bound of
visibleᵂᴹ unit.fractions
would be greater.than.greatest
and also gibberish.
Therefore
the smallest unit fraction must be dark.
A smallest unit.fraction,
visibleᵂᴹ or darkᵂᴹ, is positive
and, if it weren't gibberish,
would be a positive lower.bound of
visibleᵂᴹ unit.fractions.
However,
a positive lower.bound, visibleᵂᴹ or darkᵂᴹ,
of visibleᵂᴹ unit.fractions
is gibberish.
Therefore,
a smallest unit.fraction, visibleᵂᴹ or darkᵂᴹ,
is gibberish.
No positive point, unit.fraction or otherwise,
is NOT undercut by some visible unit fraction.
>
Wrong.
NUF(x) grows from 0 to more.
This increase cannot avoid 1
How to avoid 1
0 < ⅟⌈2+⅟x⌉ < ⅟⌈1+⅟x⌉ < x
because of ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
∀n ∈ ℕ: 1/n ≠ 0
0 is the point between
points with no unit.fractions below and
points r which,
for each countable.to k
there are more.than.k below
⎛ ∀ᴿr > 0:
⎜ ∀k ∈ Nᵈᵉᶠ:
⎝ 0 < ⅟⌈k+1+⅟x⌉ < ⅟⌈1+⅟x⌉ < x