No "First" init fraction

[William]

we have
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
note the universal quantifier. This holds for every unit fraction. Thus for every unit fraction (unrevealed or invisible or ...) there is another unit fraction closer to zero (the gap is finite.),
Thus the is no first unit fraction.

[Chris M. Thomasson]

1/1, 1/2, 1/3, ... on its way to its limit at zero.

[FredJeffries]

No.

Not 'to its limit'.

Only TOWARDS its limit.

It never arrives 'at zero'.'

[Chris M. Thomasson]

Right. Well, its on its way to its natural limit on a per iteration
basis, right?

i[0] = 1/1
i[1] = 1/2
i[2] = 1/3
i[3] = 1/4
i[4] = 1/5
....

[Chris M. Thomasson]

None of the iterates will ever equal zero, however, zero is its limit...

Fair enough?

[Archimedes Plutonium]

Moscow,Beijing▂▄▅█████████▅▄▃▂ mushroom cloud, Xi as Putin's stooge when Russia vaporizes Shanghai with its RS-28 Sarmat "Satan II", all because Xi was too dumb to realize you can never trust an insane person

Chris, which of these is Shangai ??? Is it 1/2

> i[0] = 1/1
> i[1] = 1/2
> i[2] = 1/3
> i[3] = 1/4
> i[4] = 1/5

So spamming William, a Sarmat never reaches its exact target but sends shock waves from a blast site.

> No.
>
> Not 'to its limit'.
>
> Only TOWARDS its limit.
>
> It never arrives 'at zero'.'

William, infinite spamming nutjob comparing unit fractions of Moscow compared to Beijing. William spamming nutjob is Moscow 0/1 and Beijing fraction is 1/0 ??

yan wyck in his daily spam says Beijing was nothing in 1999 and apparently a Russian Sarmat returned Beijing to "nothingness once again".


> Shanghai, Beijing, Shenzhen, Guangzhou, Chongqing, Tianjin, Chengdu, Hangzhou, Nanjing, Wuhan, Xi'An, Suzhou, Harbin, Shenyang, Qingdao, Zhengzhou, Dongguan, Foshan, Dalian, Jinan, Changchun, Hefei
>
> Shenzhen▂▄▅█████████▅▄▃▂ Xi as Putin's stooge when Russia vaporizes Shenzhen with its RS-28 Sarmat "Satan II", all because Xi was too dumb to realize you can never trust an insane person
>
>
>
> Is Pete Olcott in his Halting Problem, halting the vaporization of Wuhan by Putin's Russia???
>
>
> On Thursday, March 30, 2023 at 11:56:20 PM UTC-5, Volney wrote:
> > Botfly of Math and Blowfly of Physics "Putin's stooge"
> >"wasn't bolted down too tight in the first place"
> 
> On Friday, September 9, 2022 at 1:16:55 AM UTC-5, Michael Moroney wrote:
> > "Imp of Science"
> >"not one single marble of commonsense in my entire brain"
> 
> Moscow█۞█ blackout, knock out Moscow electric power lines█۞█







> & wrote:
> > _And as the Baby Xi grew up from the rice paddies and reeds of Outer
> > Manchuria, stolen by the Naxi and Zani Dictator Putin in Moscow, Xi
> > learned in school in chemical engineering that Taiwan was 1/28 the size
> > of Outer Manchuria, Emperor Qing's homeland, now occupied by homeless Russians drinking vodka, as Putin bombs Ukraine. And the nascent Xi orders
> > 1,000 divisions to the Outer Manchuria border to regain back the stolen
>
>
> > > Why Putin is 2X smarter than Xi as dictators// SCIENCE COUNCIL RULES EARTH, not petty dictators
> > > 2m views
> > >
> > >
> > 2> If Putin pushes nuclear buttons, he drags down China along with Russia into a nuclear ash waste pile, and this means Xi is a inferior junior partner to Putin. Putin will drag down Xi's China, never the reverse.
> > >
> > 2> So, one can look at the present situation on Earth and ask several logical questions about the 2 dictators of Putin's Russia and China's Xi.
> > >
> > > It is little wonder that both Russia and China dictators are combative towards the West. Because dictators never want to give up on power but stay in power all their life long. So they oppose the West because the West has grown up to democracy-- let the people have power, not one single idiot having power all his life time.
> > >
> > > Naturally, Putin will want to keep the Russian people suppressed and have Russia be a second rate government as a dictator. Same goes for China-- they never want to give up power so the people themselves choose their leader.
> > >
> > > But can we find differences in Putin and Xi themselves? Well in the West we call the Chinese inscrutable-- meaning -- little logical commonsense. And is this a valid description?? Yes of course, considering that Russia had stolen the lands of Outer Manchuria, some 28 times larger of a land mass than is Taiwan island. Yet there is Xi, spending so much time on wanting to invade Taiwan, when it is Outer Manchuria and Vladivostok (Haishenwai) that he should be focusing attention upon. And while Putin is distracted with Ukraine, is the time for Xi to recapture Outer Manchuria, the Qing dynasty empire, Qing's Manchurian homeland.
> > >
> > > What does Xi do instead??? He focuses on Taiwan and befriends Russia. Why, at this rate, if Russia takes Inner Manchuria, we can expect Xi and the Chinese Communist Party to become even more loving of Russia for stealing more land of China.
> > >
> > > And there is Xi, whose China has become rich with trading with the West, yet every day, Xi foaming at the mouth in hatred of the West.
> > >
> > > So yes, Putin is 2X smarter as a dictator than is Xi, as if Putin has Xi in his side pocket.
> > >
> > > Is there some scientific explanation as to why Xi is 2X dumber than Putin?? Perhaps, in that China is densely populated and the air pollution over all of China is worse than most countries. That Xi probably has 1/2 of his brain filled with CO and CO2 isomers and lead, and mercury and nitrous oxide and sulfur dioxide from just living in that air polluted hellhole of Beijing. Xi studied chemistry and should know this. Whereas Putin likely detox..s every evening with breathing in pure oxygen at his residence and takes oxygen breathing tanks to office and work. This easily can explain the light-headed reasoning that Xi and his foreign diplomats Wang Yi display, where Putin plays them like a chess game, --- checkmate in 7 moves.
> > >
> > > This explains why Xi hates the West for not stealing any Chinese lands and making China rich in trade, while loving Putin for stealing Outer Manchuria, and proposing having Russia push nuclear buttons, making both Russia and China a nuclear waste site after ICBMs wipe China off the map.
> > >
> > > Xi's brain is full of air pollution toxins from the nasty Chinese air. They still build a new coal fired plant in China every day. The air in China is the worst air in the entire world.
> > >
> > > Why Putin is 2X smarter than Xi as dictators// SCIENCE COUNCIL RULES EARTH, not petty dictators.
>

> > > > 2/1, AP tards:
> > > > > Give Ukraine drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/2, AP tards:
> > > > > Give Ukraine drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/3, AP tards:
> > > > > Every Russian missile fired into Ukraine met with a drone from Ukraine knocking out Moscow electric power lines
> > > > >
> > > > > Give Ukraine drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/4, AP tards:
> > > > > drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/9 (vacation?), AP tards:
> > > > > drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/9, AP tards (again):
> > > > > drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/10, AP tards:
> > > > > drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/11, AP tards:
> > > > > drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/12, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/12, AP tards again:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/13, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/14, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/15, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/16, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/17, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/18, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/19, AP tards:
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/20, AP tards:
> > > >
> > > > > Electricity out Novosibirsk &Volgograd█۞█knock out Moscow electric power lines█۞█
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/22, AP tards:
> > > > > Moscow electric blackout█۞█knock out Moscow electric power lines█۞█
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/23, AP tards:
> > > > > Moscow electric blackout█۞█knock out Moscow electric power lines█۞█
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/24, AP tards:
> > > > > Moscow electric blackout█۞█knock out Moscow electric power lines█۞█
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/25, AP tards:
> > > > > Moscow electric blackout█۞█knock out Moscow electric power lines█۞█
> > > > > _drones █۞█knock out Moscow electric power lines█۞█ Moscow, St.Petersburg, Volgograd, Vladivostok no electricity
> > > >
> > > > 2/26, AP tards:

[Chris M. Thomasson]

On 5/28/2023 2:01 PM, Archimedes Plutonium wrote:
>
>
> Moscow,Beijing▂▄▅█████████▅▄▃▂ mushroom cloud, Xi as Putin's stooge when Russia vaporizes Shanghai with its RS-28 Sarmat "Satan II", all because Xi was too dumb to realize you can never trust an insane person
>
> Chris, which of these is Shangai ??? Is it 1/2
>> i[0] = 1/1
>> i[1] = 1/2
>> i[2] = 1/3
>> i[3] = 1/4
>> i[4] = 1/5

[...]

i[1] = 1/2

[FromTheRafters]

on 5/28/2023, Chris M. Thomasson supposed :

This sequence does not converge.

[Chris M. Thomasson]

It certainly get closer and closer to zero? Right?

[Fritz Feldhase]

On Sunday, May 28, 2023 at 11:52:36 PM UTC+2, FromTheRafters wrote:
> on 5/28/2023, Chris M. Thomasson supposed :
> >

> > 1/1, 1/2, 1/3, ... on its way to its limit [...] zero.

> >
> This sequence does not converge.

Huh?!

The sequence (1/n)_(n e N) does not converge?

0 is not its limit? (i.e. lim_(n -> oo) 1/n =/= 0 ???).

Well... Interesting (sort of).

[FromTheRafters]

Chris M. Thomasson pretended :

Yes, and you can have a sequence converging to any element of the set.
Some sequences will not converge (they might converge in the reals) to
an element of the set of unit fractions. To me, nothing up there says
that this set is not discrete.

[Fritz Feldhase]

On Monday, May 29, 2023 at 3:03:47 AM UTC+2, FromTheRafters wrote:
> Chris M. Thomasson pretended :
> > On 5/28/2023 2:50 PM, FromTheRafters wrote:
> >> on 5/28/2023, Chris M. Thomasson supposed:
> >>>

> >>> 1/1, 1/2, 1/3, ... on its way to its limit at zero.
> >>>
> >> This sequence does not converge.
> >

> > It certainly [converges].
> >
> Yes, and <bla>

The sequence does not converge and it converges? Are you doing the Mückenheim her?

[FromTheRafters]

Fritz Feldhase used his keyboard to write :

Sure, when you misrepresent (strawman) what I said it sure looks that
way. A sequence of rationals which approaches the value of Pi does not
converge in the rationals because Pi is not a rational number.

Sure, it gets closer and closer to Pi, but Pi is not actually in the
set. Zero is not in the set of unit fractions, so how can a sequence of
unit fractions converge to it in the rationals?

[Fritz Feldhase]

On Monday, May 29, 2023 at 11:31:10 AM UTC+2, FromTheRafters wrote:

> Zero is not in the set of unit fractions, so how can a sequence of
> unit fractions converge to it in the rationals?

Holy shit! 0 is not a rational number in your book?

Trying to do the Mückenheim?

You are talking nonsense, man.

(Another hint: We usually don't restrict our consideration to the rational numbers when talking about the convergence of a sequence of real numbers. The sequence (1/1, 1/2, 1/3, ...) concerges, and its limit is 0.)

EOD

[WM]

This is contradicted by the following: NUF(0) = 0. NUF(1) = many. Therefore the unit fractions start between 0 and 1. Never more than one occupy a point. This enforces a first one. No way to circ umvent this conclusion.

Two contradicting results cannot exist in mathematics. The second result is pure logic. The first one is not enforced by logic and avoidable by dark numbers..

Regards, WM

[FredJeffries]

On Monday, May 29, 2023 at 10:28:17 AM UTC-7, WM wrote:

> NUF(0) = 0. NUF(1) = many. Therefore the unit fractions start between 0 and 1.

Thus our beloved professor reveals himself to be a devotee of the Kalam cosmological argument.

https://en.wikipedia.org/wiki/Kalam_cosmological_argument
https://en.wikipedia.org/wiki/The_Kal%C4%81m_Cosmological_Argument

[FromTheRafters]

Fritz Feldhase expressed precisely :

> On Monday, May 29, 2023 at 11:31:10 AM UTC+2, FromTheRafters wrote:
>
>> Zero is not in the set of unit fractions, so how can a sequence of
>> unit fractions converge to it in the rationals?
>
> Holy shit! 0 is not a rational number in your book?

It is not a unit fraction in my book. It is not a positive real in my
book.

[William]

On Monday, May 29, 2023 at 2:28:17 PM UTC-3, WM wrote:
> William schrieb am Sonntag, 28. Mai 2023 um 17:44:28 UTC+2:
> > we have
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
> > note the universal quantifier. This holds for every unit fraction.

>...and avoidable by dark numbers..
Hence, unit fractions cannot be "dark numbers".

[William]

On Monday, May 29, 2023 at 3:14:01 PM UTC-3, FromTheRafters wrote:
> Fritz Feldhase expressed precisely :
> > On Monday, May 29, 2023 at 11:31:10 AM UTC+2, FromTheRafters wrote:
> >
> >> Zero is not in the set of unit fractions, so how can a sequence of
> >> unit fractions converge to it in the rationals?
> >
> > Holy shit! 0 is not a rational number in your book?
> It is not a unit fraction in my book. It is not a positive real in my
> book.

https://xkcd.com/169/

[Chris M. Thomasson]

On 5/29/2023 2:29 AM, FromTheRafters wrote:
> Fritz Feldhase used his keyboard to write :
>> On Monday, May 29, 2023 at 3:03:47 AM UTC+2, FromTheRafters wrote:
>>> Chris M. Thomasson pretended :
>>>> On 5/28/2023 2:50 PM, FromTheRafters wrote:
>>>>> on 5/28/2023, Chris M. Thomasson supposed:
>>>>>>
>>>>>> 1/1, 1/2, 1/3, ... on its way to its limit at zero.
>>>>> This sequence does not converge.
>>>>
>>>> It certainly [converges].
>>>>
>>> Yes, and <bla>
>>
>> The sequence does not converge and it converges? Are you doing the
>> Mückenheim her?
>
> Sure, when you misrepresent (strawman) what I said it sure looks that
> way. A sequence of rationals which approaches the value of Pi does not
> converge in the rationals because Pi is not a rational number.

The iterates get closer and closer to zero... Their limit.

[Ben Bacarisse]

"Chris M. Thomasson" <chris.m.t...@gmail.com> writes:

>>>>>> on 5/28/2023, Chris M. Thomasson supposed:
>>>>>>> 1/1, 1/2, 1/3, ... on its way to its limit at zero.

> The iterates get closer and closer to zero... Their limit.

You know there are books and online courses/tutorials from which you
could learn what a limit is? Getting "closer and closer to zero" does
not mean that zero is the limit. Don't get me wrong -- zero /is/ the
limit in this case, but if you were a student of mine I'd invite you to
define a sequence that gets forever "closer and closer to zero" but
which does /not/ converge to a limit of zero.

--
Ben.

[WM]

Wrong conclusion. The unit fractions start between 0 and 1. Never more than one occupy a point. This enforces a first one.

Regards, WM

[Fritz Feldhase]

On Monday, May 29, 2023 at 8:14:01 PM UTC+2, FromTheRafters wrote:
> Fritz Feldhase expressed precisely :
> > On Monday, May 29, 2023 at 11:31:10 AM UTC+2, FromTheRafters wrote:
> >
> >> Zero is not in the set of unit fractions, so how can a sequence of
> >> unit fractions converge to it in the rationals?
> >
> > Holy shit! 0 is not a rational number in your book?
> It is not a unit fraction in my book. It is not a positive real in my
> book.

??? Ok, you are an idiot, I see.

*plonk*

> > Trying to do the Mückenheim?
> >
> > You are talking nonsense, man.
> >
> > (Another hint: We usually don't restrict our consideration to the rational
> > numbers when talking about the convergence of a sequence of real numbers. The
> > sequence (1/1, 1/2, 1/3, ...) concerges, and its limit is 0.)
> >
> > EOD

i

[Chris M. Thomasson]

Humm... Perhaps, something like:

1/2 + 1/3 + 1/4 + 1/5 ... ?

The terms get closer to zero, but the sum always gets bigger?

i[0] = 1/2
i[1] = i[0] + 1/3
i[2] = i[1] + 1/4
i[3] = i[2] + 1/5

...

?

[Chris M. Thomasson]

Not between, 1/1 is a unit fraction 0/1 is not. 1/1 is the first unit
fraction, 1/2, 1/3, ..., is their journey down to getting closer and
closer to their limit at 0/1...

[Ben Bacarisse]

"Chris M. Thomasson" <chris.m.t...@gmail.com> writes:

> On 5/29/2023 1:35 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>
>>>>>>>> on 5/28/2023, Chris M. Thomasson supposed:
>>>>>>>>> 1/1, 1/2, 1/3, ... on its way to its limit at zero.
>>
>>> The iterates get closer and closer to zero... Their limit.
>> You know there are books and online courses/tutorials from which you
>> could learn what a limit is? Getting "closer and closer to zero" does
>> not mean that zero is the limit. Don't get me wrong -- zero /is/ the
>> limit in this case, but if you were a student of mine I'd invite you to
>> define a sequence that gets forever "closer and closer to zero" but
>> which does /not/ converge to a limit of zero.
>>
>
> Humm... Perhaps, something like:
>
> 1/2 + 1/3 + 1/4 + 1/5 ... ?
>
> The terms get closer to zero, but the sum always gets bigger?

No. I meant literally what I said. There no games or word trickery
going on. Find a sequence whose terms get closer and closer to zero but
which does not converge to zero. It's much simpler than you think.

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Hint: for the terms of s(n) to get "closer and closer to zero" all we
need is that

|s(n+1) - 0| < |s(n) - 0|

--
Ben.

[Chris M. Thomasson]

On 5/29/2023 1:59 PM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>
>> On 5/29/2023 1:35 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>
>>>>>>>>> on 5/28/2023, Chris M. Thomasson supposed:
>>>>>>>>>> 1/1, 1/2, 1/3, ... on its way to its limit at zero.
>>>
>>>> The iterates get closer and closer to zero... Their limit.
>>> You know there are books and online courses/tutorials from which you
>>> could learn what a limit is? Getting "closer and closer to zero" does
>>> not mean that zero is the limit. Don't get me wrong -- zero /is/ the
>>> limit in this case, but if you were a student of mine I'd invite you to
>>> define a sequence that gets forever "closer and closer to zero" but
>>> which does /not/ converge to a limit of zero.
>>>
>>
>> Humm... Perhaps, something like:
>>
>> 1/2 + 1/3 + 1/4 + 1/5 ... ?
>>
>> The terms get closer to zero, but the sum always gets bigger?
>
> No. I meant literally what I said. There no games or word trickery
> going on. Find a sequence whose terms get closer and closer to zero but
> which does not converge to zero. It's much simpler than you think.

[...]

Humm... For some reason I am thinking about alternating the signs of the
terms?

1/1, -1/2, 1/3, -1/4, 1/5, -1/6, ?

Damn, can't be right. I need to work on some other stuff right now, but
I will get back to you. Thanks Ben! :^)

[Chris M. Thomasson]

For some reason I also thought about alternating the roots

1, -1, 1, -1

Not sure exactly why I thought of it.

[Chris M. Thomasson]

On 5/29/2023 2:04 PM, Chris M. Thomasson wrote:

They do get closer and closer to zero if we used their absolute values,
so damn. Sorry.

[Ben Bacarisse]

This is why one can't do mathematics with vague metaphors. You have
picked one meaning for getting closer and closer to zero, and you can't
see any other.

--
Ben.

[William]

And since this "fist one" is a unit fraction, anything that is true for all unit fractions must be true for this putative "fist one". In particular
it is true for every unit fraction that there is a second unit fraction between it and zero

[Chris M. Thomasson]

I can see many ways to get closer and closer to zero. Say divide by 2:

i[0] = 10
i[1] = i[0] / 2
i[2] = i[1] / 2
i[3] = i[2] / 2
i[4] = i[3] / 2
i[5] = i[4] / 2
...

It gets closer and closer to zero. But, its not what you asked for.

[Ben Bacarisse]

How about 2, 3/2, 4/3, 5/4, 6/5, 7/6, ... 1+1/n, ... Do the terms get
closer and closer to zero, forever? What is the limit?

--
Ben.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> William schrieb am Sonntag, 28. Mai 2023 um 17:44:28 UTC+2:
>> we have
>> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
>> note the universal quantifier. This holds for every unit fraction. Thus for every unit fraction (unrevealed or invisible or ...) there is another unit fraction closer to zero (the gap is finite.),
>> Thus the is no first unit fraction.
>
> This is contradicted by the following: NUF(0) = 0. NUF(1) =
> many. Therefore the unit fractions start between 0 and 1. Never more
> than one occupy a point. This enforces a first one. No way to circ
> umvent this conclusion.

Oh dear. Then, according to you, there is another error in your book.
You give the set of rationals B = {x > sqrt(2)} and say that it has no
smallest element. Define NR(z) as the number of elements of B less than
z. Now NR(sqrt(2)) = 0 and NR(2) = many so the elements of your set B
start between sqrt(2) and 2. Never more than one occupy a point. This
enforces a first one, yet you say that B has no first element.

--
Ben.

[WM]

My case is this: If the number of unit fractions is 0 at 0, then it must have decreased before. That is dictated by pure logic. The case of ZFC is that every unit fraction is succeeded by another one. That is Peano, obtained from the visible natnumbers, but not dictated by logic. Logic is stronger than maths. Therefore my case prevails.

Regards, WM

[WM]

Ben Bacarisse schrieb am Dienstag, 30. Mai 2023 um 03:49:02 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> > William schrieb am Sonntag, 28. Mai 2023 um 17:44:28 UTC+2:
> >> we have
> >> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
> >> note the universal quantifier. This holds for every unit fraction. Thus for every unit fraction (unrevealed or invisible or ...) there is another unit fraction closer to zero (the gap is finite.),
> >> Thus the is no first unit fraction.
> >
> > This is contradicted by the following: NUF(0) = 0. NUF(1) =
> > many. Therefore the unit fractions start between 0 and 1. Never more
> > than one occupy a point. This enforces a first one. No way to circ
> > umvent this conclusion.
> Oh dear. Then, according to you, there is another error in your book.
> You give the set of rationals B = {x > sqrt(2)} and say that it has no
> smallest element.

In my book there are no dark numbers.

> Define NR(z) as the number of elements of B less than
> z. Now NR(sqrt(2)) = 0 and NR(2) = many so the elements of your set B
> start between sqrt(2) and 2. Never more than one occupy a point. This
> enforces a first one, yet you say that B has no first element.

Probably my book refering correct mats is better than set theory with its completed sets. There is however the problem that every definable real number sits in a vaccum.

Regards, WM

[Gus Gassmann]

On Tuesday, 30 May 2023 at 12:07:15 UTC-3, WM wrote:
[...]

> My case is this: If the number of unit fractions is 0 at 0, then it must have decreased before.

Holy shit! Even this you manage to get wrong!!!

[Fritz Feldhase]

On Tuesday, May 30, 2023 at 5:07:15 PM UTC+2, WM wrote:

> If the number of unit fractions [smaller than x] is 0 at x = 0, then it must have decreased before. That is dictated by pure logic.

There are infinitely many unit fractions in (0, 1]. But for each and every real number x > 0, there are only finitely many unit fractions >= x. HENCE for each and every real number x > 0, there are infinitely many unit fractions smaller than x.

That is dictated by classical mathematics (which is based on classical logic and set theory).

See: https://en.wikipedia.org/wiki/Classical_mathematics

[William]

[WM]

Fritz Feldhase schrieb am Dienstag, 30. Mai 2023 um 18:28:54 UTC+2:
> On Tuesday, May 30, 2023 at 5:07:15 PM UTC+2, WM wrote:
>
> > If the number of unit fractions [smaller than x] is 0 at x = 0, then it must have decreased before. That is dictated by pure logic.
>
> There are infinitely many unit fractions in (0, 1]. But for each and every real number x > 0, there are only finitely many unit fractions >= x. HENCE for each and every real number x > 0, there are infinitely many unit fractions smaller than x.

Infinite means ordered and without visible end. Yes, ∀x ∈ (eps, 1]: SBZ(x) = ℵo.

>
> That is dictated by classical mathematics (which is based on classical logic and set theory).

It contradicts logic and therefore has to be disposed of. When different points are passed such that all have been passed at 0, then a last one before 0 has been passed. That is the more obvious since all have positive internal distances.

Regards, WM

[Fritz Feldhase]

On Tuesday, May 30, 2023 at 7:33:53 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 30. Mai 2023 um 18:28:54 UTC+2:
> >
> > There are infinitely many unit fractions in (0, 1]. But for each and every real number x > 0, there are only finitely many unit fractions >= x. HENCE for each and every real number x > 0, there are infinitely many unit fractions smaller than x.
> >
> Infinite means ordered

No, /infinite/ does not "mean ordered".

> and without visible end.

Fascinating.

Please define *visible*. Then proof your claim.

> Yes, ∀x ∈ (eps, 1]: SBZ(x) = ℵo.

*sigh* You are dumb like shit, man.

Without "specifying" eps, this is just nonsense.

On the other hand, with eps = 0, we get: ∀x ∈ (0, 1]: SBZ(x) = ℵo (which is indeed true).

[Chris M. Thomasson]

Afaict, they get closer and closer to one, so, I would say the limit is
one? Is that a wrong way of thinking? Thanks, Ben. :^)

[Chris M. Thomasson]

The terms get closer and closer on both ends. From positive to zero, and
from negative to zero.

[Chris M. Thomasson]

How about,

(1/1 + 1), (1/2 + 1), (1/3 + 1), (1/4 + 1)...

That should get closer and closer to one simply because I offset it by one?

[Chris M. Thomasson]

Since the unit fractions get closer and closer to zero, I can offset it
by anything I want to and that offset becomes the limit? Is that kook
shit, Ben? Any sequence that limits out on zero can be offset such that
it limits out on another number? Fair enough, or kooky! ;^o

[Chris M. Thomasson]

On 5/30/2023 8:07 AM, WM wrote:
> William schrieb am Dienstag, 30. Mai 2023 um 00:38:30 UTC+2:
>> On Monday, May 29, 2023 at 5:40:13 PM UTC-3, WM wrote:
>>> William schrieb am Montag, 29. Mai 2023 um 21:50:45 UTC+2:
>>>> On Monday, May 29, 2023 at 2:28:17 PM UTC-3, WM wrote:
>>>>> William schrieb am Sonntag, 28. Mai 2023 um 17:44:28 UTC+2:
>>>>>> we have
>>>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
>>>>>> note the universal quantifier. This holds for every unit fraction.
>>>>> ...and avoidable by dark numbers..
>>>> Hence, unit fractions cannot be "dark numbers".
>>> Wrong conclusion. The unit fractions start between 0 and 1. Never more than one occupy a point. This enforces a first one.
>> And since this "fist one" is a unit fraction, anything that is true for all unit fractions must be true for this putative "fist one". In particular
>> it is true for every unit fraction that there is a second unit fraction between it and zero
>
> My case is this: If the number of unit fractions is 0 at 0,

Unit fractions go from one to zero, although they will never equal zero.

You seem to be going the other way around from zero to one. What about:

(1 - 1/1), (1 - 1/2), (1 - 1/3), (1 - 1/4), ...

Now, the go from zero to one, where one is their limit? Fair enough?

[Chris M. Thomasson]

On 5/30/2023 12:56 PM, Chris M. Thomasson wrote:
> On 5/30/2023 8:07 AM, WM wrote:
>> William schrieb am Dienstag, 30. Mai 2023 um 00:38:30 UTC+2:
>>> On Monday, May 29, 2023 at 5:40:13 PM UTC-3, WM wrote:
>>>> William schrieb am Montag, 29. Mai 2023 um 21:50:45 UTC+2:
>>>>> On Monday, May 29, 2023 at 2:28:17 PM UTC-3, WM wrote:
>>>>>> William schrieb am Sonntag, 28. Mai 2023 um 17:44:28 UTC+2:
>>>>>>> we have
>>>>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
>>>>>>> note the universal quantifier. This holds for every unit fraction.
>>>>>> ...and avoidable by dark numbers..
>>>>> Hence, unit fractions cannot be "dark numbers".
>>>> Wrong conclusion. The unit fractions start between 0 and 1. Never
>>>> more than one occupy a point. This enforces a first one.
>>> And since this "fist one" is a unit fraction, anything that is true
>>> for all unit fractions must be true for this putative "fist one". In
>>> particular
>>> it is true for every unit fraction that there is a second unit
>>> fraction between it and zero
>>
>> My case is this: If the number of unit fractions is 0 at 0,
>
> Unit fractions go from one to zero, although they will never equal zero.
>
> You seem to be going the other way around from zero to one. What about:
>
> (1 - 1/1), (1 - 1/2), (1 - 1/3), (1 - 1/4), ...
>
> Now, the go from zero to one, where one is their limit? Fair enough?

Thanks Ben. :^)

[Ben Bacarisse]

The limit is indeed 1. They do indeed get closer and closer to 1. They
also get closer and closer to 1/2. And they get closer and closer to
zero. We usually measure closeness by distance, and the most natural
notion of the distance between two numbers a and b is the absolute
difference |a-b|. |2-0| = 2 but |3/2 - 0| = 3/2 because 3/2 is closer
to zero than 2 is. Each member of the 1+1/n sequence is closer to zero
than all the previous ones.

A limit is (very loosely) a number that some sequence gets /arbitrarily/
close to (and, eventually, stays close to). That is not the same as
simply getting closer and closer to some number.

>> How about,
>> (1/1 + 1), (1/2 + 1), (1/3 + 1), (1/4 + 1)...
>> That should get closer and closer to one simply because I offset it by
>> one?
>
> Since the unit fractions get closer and closer to zero, I can offset it by
> anything I want to and that offset becomes the limit? Is that kook shit,
> Ben? Any sequence that limits out on zero can be offset such that it limits
> out on another number? Fair enough, or kooky! ;^o

Nothing kooky there. Books (and so on) about the limits of sequences
will give you a whole host of similar theorems.

--
Ben.

[WM]

Fritz Feldhase schrieb am Dienstag, 30. Mai 2023 um 20:03:48 UTC+2:
> On Tuesday, May 30, 2023 at 7:33:53 PM UTC+2, WM wrote:

> > Yes, ∀x ∈ (eps, 1]: SBZ(x) = ℵo.
>

> Without "specifying" eps, this is just nonsense.

Every eps > 0 that can be chosen can be inserted here.

∀x ∈ (eps, 1]: SBZ(x) = ℵo.
>

> On the other hand, with eps = 0, we get: ∀x ∈ (0, 1]: SBZ(x) = ℵo (which is indeed true).

It is wrong as wrong can be. Starting from zero (who could forbid it?) we see that there must be a first unit fraction, because two or more are separated by gaps.

Regards, WM

[WM]

Chris M. Thomasson schrieb am Dienstag, 30. Mai 2023 um 21:56:30 UTC+2:
> On 5/30/2023 8:07 AM, WM wrote:

> > My case is this: If the number of unit fractions is 0 at 0,
> Unit fractions go from one to zero, although they will never equal zero.

[Chris M. Thomasson]

Right. The first unit fraction is 1/1

[Chris M. Thomasson]

Fwiw, I have some code that tries to detect when a field line in a
vector field starts to orbit, aka spiral around a point. The field line
starts plotting itself, then its plot might end up in an orbit around a
critical point. It fun to try to identify these points. For instance,
notice how the objects in the yellow field lines converge on a point?

https://youtu.be/oVCjAaY1pOY

This point is where they are going to orbit around forevermore. A limit,
in a sense?

>>> How about,
>>> (1/1 + 1), (1/2 + 1), (1/3 + 1), (1/4 + 1)...
>>> That should get closer and closer to one simply because I offset it by
>>> one?
>>
>> Since the unit fractions get closer and closer to zero, I can offset it by
>> anything I want to and that offset becomes the limit? Is that kook shit,
>> Ben? Any sequence that limits out on zero can be offset such that it limits
>> out on another number? Fair enough, or kooky! ;^o
>
> Nothing kooky there. Books (and so on) about the limits of sequences
> will give you a whole host of similar theorems.
>

Thanks Ben. :^)

[FredJeffries]

On Wednesday, May 31, 2023 at 12:14:52 PM UTC-7, Chris M. Thomasson wrote:

> Fwiw, I have some code that tries to detect when a field line in a
> vector field starts to orbit, aka spiral around a point. The field line
> starts plotting itself, then its plot might end up in an orbit around a
> critical point. It fun to try to identify these points. For instance,
> notice how the objects in the yellow field lines converge on a point?
>
> https://youtu.be/oVCjAaY1pOY
>
> This point is where they are going to orbit around forevermore. A limit,
> in a sense?

https://en.wikipedia.org/wiki/Attractor

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Dienstag, 30. Mai 2023 um 03:49:02 UTC+2:
>> WM <askas...@gmail.com> writes:
>> > William schrieb am Sonntag, 28. Mai 2023 um 17:44:28 UTC+2:
>> >> we have
>> >> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
>> >> note the universal quantifier. This holds for every unit fraction. Thus for every unit fraction (unrevealed or invisible or ...) there is another unit fraction closer to zero (the gap is finite.),
>> >> Thus the is no first unit fraction.
>> >
>> > This is contradicted by the following: NUF(0) = 0. NUF(1) =
>> > many. Therefore the unit fractions start between 0 and 1. Never more
>> > than one occupy a point. This enforces a first one. No way to circ
>> > umvent this conclusion.
>> Oh dear. Then, according to you, there is another error in your book.
>> You give the set of rationals B = {x > sqrt(2)} and say that it has no
>> smallest element.
>
> In my book there are no dark numbers.

So in WMaths there is no smallest unit fraction, and SUF(x) is
discontinuous at 0?

>> Define NR(z) as the number of elements of B less than
>> z. Now NR(sqrt(2)) = 0 and NR(2) = many so the elements of your set B
>> start between sqrt(2) and 2. Never more than one occupy a point. This
>> enforces a first one, yet you say that B has no first element.
>
> Probably my book refering correct mats is better than set theory with
> its completed sets. There is however the problem that every definable
> real number sits in a vaccum.

You seem to have missed the point. The argument you give for there
being a "first" (smallest) unit fraction does not involve anything
"dark" yet you don't apply it to B in your book.

--
Ben.

[WM]

When starting from 0, it is the last.

Regards, WM

[WM]

Ben Bacarisse schrieb am Donnerstag, 1. Juni 2023 um 02:34:27 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des

> Unendlichen" at Technische Hochschule Augsburg.)

> > In my book there are no dark numbers.
> So in WMaths there is no smallest unit fraction, and SUF(x) is
> discontinuous at 0?

SUF(x) = oo for every x = eps > 0.

> > Probably my book refering correct maths is better than set theory with

> > its completed sets. There is however the problem that every definable
> > real number sits in a vaccum.
> You seem to have missed the point. The argument you give for there
> being a "first" (smallest) unit fraction does not involve anything
> "dark"

The argument is that all unit fractions are separated by gaps. Therefore when staring from zero, there can only one be met first. You believe that infinitely many are met together. That is excluded by ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.

An x > 0 with |SUF(x)| = finite however, cannot be defined. Undefinable numbers are the unavoidable result.

Regards, WM

[Gus Gassmann]

On Thursday, 1 June 2023 at 10:32:30 UTC-3, WM wrote:
> Ben Bacarisse schrieb am Donnerstag, 1. Juni 2023 um 02:34:27 UTC+2:
> > WM <askas...@gmail.com> writes:
> > (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> > Unendlichen" at Technische Hochschule Augsburg.)
> > > In my book there are no dark numbers.
> > So in WMaths there is no smallest unit fraction, and SUF(x) is
> > discontinuous at 0?
> SUF(x) = oo for every x = eps > 0.

Get your own notation straight, idiot. SUF is a *SET* of unit fractions.

[FromTheRafters]

WM brought next idea :

Unit fractions don't start at zero.

[Gus Gassmann]

On Thursday, 1 June 2023 at 12:32:44 UTC-3, FromTheRafters wrote:
[..]

> Unit fractions don't start at zero.

I'd say you are right. Even WM would agree that unit fractions start at 1, followed by a forward slash, etc...

[FromTheRafters]

Gus Gassmann used his keyboard to write :

They all happen at once, and zero is still not one of them.

[Jim Burns]

I'm sure that you don't mean all of them.

May I suggest:
almost all happen when crossing 0

Step from before 0 to after 0
and cross almost all of them.
For each split of (0,1]
almost all are in the fore segment.

⟨...,⅟n⁺⁺⟩,    ⟨⅟n,...,⅟1⟩
⟨1×1 1-ended⟩, ⟨1×1 2-ended⟩
⟨ ℵ₀-many ⟩,   ⟨ < ℵ₀-many ⟩
⟨almost all⟩,  ⟨almost none⟩

And zero is still not one of them.

[Gus Gassmann]

Quite. But there are people who use "start" and "end" to imply a first and a last element (or a largest and smallest) without as much as a nod that this requires an explicit order. And it so happens that in the usual order 1/1 > 1/2 > 1/3 > ... there is a largest unit fraction but not a last. (WM: Don't bother commenting on this; we know that all you can contribute is SHIT.)

[WM]

FromTheRafters schrieb am Donnerstag, 1. Juni 2023 um 17:32:44 UTC+2:
> WM brought next idea :
> > Chris M. Thomasson schrieb am Mittwoch, 31. Mai 2023 um 20:59:15 UTC+2:
> >> On 5/31/2023 7:32 AM, WM wrote:
> >
> >>> Starting from zero (who could forbid it?) we see that there must be a first
> >>> unit fraction, because two or more are separated by gaps.
> >> Right. The first unit fraction is 1/1
> >
> > When starting from 0, it is the last.
> >

> Unit fractions don't start at zero.

But SUF(x) does.

Regards, WM

[WM]

They do not happen, but they are there. And the function NUF(x) counts them. Since they all are separated, NUF starts with 1.

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 1. Juni 2023 um 21:19:27 UTC+2:
> On 6/1/2023 2:22 PM, FromTheRafters wrote:

> > They all happen at once,
> > and zero is still not one of them.
> I'm sure that you don't mean all of them.
>
> May I suggest:
> almost all happen when crossing 0

No. The gaps are there and separate all of them from each other. A good mathematician should not overlook that.

>
> Step from before 0 to after 0
> and cross almost all of them.

They are not separated by uncountably many points each?

Regards, WM

[Fritz Feldhase]

On Friday, June 2, 2023 at 2:59:03 PM UTC+2, WM wrote:

> NUF starts with 1.

Nope. NUF "starts" with 0 and has a "jump" at x = 0.

In detail: NUF(0) = 0. And NUF(x) = aleph_0 for all x e IR, x > 0.

[Fritz Feldhase]

On Friday, June 2, 2023 at 3:02:14 PM UTC+2, WM wrote:
> Jim Burns schrieb am Donnerstag, 1. Juni 2023 um 21:19:27 UTC+2:

> > Step from before 0 [or from 0] to after 0 and cross almost all of them.
> >
> They are not separated[?]
> [After all,] the gaps [between them] are there and separate all of them from each other.

Sure, so what?

[WM]

Fritz Feldhase schrieb am Freitag, 2. Juni 2023 um 15:07:40 UTC+2:
> On Friday, June 2, 2023 at 2:59:03 PM UTC+2, WM wrote:
>
> > NUF starts with 1.
>
> Nope. NUF "starts" with 0 and has a "jump" at x = 0.

Nonsense. 0 is not a unit fraction. The first one lies at x > 0.

> And NUF(x) = aleph_0 for all x e IR, x > 0.

In detail: ℵ₀ unit fractions and their gaps cover a distance > 0.

Regards, WM

[WM]

Therefore your ""jump" at x = 0" concerns only what you know, not what there is, unknown to you. Of course I know what you mean: There is no definable x > 0 with NUF(x) < ℵ₀. That is right. But that's not all there is.

Regards, WM

[Gus Gassmann]

On Friday, 2 June 2023 at 10:39:00 UTC-3, WM wrote:
> Fritz Feldhase schrieb am Freitag, 2. Juni 2023 um 15:07:40 UTC+2:
> > On Friday, June 2, 2023 at 2:59:03 PM UTC+2, WM wrote:
> >
> > > NUF starts with 1.
> >
> > Nope. NUF "starts" with 0 and has a "jump" at x = 0.
> Nonsense. 0 is not a unit fraction. The first one lies at x > 0.

NUF can be defined on the entire set of real numbers, in which case NUF(0) = 0 --- in fact NUF(x) = 0 for x <= 0 and = ℵ₀ if x > 0.

[WM]

In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side. This means ℵ₀ points (the unit fractions) and ℵ₀ distances between them with uncountably many points each. Alltogether this is larger than 0. Are you really too fanatized to understand this?

Regards, WM

[V õ l u r]

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On Sunday, May 28, 2023 at 10:05:32 PM UTC+2, Chris M. Thomasson wrote:

> On 5/28/2023 8:44 AM, William wrote:
> > we have
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
> > note the universal quantifier. This holds for every unit fraction. Thus for every unit fraction (unrevealed or invisible or ...) there is another unit fraction closer to zero (the gap is finite.),
> > Thus the is no first unit fraction.

[FromTheRafters]

Gus Gassmann used his keyboard to write :

Yes, his reversed sequence of inverted naturals (unit fractions)
converge to zero in the reals. One of his problems is that he expects
the discrete set of (isolated) unit fractions to converge to zero even
when there is no zero in the discrete set for it to converge to.

If he wants convergence, he must accept how the reals are represented.
No dark numbers in the reals or any of its subsets. No need to use the
reals (or complex, or extended reals or hyperreals, or surreals) just
to say something about dark natural numbers.

[Gus Gassmann]

On Friday, 2 June 2023 at 11:37:43 UTC-3, WM wrote:
[...]

> In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side.

That's all of them, silly!

And now FUCK OFF.

[Gus Gassmann]

On Friday, 2 June 2023 at 12:51:38 UTC-3, FromTheRafters wrote:
> Gus Gassmann used his keyboard to write :
> > On Friday, 2 June 2023 at 10:39:00 UTC-3, WM wrote:
> >> Fritz Feldhase schrieb am Freitag, 2. Juni 2023 um 15:07:40 UTC+2:
> >>> On Friday, June 2, 2023 at 2:59:03 PM UTC+2, WM wrote:
> >>>
> >>>> NUF starts with 1.
> >>>
> >>> Nope. NUF "starts" with 0 and has a "jump" at x = 0.
> >> Nonsense. 0 is not a unit fraction. The first one lies at x > 0.
> >
> > NUF can be defined on the entire set of real numbers, in which case NUF(0) =
> > 0 --- in fact NUF(x) = 0 for x <= 0 and = ℵ₀ if x > 0.
> Yes, his reversed sequence of inverted naturals (unit fractions)
> converge to zero in the reals. One of his problems is that he expects
> the discrete set of (isolated) unit fractions to converge to zero even
> when there is no zero in the discrete set for it to converge to.

I have no idea what he expects the sets to converge to. And neither does he. In my case it is through a lack of definition, in his case it's a matter of having no neurons left to fire.

> If he wants convergence, he must accept how the reals are represented.
> No dark numbers in the reals or any of its subsets. No need to use the
> reals (or complex, or extended reals or hyperreals, or surreals) just
> to say something about dark natural numbers.

Good luck convincing him of that!!!

[Jim Burns]

On 6/2/2023 9:02 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 1. Juni 2023 um 21:19:27 UTC+2:
>> On 6/1/2023 2:22 PM, FromTheRafters wrote:

>>> They all happen at once,
>>> and zero is still not one of them.
>>
>> I'm sure that you don't mean all of them.
>> May I suggest:
>> almost all happen when crossing 0
>
> No.
> The gaps are there and
> separate all of them from each other.

That does not have the consequences
to which you feel entitled.

> A good mathematician should not
> overlook that.

and also not-overlook

| u is a unit fraction
| if and only if
| exists n ∈ ℕ⁺: n⋅u = 1

| ⅟n⁺⁺ < ⅟n ⟺ n < n⁺⁺

| ℕ⁺ ∋ 1 ∧ ∀n ∈ ℕ⁺: ℕ⁺ ∋ n⁺⁺
| (M ∋ 1 ∧ ∀m ∈ M: M ∋ m⁺⁺) ⟹ M ⊇ ℕ⁺

| A successor j=i⁺⁺ is
| non-0 non-doppelgänger non-final
| ∀i: ∃j:i⁺⁺=j ⟹
| i⁺⁺≠0 ∧ ¬∃h≠i:i⁺⁺=h⁺⁺ ∧ ∃k:j⁺⁺=k

>> Step from before 0 to after 0
>> and cross almost all of them.
>
> They are not separated by
> uncountably many points each?

The unit fractions are separated from each other.
That does not have the consequences
to which you feel entitled.

Each step across 0
splits the unit fractions so that

almost all are in the fore segment.

Almost all are in the fore segment
because
each step across splits the unit fractions,
and each split has
a 1×1 1-ended fore segment and
a 1×1 2-ended hind segment.
and
a 1×1 1-ended is always
infinitely more than
a 1×1 2-ended.

[WM]

FromTheRafters schrieb am Freitag, 2. Juni 2023 um 17:51:38 UTC+2:
> Gus Gassmann used his keyboard to write :

> > NUF can be defined on the entire set of real numbers, in which case NUF(0) =
> > 0 --- in fact NUF(x) = 0 for x <= 0 and = ℵ₀ if x > 0.
> Yes,

No. Even two unit fractions have a distance > 0 which makes NUF(x > 0) = ℵ₀ blatantly wrong.

> his reversed sequence of inverted naturals (unit fractions)
> converge to zero in the reals. One of his problems is that he expects
> the discrete set of (isolated) unit fractions to converge to zero even
> when there is no zero in the discrete set for it to converge to.

Convergence is irrelevant here. We use only isolated unit fractions > 0.

Regards, WM

[WM]

Gus Gassmann schrieb am Freitag, 2. Juni 2023 um 19:19:42 UTC+2:
> On Friday, 2 June 2023 at 11:37:43 UTC-3, WM wrote:
> [...]
> > In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side.
> That's all of them,

That's only those x which have infinitely times uncountable many points at their left-hand side.

Regards, WM

[WM]

Jim Burns schrieb am Freitag, 2. Juni 2023 um 20:08:28 UTC+2:
> On 6/2/2023 9:02 AM, WM wrote:

> > The gaps are there and
> > separate all of them from each other.
> That does not have the consequences
> to which you feel entitled.

I am entitled to this: If a sequence of reals has no term before a but has terms before b > a, then there is a beginning between a and b. If all terms have finite distances, then the beginning consists of one term. If you can't understand or won't accept this, then it is an unshakeable truth neveretheless.

> > A good mathematician should not
> > overlook that.
> and also not-overlook
>
> | u is a unit fraction
> | if and only if
> | exists n ∈ ℕ⁺: n⋅u = 1
>
> | ⅟n⁺⁺ < ⅟n ⟺ n < n⁺⁺

This does not solve the contradiction. One of the basics must fall. Mine is indispensable logic. Yours is an axiom derived from and modelling visible numbers.

Regards, WM

[Fritz Feldhase]

On Saturday, June 3, 2023 at 4:18:14 PM UTC+2, WM wrote:

> If you can't understand or won't accept this, then it is an unshakeable truth neveretheless.

Right. It is an instance of the well known axiom schema "Because I said so".

[Fritz Feldhase]

On Saturday, June 3, 2023 at 4:11:56 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Freitag, 2. Juni 2023 um 19:19:42 UTC+2:
> > On Friday, 2 June 2023 at 11:37:43 UTC-3, WM wrote:
> > >
> > > In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side.
> > >
> > That's all of them,
> >
> That's

all of them [with > 0]. Right.

Using symbols: Ax e IR: x > 0 -> NUF(x) = ℵ₀.

[Gus Gassmann]

And that, my unlearned friend, is all of them. For every x > 0 the interval (0,x) has the same cardinality as the positive reals.

[Jim Burns]

On 6/3/2023 10:18 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 2. Juni 2023 um 20:08:28 UTC+2:
>> On 6/2/2023 9:02 AM, WM wrote:

>>> The gaps are there and
>>> separate all of them from each other.
>>
>> That does not have the consequences
>> to which you feel entitled.
>
> I am entitled to this:
> If
> a sequence of reals
> has no term before a but
> has terms before b > a,
> then
> there is a beginning between a and b.

You are entitled to this:
If
a 1×1 2-ended total order
of (some) reals

has no term before a but
has terms before b > a,
then
there is a beginning between a and b.

Some things have
a 1×1 2-ended total order.

The corners of right triangle ABC have
a 1×1 2-ended total order.
The have six 1×1 2-ended total orders
ABC ACB BAC BCA CAB CBA
_All_ of their total orders are
1×1 2-ended.

Some things do not have
a 1×1 2-ended total order.

Right triangles do not have
a 1×1 2-ended total order.
_None of their total orders_ are
1×1 2-ended.

You are entitled to this:
For a fixed set,
either
each of its total orders is 1×1 2-ended
or
none of its total orders is 1×1 2-ended.

...because
'<₁' 1×1 2-ended total and '<₂' total
is enough to prove '<₂' 1×1 2-ended.
Provable: if any then all.

You are entitled to this:
A 1×1 2-ended total order
starts somewhere.

Order the corners of right triangle ABC
It could start with A or start with B
or start with C, but it starts.

Things which do not have
a 1×1 2-ended total order
might start somewhere and might not.

Consider (0,1)∩ℚ the rationals in (0,1)

In their standard order,
each rational is preceded by and
followed by other rationals in (0,1)
0 and 1 are not in (0,1)
That order of (0,1)∩ℚ is
0-ended and not 1×1.

There is also a Cantor-order of (0,1)∩ℚ
⟨ 1/2 1/3 2/3 1/4 2/4 3/4 1/5 ... ⟩

In this order, (0,1)∩ℚ is 1×1
1/2 is an end.
A second end not-exists.

If a second end existed,
it would be a 1×1 2-ended order,
and
_all_ total orders would be 1×1 2-ended
and
the standard order would be 1×1 2-ended
and
the standard order isn't.

> If all terms have finite distances,
> then the beginning consists of one term.

If there is beginning, then
there is a unit fraction ⅟n such that
⅟(4n) is NOT a unit fraction.
and
there isn't such a ⅟n

> If you can't understand or won't accept this,
> then it is an unshakeable truth neveretheless.

| There's an old legal aphorism that goes,
|| If you have the facts on your side,
|| pound the facts.
|| If you have the law on your side,
|| pound the law.
|| If you have neither on your side,
|| pound the table.
|
-- Wiktionary.

You (WM) are pounding the table.
You have neither facts nor law.

[FromTheRafters]

WM wrote :

Then your 'cursor' cannot pass anything between elements.

[Chris M. Thomasson]

On 5/31/2023 2:11 PM, FredJeffries wrote:
> On Wednesday, May 31, 2023 at 12:14:52 PM UTC-7, Chris M. Thomasson wrote:
>
>> Fwiw, I have some code that tries to detect when a field line in a
>> vector field starts to orbit, aka spiral around a point. The field line
>> starts plotting itself, then its plot might end up in an orbit around a
>> critical point. It fun to try to identify these points. For instance,
>> notice how the objects in the yellow field lines converge on a point?
>>
>> https://youtu.be/oVCjAaY1pOY
>>
>> This point is where they are going to orbit around forevermore. A limit,
>> in a sense?
>
> https://en.wikipedia.org/wiki/Attractor

Exactly. I have some code that can detect when a field line plotting in
real time gets into an attractive critical point. Its a tortoise and
hare type of analysis. The problem I have is that some of the field
lines might spiral a couple of times then break out and not be totally
captured and spiral forever. Other field lines will spiral forever. Some
of them kind of curve toward an attractive point, but will go off and
get locked into another attractor.

Check some of these out:

https://i.ibb.co/X51qM3c/image.png

https://i.ibb.co/2gYmT1C/image.png

https://i.ibb.co/GPKvYTx/image.png

https://i.ibb.co/r2qxKWS/image.png
(this one is easy to detect because the main attractor is very strong)...

[Chris M. Thomasson]

On 5/31/2023 2:11 PM, FredJeffries wrote:
> On Wednesday, May 31, 2023 at 12:14:52 PM UTC-7, Chris M. Thomasson wrote:
>
>> Fwiw, I have some code that tries to detect when a field line in a
>> vector field starts to orbit, aka spiral around a point. The field line
>> starts plotting itself, then its plot might end up in an orbit around a
>> critical point. It fun to try to identify these points. For instance,
>> notice how the objects in the yellow field lines converge on a point?
>>
>> https://youtu.be/oVCjAaY1pOY
>>
>> This point is where they are going to orbit around forevermore. A limit,
>> in a sense?
>
> https://en.wikipedia.org/wiki/Attractor

Also detecting the fractal attractors in this type of dynamic field can
get complicated.

https://youtu.be/TLd64a4gdZQ

Look for the "holes" in the mutated tori. See the points of attraction?

[Chris M. Thomasson]

There are attractive basins in here as well:

https://youtu.be/tg03Ti3qQ2s

[WM]

Pure logic is applied: If a sequence of terms with intervals starts, then there is a first term.

Regards, WM

[WM]

Fritz Feldhase schrieb am Samstag, 3. Juni 2023 um 17:22:02 UTC+2:
> On Saturday, June 3, 2023 at 4:11:56 PM UTC+2, WM wrote:
> > Gus Gassmann schrieb am Freitag, 2. Juni 2023 um 19:19:42 UTC+2:
> > > On Friday, 2 June 2023 at 11:37:43 UTC-3, WM wrote:
> > > >
> > > > In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side.
> > > >
> > > That's all of them,
> > >
> > That's
> all of them [with > 0]. Right.

Not even two unit fractions with their internal distance will fit between 0 and every x > 0.

Regards, WM

[Python]

Le 04/06/2023 à 15:51, Crank Pr. Wolfgang Mückenheim, aka WM wrote:
> Fritz Feldhase schrieb am Samstag, 3. Juni 2023 um 17:17:20 UTC+2:
>> On Saturday, June 3, 2023 at 4:18:14 PM UTC+2, WM wrote:
>>
>>> If you can't understand or won't accept this, then it is an unshakeable truth neveretheless.
>> Right. It is an instance of the well known axiom schema "Because I said so".
>
> Pure logic is applied:

Pure dementia, on your side, you mean, Crank Pr. Wolfgang Mückenheim
from Hochschule Augsburg.

> If a sequence of terms with intervals starts, then there is a first term.

If it is, but it is not a sequence, if it is, but it doesn't start, then
it may not have a first term.

[WM]

Gus Gassmann schrieb am Samstag, 3. Juni 2023 um 18:51:01 UTC+2:
> On Saturday, 3 June 2023 at 11:11:56 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Freitag, 2. Juni 2023 um 19:19:42 UTC+2:
> > > On Friday, 2 June 2023 at 11:37:43 UTC-3, WM wrote:
> > > [...]
> > > > In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side.
> > > That's all of them,
> > That's only those x which have infinitely times uncountable many points at their left-hand side.

> And that is all of them. For every x > 0 the interval (0,x) has the same cardinality as the positive reals.

This nonsense is irrelevant here. Mathematics proves that not even two unit fractions with their internal distance can fit between 0 and every x > 0. Their distance D is finite and not empty. Therefore NUF(D) < 3.

Regards, WM

[WM]

Jim Burns schrieb am Samstag, 3. Juni 2023 um 20:20:47 UTC+2:
> On 6/3/2023 10:18 AM, WM wrote:

> > If
> > a sequence of reals
> > has no term before a but
> > has terms before b > a,
> > then
> > there is a beginning between a and b.
> You are entitled to this:
> If

My statement is derived from pure logic. It cannot be violated.

> Some things do not have
> a 1×1 2-ended total order.

Then explain how you imagine the beginning of the sequence of unit fractions without violating mathematics and logic.
>
> Right triangles do not

serve this purpose.

> Things which do not have
> a 1×1 2-ended total order
> might start somewhere and might not.

Somewhere, yes. That's logic. But mathematics also has to be satisfied. Never vanishing internal distances are existing. Between the first two unit fractions, there is the distance D.
NUF(D) < 3.

>
> Consider (0,1)∩ℚ the rationals in (0,1)
>
> In their standard order,
> each rational is preceded by and
> followed by other rationals in (0,1)
> 0 and 1 are not in (0,1)
> That order of (0,1)∩ℚ is
> 0-ended and not 1×1.

It is not as obvious as the unit fractions with their internal distances although the principle of dark nunbers is the same.

> > If you can't understand or won't accept this,
> > then it is an unshakeable truth neveretheless.
> | There's an old legal aphorism that goes,
> || If you have the facts on your side,
> || pound the facts.

I do. Logic and mathematics.
Something existing in linear order has a beginning in that one dimension.
And mathematics: Two unit fractions have a non-vanishing distance which is less than the distances of 10^10^100 unit fractions.

Regards, WM

[WM]

FromTheRafters schrieb am Samstag, 3. Juni 2023 um 20:24:10 UTC+2:
> WM wrote :

> > Convergence is irrelevant here. We use only isolated unit fractions > 0.
> Then your 'cursor' cannot pass anything between elements.

I does. NUF(x) passes all x.

Regards, WM

[Gus Gassmann]

On Sunday, 4 June 2023 at 10:51:43 UTC-3, WM wrote:

> Pure logic is applied: If a sequence of terms with intervals starts, then there is a first term.

Congratulations on coming up with this profound insight. I presume by "terms with intervals", you mean a strictly decreasing sequence like {1/1/, 1/2, 1/3, ...}

This sequence does indeed have a first term (are you still able to figure out what that is??), but it does not have a last term.

In other words, you are full of shit, again. And now FUCK OFF.

[Gus Gassmann]

On Sunday, 4 June 2023 at 10:58:31 UTC-3, WM wrote:
> Gus Gassmann schrieb am Samstag, 3. Juni 2023 um 18:51:01 UTC+2:
> > On Saturday, 3 June 2023 at 11:11:56 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Freitag, 2. Juni 2023 um 19:19:42 UTC+2:
> > > > On Friday, 2 June 2023 at 11:37:43 UTC-3, WM wrote:
> > > > [...]
> > > > > In fact NUF(x) = ℵ₀ can be valid only for an x which has ℵ₀ unit fractions at its left-hand side.
> > > > That's all of them,
> > > That's only those x which have infinitely times uncountable many points at their left-hand side.

[...]
>
> This nonsense is...
all yours. You are foaming at the mouth, but you are not producing any meaningful sounds

[Jim Burns]

On 6/4/2023 9:58 AM, WM wrote:
> Gus Gassmann schrieb am Samstag,
> 3. Juni 2023 um 18:51:01 UTC+2:

[...]

> Mathematics proves that
> not even two unit fractions with
> their internal distance
> can fit between 0 and every x > 0.
> Their distance D is finite and not empty.
> Therefore NUF(D) < 3.

A unit fraction between 0 and each x > 0
not-exists.
¬∃n ∈ ℕ⁺:
∀x ∈ (0,1]:
0 < ⅟n < x

| Assume otherwise.
| Assume ∃n₀ ∈ ℕ⁺
| ∀x ∈ (0,1]:
| 0 < ⅟n₀ < x
|
| However,
| n₀ < n₀+1
| for xₙ₀ := ⅟(n₀+1)
| ¬(0 < ⅟n₀ < xₙ₀)
| Contradiction.

However,
for each x > 0
a unit fraction between that x and 0
exists.
∀x ∈ (0,1]:
∃n ∈ ℕ⁺:
0 < ⅟n < x

⌊⅟x⌋ ≤ ⅟x < ⌊⅟x⌋+1 ∈ ℕ⁺
for nₓ := ⌊⅟x⌋+1
0 < 1/nₓ < x

A note about "quantifier nonsense".

If
quantifier shifts were ¬#1◇⊥ (valid)
then
∀x ∈ (0,1]: ∃n ∈ ℕ⁺: 0 < ⅟n < x
and
¬∃n ∈ ℕ⁺: ∀x ∈ (0,1]: 0 < ⅟n < x
would be incorrect.

However,
because n₀ < n₀+1 and ⅟x < ⌊⅟x⌋+1 ∈ ℕ⁺
we know that they are correct.

Which is how we know that
quantifier shifts aren't ¬#1◇⊥ (valid)

> Therefore NUF(D) < 3.

No.

For x > 0
0 < 1/nₓ < x

For 1/nₓ > 0
0 < 1/nₙₓ < 1/nₓ < x

For 1/nₙₓ > 0
0 < 1/nₙₙₓ < 1/nₙₓ < 1/nₓ < x

Therefore NUF(x) ≥ 3

[Tom Bola]

The idiotic clown WM drivels:

> Pure logic

There is no such a thing, idiot.

[FromTheRafters]

Not when they are isolated.