_The rule of subset_ proves that
every proper subset has
less elements than its superset.
_The rule of subset_ proves that
every proper subset has
less elements than its superset.
sci.logic and sci.math
On 7/26/2024 12:31 PM, WM wrote:
_The rule of subset_ proves that [bla bla bla]
Le 31/07/2024 à 18:20, joes a écrit :How is this different from two consecutive infinities?
Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:Dark natnumbers are larger than defined natnumbers. Even dar natnumbers
Le 31/07/2024 à 03:28, Richard Damon a écrit :That is a bit imprecise. Even though you keep on talking about
On 7/30/24 1:37 PM, WM wrote:In the midst, far beyond all definable numbers, far beyond ω/10^10.
Le 30/07/2024 à 03:18, Richard Damon a écrit :And where is that in {1, 2, 3, ... w} ?
On 7/29/24 9:11 AM, WM wrote:ω/2
But what number became ω when doubled?
consecutive infinities, you can't compare natural and "dark" numbers.
can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.
There is no "before".What is immediately before ω? Is it a blasphemy to ask such questions?ω/10^10 and ω/10 are dark natural numbers.Completeness of N? No number n reaches omega.
I assume completness.But you combined two different sets, so why can't there be a gap?Because otherwise there was a gap below ω.If all natural numbers exist, then ω-1 exists.Why?
... to infinitely many unit fractions.That is not a contradiction.No. My formula says ∀n ∈ ℕ.∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every >>>> unit fraction 1/n, there exists another unit fraction smaller than
itself.
Am Thu, 01 Aug 2024 12:02:52 +0000 schrieb WM:
ω/10^10 < ω/10 < ω/2, < ω-1.
How is this [...]
ω/10^10 and ω/10 are dark [...] numbers.
If all natural numbers exist, then ω-1 exists.
What is immediately before ω? Is it <bla>
Am Thu, 01 Aug 2024 12:02:52 +0000 schrieb WM:
Dark natnumbers are larger than defined natnumbers. Even dark natnumbersHow is this different from two consecutive infinities?
can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.
What is immediately before ω? Is it a blasphemy to ask such questions?There is no "before".
Am 01.08.2024 um 14:10 schrieb joes:
Am Thu, 01 Aug 2024 12:02:52 +0000 schrieb WM:
ω/10^10 < ω/10 < ω/2, < ω-1.
Keiner dieser Ausdrücke ist definiert.
What is immediately before ω?
NOTHING is "immediately before ω".
Genauer: Keine Ordinalzahl.
Le 01/08/2024 à 14:24, Moebius a écrit :Do you imagine N as symmetrically countable from either end, with
Am 01.08.2024 um 14:10 schrieb joes:They are not defined by FISONs. Their relative sizes do not need FISONs
Am Thu, 01 Aug 2024 12:02:52 +0000 schrieb WM:Keiner dieser Ausdrücke ist definiert.
ω/10^10 < ω/10 < ω/2, < ω-1.
to be comparable.
Not at all. They are all infinitely far away, or rather omega isHow close do the ordinal numbers 1, 2, 3, ... come to their limit ω?NOTHING is "immediately before ω".What is immediately before ω?
Genauer: Keine Ordinalzahl.
On 8/1/2024 8:02 AM, WM wrote:
It is not a contradiction to my formula
if some n has no n+1.
No, it literally contradicts your formula
for some n e N to not.have n+1
Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:
separated from 0 by any eps. Therefore your claim is wrong.No. There is ALWAYS an epsilon.
What is the reason for the gap before omega? How large is it? Are theseA "gap" implies some sort of space that is not filled. There is no such
questions a blasphemy?
space (it would be filled with infinitely many natural numbers).
We just condense the whole of N into one concept and call that omega,
or add it on the next level of infinity.
Your questions are only a display of your unwillingness to understand infinity,
If k did not have a successor, what would k+1 be?
Le 01/08/2024 à 18:04, joes a écrit :Well, there's no epsilon that separates all positive numbers from zero.
Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:
Failing to separate almost all unit fractions.Every eps interval around 0 contains unit fractions which cannot beNo. There is ALWAYS an epsilon.
separated from 0 by any eps. Therefore your claim is wrong.
Don't claim the contrary. Define (separate by an eps from 0) all unit fractions. Fail.
And no sequence of omega - k.Hence there is only the sequence of natnumbers.What is the reason for the gap before omega? How large is it? AreA "gap" implies some sort of space that is not filled. There is no such
these questions a blasphemy?
space (it would be filled with infinitely many natural numbers).
Following the WHOLE of the natural numbers. The successor of a naturalWe just condense the whole of N into one concept and call that omega,That is nonsense. ω is the first number following upon all natural
numbers.
Ah, a natural number.If k did not have a successor, what would k+1 be?ω
Am Thu, 01 Aug 2024 12:46:19 +0000 schrieb WM:
Do you imagine N as symmetrically countable from either end, with
something in between?
Not at all. They are all infinitely far away, or rather omega isHow close do the ordinal numbers 1, 2, 3, ... come to their limit ω?NOTHING is "immediately before ω".What is immediately before ω?
Genauer: Keine Ordinalzahl.
infinitely far away from all of them. The infinity is at infinity.
Being a limit, this distance does not shrink. In the same sense as cardinality, none of them get any closer.
The limit is not a part of
the sequence, i.e. the natural numbers. It lies outside and beyond.
It is not even a proper value: the sequence diverges, it transcends
all bounds.
WM brought next idea :
How close do the ordinal numbers 1, 2, 3, ... come to their limit ω?
It is merely the symbol for the countably infinite ordered set of
naturals. It represents all of the ordinals which came before it, and
yet it is the first of its kind, the first transfinite ordinal.
You might as well be asking what natural number comes before the least natural number.
Am Fri, 02 Aug 2024 11:38:33 +0000 schrieb WM:
Don't claim the contrary. Define (separate by an eps from 0) all unitWell, there's no epsilon that separates all positive numbers from zero.
fractions. Fail.
But every fraction has an epsilon that is smaller.
Hence there is only the sequence of natnumbers.And no sequence of omega - k.
Following the WHOLE of the natural numbers. The successor of a naturalWe just condense the whole of N into one concept and call that omega,That is nonsense. ω is the first number following upon all natural
numbers.
is also one.
Ah, a natural number.If k did not have a successor, what would k+1 be?ω
Le 02/08/2024 à 19:31, Moebius a écrit :It is the definition you have previously used.
For each and every of these points [here referred to with the variableI recognized lately that you use the wrong definition of NUF.
"x"]: NUF(x) = ℵ₀ .
Here is the correct definition:You are specifying an exact number, not only at least one.
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already is wrong since there is no unit fraction smaller than all unit fractions.New sig.
ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.
Am Sat, 03 Aug 2024 14:25:03 +0000 schrieb WM:
Here is the correct definition:You are specifying an exact number, not only at least one.
There exist NUF(x) unit fractions u, such that for all y >= x: u < y.
Note that the order is ∃ u ∀ y.
NUF(x) = ℵ₀ for all x > 0 is wrong. NUF(x) = 1 for all x > 0 already is >> wrong since there is no unit fraction smaller than all unit fractions.New sig.
On 8/4/2024 11:44 AM, WM wrote:
Le 03/08/2024 à 19:42, joes a écrit :
[...]
The reversed quantification is nonsense
...because from ∀x∃u to ∃u∀x is unreliable.
From ∃u∀x to ∀x∃u is reliable.
From ∃u∃x to ∃x∃u is reliable.
From ∀u∀x to ∀x∀u is reliable.
However,
from ∀x∃u to ∃u∀x is unreliable.
In some cases,
from ∀x∃u to ∃u∀x is not.only.unreliable,
it is incorrect.
Le 03/08/2024 à 19:42, joes a écrit :
[...]
The reversed quantification is nonsense
because
ℵ₀ unit fractions need ℵ₀*2^ℵ₀ points above zero.
Le 04/08/2024 à 21:18, Richard Damon a écrit :What is half of an infinity? Still infinite? Can you make that precise?
On 8/4/24 11:23 AM, WM wrote:Always.
Le 03/08/2024 à 17:56, Richard Damon a écrit :Not with infinities.
On 8/3/24 10:30 AM, WM wrote:If there are all, then there is half of all.
But there is not even an eps that separates half of all unitBecause such a question is meaningles, as there isn't a finite number
fractions.
that is half of the count of unit fractions.
Le 04/08/2024 à 22:16, Jim Burns a écrit :In natural language those mean the same thing. The reversed quantifier
On 8/4/2024 2:13 PM, WM wrote:
More of interest are these two claims which are not both true or both
false:
For every x there is u < x.
There is u < x for every x.
I don't know how to formalise that.The latter is close to my function:
There are NUF(x) u < x.
You were confusing the quantifiers.From ∀x∃U to ∃U∀x is unreliable,There is no from to. NUF(x) is so defined.
For every number of unit fractions NUF(x) gives the smallest intervalAre you talking about the inverse function?
(0, x).
Am Mon, 05 Aug 2024 18:58:40 +0000 schrieb WM:
Le 04/08/2024 à 21:18, Richard Damon a écrit :What is half of an infinity? Still infinite? Can you make that precise?
On 8/4/24 11:23 AM, WM wrote:Always.
Le 03/08/2024 à 17:56, Richard Damon a écrit :Not with infinities.
On 8/3/24 10:30 AM, WM wrote:If there are all, then there is half of all.
But there is not even an eps that separates half of all unitBecause such a question is meaningles, as there isn't a finite number >>>>> that is half of the count of unit fractions.
fractions.
Am Mon, 05 Aug 2024 19:04:14 +0000 schrieb WM:
Le 04/08/2024 à 22:16, Jim Burns a écrit :In natural language those mean the same thing. The reversed quantifier
On 8/4/2024 2:13 PM, WM wrote:
More of interest are these two claims which are not both true or both
false:
For every x there is u < x.
There is u < x for every x.
order would be: There is a single u<x *such that for all x* ...
That is different from the first: Every x has a *corresponding* u ...
I don't know how to formalise that.The latter is close to my function:
There are NUF(x) u < x.
You were confusing the quantifiers.From ∀x∃U to ∃U∀x is unreliable,There is no from to. NUF(x) is so defined.
For every number of unit fractions NUF(x) gives the smallest intervalAre you talking about the inverse function?
(0, x).
Le 06/08/2024 à 00:19, Jim Burns a écrit :
NUF(1) = ℵ₀
NUF(x) = ℵ₀
NUF(x) gives
the number of unit fractions smaller than x.
Following unreadable symbols.
For each x > 0
⅟ℕᵈᵉᶠ∩(0,x) is not finite.
For NUF(x) = 3
⅟ℕᵈᵉᶠ∩(0,x) is finite, namely 3.
Le 06/08/2024 à 14:38, Jim Burns a écrit :
NUF(x) ≠ 1
is true everywhere
NUF(x) = 1 ⇒ INVNUF(1) = x
is true everywhere
However,
its truth doesn't imply INVNUF(1) exists.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
implies its existence.
Le 09/08/2024 à 05:34, Jim Burns a écrit :"Before" is not a number. There are still infinitely many of them,
and he thinks that a set ordered with two ends is more complete thanThe set of unit fractions has two ends, namely at 1 and before 0.
the same set with one or zero ends.
Am Sat, 10 Aug 2024 15:59:05 +0000 schrieb WM:
The set of unit fractions has two ends, namely at 1 and before 0."Before" is not a number. There are still infinitely many of them,
because their distances (differences, rather) decrease. They have
only one end that you can count from.
Le 12/08/2024 à 15:27, joes a écrit :
Am Sat, 10 Aug 2024 15:59:05 +0000 schrieb WM:
The set of unit fractions has two ends, namely at 1 and before 0."Before" is not a number. There are still infinitely many of them,
because their distances (differences, rather) decrease. They have
only one end that you can count from.
There is a first one because all have finite gaps.
A simple picture: Let a cursor run from 1 to 0. Every passed accessible
unit fraction has infinitely many smaller unit fractions as successors.
When the cursor passes 0, all unit fractions have been passed. None
remains, not even the infinitely many successors of every accessible
unit fraction. They are not accessible. They are dark. This prevents
that the last unit fractions passed by the cursor can be determined and
put in order. The gaps and the linearity of the problem require that the cursor never passes two or more unit fractions at one position.
Therefore a last one must have been passed when arriving at zero. But we don't know about the structure of dark points.
If all unit fractions had ℵo smaller unit fractions as successors the cursor could never diminish the number of unit fractions between itself
and zero to fewer than ℵo and could never reach zero. Even if every unit fraction had only one successor, the physical movement would be hampered
by this philosophical assumption. That is impossible.
Am Mon, 12 Aug 2024 14:07:34 +0000 schrieb WM:
[...] a last one must have been passed when arriving at zero.
Le 12/08/2024 à 15:27, joes a écrit :Then it can never reach 0.
Am Sat, 10 Aug 2024 15:59:05 +0000 schrieb WM:
The set of unit fractions has two ends, namely at 1 and before 0."Before" is not a number. There are still infinitely many of them,
because their distances (differences, rather) decrease. They have only
one end that you can count from.
A simple picture: Let a cursor run from 1 to 0. Every passed accessible
unit fraction has infinitely many smaller unit fractions as successors.
When the cursor passes 0, all unit fractions have been passed. None
remains, not even the infinitely many successors of every accessible
unit fraction. They are not accessible. They are dark. This prevents
that the last unit fractions passed by the cursor can be determined and
put in order. The gaps and the linearity of the problem require that the cursor never passes two or more unit fractions at one position.
Therefore a last one must have been passed when arriving at zero. But we don't know about the structure of dark points.This is the case.
If all unit fractions had ℵo smaller unit fractions as successors the cursor could never diminish the number of unit fractions between itself
and zero to fewer than ℵo and could never reach zero.
Even if every unitThey do have exactly one successor. Anyway this is not physical.
fraction had only one successor, the physical movement would be hampered
by this philosophical assumption. That is impossible.
Le 12/08/2024 à 16:07, WM a écrit :
Le 12/08/2024 à 15:27, joes a écrit :
Am Sat, 10 Aug 2024 15:59:05 +0000 schrieb WM:
The set of unit fractions has two ends, namely at 1 and before 0."Before" is not a number. There are still infinitely many of them,
because their distances (differences, rather) decrease. They have
only one end that you can count from.
There is a first one because all have finite gaps.
A simple picture: Let a cursor run from 1 to 0. Every passed accessible
unit fraction has infinitely many smaller unit fractions as successors.
When the cursor passes 0, all unit fractions have been passed. None
remains, not even the infinitely many successors of every accessible
unit fraction. They are not accessible. They are dark. This prevents
that the last unit fractions passed by the cursor can be determined and
put in order. The gaps and the linearity of the problem require that the
cursor never passes two or more unit fractions at one position.
Therefore a last one must have been passed when arriving at zero. But we
don't know about the structure of dark points.
If all unit fractions had ℵo smaller unit fractions as successors the
cursor could never diminish the number of unit fractions between itself
and zero to fewer than ℵo and could never reach zero. Even if every unit >> fraction had only one successor, the physical movement would be hampered
by this philosophical assumption. That is impossible.
When Zeno proposed his well known paradox, he didn't attempt to prove
that motion was impossible,
On 8/13/2024 10:21 AM, WM wrote:
either in a step of size 1
or in a step of size more than 1.
But increase by more than 1 is excluded by
the gaps between unit fractions.
Le 13/08/2024 à 19:32, Moebius a écrit :What is "seeing"?
Am 13.08.2024 um 19:02 schrieb Jim Burns:Numbers can be seen.
On 8/13/2024 10:21 AM, WM wrote:No number can be seen, dark or not.
Le 12/08/2024 à 19:44, Jim Burns a écrit :
There is no ⅟nₓ before the end of the positive axis without ⅟(nₓ+1)You cannot see it. It is dark.
before the end of the positive axis.
That's one way to resolve it. But how do you do mathematics without thatTherefore 1/(n0 + 1) does not exist.Assume NUF(x0) = 1 with x0 e IR. This means that there is exactly oneresulting in a real coordinate x with NUF(x) = 1.
unit fraction u such that u < x0. Let's call this unit fraction u0.
Then (by definition) there is a (actually exactly one) natural number n
such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by
definition) 1/(n0 + 1) is a unit fraction which is smaller than u0 and
hence smaller than x0. Hence NUF(x0) > 1. Contradiction!
Am Wed, 14 Aug 2024 12:26:27 +0000 schrieb WM:
Le 13/08/2024 à 19:32, Moebius a écrit :
Am 13.08.2024 um 19:02 schrieb Jim Burns:
On 8/13/2024 10:21 AM, WM wrote:
Le 12/08/2024 à 19:44, Jim Burns a écrit :
There is no ⅟nₓ before the end of the positive axis without ⅟(nₓ+1)
before the end of the positive axis.
You cannot see it. It is dark.
No number can be seen, dark or not.
Numbers can be seen.
What is "seeing"?
clairvoyance, delusion
resulting in a real coordinate x with NUF(x) = 1.
Assume NUF(x0) = 1 with x0 e IR. This means that there is exactly one
unit fraction u such that u < x0. Let's call this unit fraction u0.
Then (by definition) there is a (actually exactly one) natural number n
such that u0 = 1/n. Let n0 e IN such that u0 = 1/n0. But then (again by
definition) 1/(n0 + 1) is a unit fraction which is smaller than u0 and
hence smaller than x0. Hence NUF(x0) > 1. Contradiction!
Therefore 1/(n0 + 1) does not exist.
That's one way to resolve it.
Am Wed, 14 Aug 2024 12:26:27 +0000 schrieb WM:
Numbers can be seen.What is "seeing"?
That's one way to resolve it.But then (again byTherefore 1/(n0 + 1) does not exist.
definition) 1/(n0 + 1) is a unit fraction which is smaller than u0 and
hence smaller than x0. Hence NUF(x0) > 1. Contradiction!
But how do you do mathematics without that
axiom?
Le 08/08/2024 à 12:30, FromTheRafters a écrit :They are countably infinite.
on 8/8/2024, WM supposed :
No, that is nonsense. There are not finitely many unit fractions.
Then stop assuming that there is a first and last element.
Finitely many means that you can count from first to last. You cannot
count through the unit fractions.
Le 10/08/2024 à 19:16, FromTheRafters a écrit :0 is not a unit fraction, so not an end; there is no smallest.
WM expressed precisely :
Le 09/08/2024 à 05:34, Jim Burns a écrit :
and he thinks that a set ordered with two ends is more complete than
the same set with one or zero ends.
The set of unit fractions has two ends, namely at 1 and before 0.
Wrong,
Name unit fractions larger than 1 or smaller than 0.
Note: Before domains without unit fractions the set has ended.
Le 10/08/2024 à 19:28, Jim Burns a écrit :What is the largest real number in (0, 1)?
On 8/10/2024 11:54 AM, WM wrote:
Le 09/08/2024 à 02:32, Jim Burns a écrit :If an interval contains unit fractions, then it contains a first one.
Surely, a lawyer wouldn't think that "Boom! Here's the conclusion"∀n ∈ ℕ: 1/n - 1/(n+1) > 0
is an _argument_ ?
No.The existence of unit frations enforces one or more first unitTherefore there can only be a single first unit fraction.No one has said there are two first unit.fractions.
What forbids zero first unit.fractions?
fractions.
What fucking endWhat causes an exception: nₓ ∈ ℕ without ⅟(nₓ+1) ?The end of the positivee axis.
Am 16.08.2024 um 07:05 schrieb Jim Burns:
On 8/15/2024 7:16 PM, Moebius wrote:
Am 16.08.2024 um 00:51 schrieb Jim Burns:
On 8/15/2024 3:37 PM, Moebius wrote:
Am 15.08.2024 um 21:28 schrieb Jim Burns:
Defining ⅟𝔊 to be the smallest unit fraction
isn't a claim that ⅟𝔊 exists.
You need an existence proof [...]
BEFORE stating a/the proper definition.
Half or more of my proofs to WM say
"Assume otherwise... However... Contradiction."
Yeah, to be precise a proof by contradiction
assumes a STATEMENT/CLAIM.
Many times, a false existence claim.
Right.
√2 is irrational.
This statement is just nonsense,
_if_ "√2" is not already defined.*)
⎛ Assume otherwise.
Nope. You clearly don't assume
√2 is rational ,
but:
⎜ Assume p₃,q₃ ∈ ℕ₁: p₃⋅p₃ = 2⋅q₃⋅q₃:
⎝ Contradiction.
________________________________
*) Of course,
_if_ we already have introduced the real numbers
(i.e. IR)
we may define
√2 = the real number x such that x*x = 2 , (*)
_after_ we have shown that that
there is exactly one x e IR such that x*x = 2.
From (*) we get immediately: √2*√2 = 2.
0 is not a unit fraction, so not an end; there is no smallest.
The existence of unit frations enforces one or more first unitNo.
fractions.
Le 15/08/2024 à 19:01, Moebius a écrit :It does not diminish, there are always infinitely many.
Assume that there is an x e IR such that NUF(x) = 1. Let x0 e IR suchWe can reduce the interval (0, x) c [0, 1] such that x converges to 0.
that NUF(x0) = 1. This means that there is exactly one unit fraction u
such that u < x0. Let's call this unit fraction u0. Then (by
definition)
there is a (actually exactly one) natural number n such that u0 = 1/n.
Let n0 e IN such that u0 = 1/n0. But then (again by definition) 1/(n0 +
1) is an unit fraction which is smaller than u0 and hence smaller than
x0. Hence NUF(x0) > 1. Contradiction!
Then the number of unit fractions diminishes. Finally there is none remaining. But never, for no interval (0, x), more than one unit
fraction is lost. Therefore there is only one last unit fraction.
Le 15/08/2024 à 18:17, joes a écrit :
0 is not a unit fraction, so not an end; there is no smallest.
We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
Then the number of unit fractions diminishes. Finally there is none remaining. But never, for no interval (0, x), more than one unit
fraction is lost. Therefore there is only one last unit fraction.
Regards, WM
Le 15/08/2024 à 18:16, Jim Burns a écrit :Taking steps only until the next unit fraction, you never reach 0.
On 8/15/2024 9:52 AM, WM wrote:
⎜ Assume NUF(x) = 0 and x > 0We assume that NUF(0) = 0 and many unit fractions are within (0, 1].
We can reduce the interval to (0, x) c [0, 1]. Let x converge to 0.
Then the number of unit fractions diminishes. Finally there is none remaining. But never, for no interval (0, x), more than one unit
fraction is lost. Therefore there is only one last unit fraction.
On 8/16/24 12:21 PM, WM wrote:
Le 15/08/2024 à 18:17, joes a écrit :
0 is not a unit fraction, so not an end; there is no smallest.
We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
Then the number of unit fractions diminishes. Finally there is none
remaining. But never, for no interval (0, x), more than one unit
fraction is lost. Therefore there is only one last unit fraction.
Regards, WM
No, you can't because the count of the unit fractions in (0, x) is
always aleph_0. When you reduce the size to get the next smaller one,
the count is still aleph_0, as that is how arithmetic on infinite values work.
The value never converges to 0, as you can never get off aleph_0. To converge, you first need to find a value of x where the count of unit fractions in (0, x) is a finite number, but no finite number x has that property.
Am Fri, 16 Aug 2024 16:19:17 +0000 schrieb WM:
Taking steps only until the next unit fraction, you never reach 0.
Am Fri, 16 Aug 2024 16:45:29 +0000 schrieb WM:
It does not diminish, there are always infinitely many.
Le 16/08/2024 à 18:49, joes a écrit :
Am Fri, 16 Aug 2024 16:45:29 +0000 schrieb WM:
It does not diminish, there are always infinitely many.
Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Regards, WM
Le 16/08/2024 à 18:49, joes a écrit :
It does not diminish,
there are always infinitely many.
Not according to mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Le 16/08/2024 à 18:50, joes a écrit :
Taking steps only until the next unit fraction,
you never reach 0.
Start from x = 0.
The increase cannot be more than 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Le 16/08/2024 à 18:48, Richard Damon a écrit :
On 8/16/24 12:21 PM, WM wrote:
Le 15/08/2024 à 18:17, joes a écrit :
0 is not a unit fraction, so not an end; there is no smallest.
We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
Then the number of unit fractions diminishes. Finally there is none
remaining. But never, for no interval (0, x), more than one unit
fraction is lost. Therefore there is only one last unit fraction.
Regards, WM
No, you can't because the count of the unit fractions in (0, x) is
always aleph_0. When you reduce the size to get the next smaller one,
the count is still aleph_0, as that is how arithmetic on infinite
values work.
It works so for cranks.
The value never converges to 0, as you can never get off aleph_0. To
converge, you first need to find a value of x where the count of unit
fractions in (0, x) is a finite number, but no finite number x has
that property.
Then more than one diminish simultaneously. Contradiction.
This proves a first one: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Regards, WM
Le 16/08/2024 à 18:49, joes a écrit :I don't see the connection.
Am Fri, 16 Aug 2024 16:45:29 +0000 schrieb WM:
It does not diminish, there are always infinitely many.Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
On 8/16/2024 11:10 AM, Jim Burns wrote:
Taking an equation and waving it around,
apparently just for the hell of it,
has a lot of parallels to
carving headphones from wood and
sitting in a fabricated control tower.
ROFL!!!
Am Fri, 16 Aug 2024 16:59:11 +0000 schrieb WM:
Le 16/08/2024 à 18:49, joes a écrit :
It does not diminish, there are always infinitely many.
Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
I don't see the connection.
Am Fri, 16 Aug 2024 16:59:11 +0000 schrieb WM:
Le 16/08/2024 à 18:49, joes a écrit :I don't see the connection.
Am Fri, 16 Aug 2024 16:45:29 +0000 schrieb WM:Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
It does not diminish, there are always infinitely many.
Le 13/08/2024 à 19:51, Moebius a écrit :Quantifier shift.
Am 13.08.2024 um 19:02 schrieb Jim Burns:
On 8/13/2024 10:21 AM, WM wrote:
Actually, for each and every real number x > 0 there are infinitelyThat is wrong because ℵo unit fractions occupy a nonvanishig real
many unit fractions smaller than x:
interval, call it y. Then not every point of y has infinitely many unit fractions smaller than it self.
In fact, Ax > 0: NUF(x) = aleph_0, while NUF(0) = 0.A very stupid statement. It claims unit fractions smaller than every
positive x. "Every positive x" is tantamount to the interval (0, oo).
Or just that you just don't believe in how
This proves a first one: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
No, that proves that there is no smallest unit fraction as it shows that 1/(n+1) exists,
Am Fri, 16 Aug 2024 16:59:11 +0000 schrieb WM:
I don't see the connection.It does not diminish, there are always infinitely many.Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
On 8/16/2024 12:59 PM, WM wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Taking an equation and waving it around,
On 8/16/2024 1:00 PM, WM wrote:
Start from x = 0.
The increase cannot be more than 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Start from x = 0
The increase cannot be less than 2
∀ᴿx > 0: x > ⅟nₓ > ⅟(nₓ+1)
nₓ = ⌊1+⅟x⌋
Am Fri, 16 Aug 2024 17:00:26 +0000 schrieb WM:
Le 16/08/2024 à 18:50, joes a écrit :No, you were counting down from 1.
Am Fri, 16 Aug 2024 16:19:17 +0000 schrieb WM:Start from x = 0. The increase cannot be more than 1.
Taking steps only until the next unit fraction, you never reach 0.
Am Wed, 14 Aug 2024 12:14:21 +0000 schrieb WM:
Every chosen ε > 0 is larger than this interval y.Then y must be 0, if epsilon can't be chosen smaller.
Am Wed, 14 Aug 2024 12:32:41 +0000 schrieb WM:
"Every positive x" is tantamount to the interval (0, oo).Quantifier shift.
Le 17/08/2024 à 13:43, joes a écrit :
Am Wed, 14 Aug 2024 12:14:21 +0000 schrieb WM:
Every chosen ε > 0 is larger than this interval y.Then y must be 0, if epsilon can't be chosen smaller.
Try to choose an eps smaller infinitely many unit fractions. Fail.
Regards, WM
Le 16/08/2024 à 19:54, Jim Burns a écrit :
On 8/16/2024 1:00 PM, WM wrote:
Start from x = 0.
The increase cannot be more than 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Start from x = 0
The increase cannot be less than 2
That violates ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
∀ᴿx > 0: x > ⅟nₓ > ⅟(nₓ+1)
nₓ = ⌊1+⅟x⌋
That holds only for definable numbers.
Regards, WM
Le 16/08/2024 à 20:11, joes a écrit :
Am Fri, 16 Aug 2024 16:59:11 +0000 schrieb WM:
I don't see the connection.It does not diminish, there are always infinitely many.Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
NUF(x) grows from 0 to more, but at no point it grows by more than 1.
Regards, WM
Richard Damon formulated on Saturday :
On 8/17/24 9:30 AM, WM wrote:
Le 16/08/2024 à 19:54, Jim Burns a écrit :
On 8/16/2024 1:00 PM, WM wrote:
Start from x = 0.
The increase cannot be more than 1.
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Start from x = 0
The increase cannot be less than 2
That violates ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
∀ᴿx > 0: x > ⅟nₓ > ⅟(nₓ+1)
nₓ = ⌊1+⅟x⌋
That holds only for definable numbers.
Regards, WM
Which is all of the Rational Numbers, and thus all of the unit fractions.
Thats your problem, you don't seem to know how to definie numbers.
Worse than that, he tries to make a set of undefined/undefinable numbers which he calls dark numbers while sets have well defined elements.
Le 16/08/2024 à 20:11, Jim Burns a écrit :
On 8/16/2024 12:59 PM, WM wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Taking an equation and waving it around,
It shows that
NUF(x) at no point grows by more than 1.
But NUF(x) grows.
On 8/17/2024 9:36 AM, WM wrote:
NUF(x) at no point grows by more than 1.
But NUF(x) grows.
At 0.
Le 17/08/2024 à 13:43, joes a écrit :Why should it be possible?
Am Wed, 14 Aug 2024 12:14:21 +0000 schrieb WM:
Try to choose an eps smaller infinitely many unit fractions. Fail.Every chosen ε > 0 is larger than this interval y.Then y must be 0, if epsilon can't be chosen smaller.
Le 16/08/2024 à 20:11, joes a écrit :How does it even reach infinity then?
Am Fri, 16 Aug 2024 16:59:11 +0000 schrieb WM:
NUF(x) grows from 0 to more, but at no point it grows by more than 1.I don't see the connection.It does not diminish, there are always infinitely many.Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
On 8/17/24 9:42 AM, WM wrote:
Try to choose an eps smaller infinitely many unit fractions. Fail.You just confuse your qualifiers
We can choose an eps smaller than ANY GIVEN value.
On 8/17/24 9:37 AM, WM wrote:
NUF(x) grows from 0 to more, but at no point it grows by more than 1.
And there is "no point" that is smaller than all unit fractions but
greater than 0, so at that point NUF(x) jumps from 0 to Aleph_0.
Your problem is NUF(x) may have a clear verbal description, but not a mathematical one, as it is based on a false assumption that there exists
a smallest unit fraction.
Le 17/08/2024 à 16:15, Richard Damon a écrit :
On 8/17/24 9:42 AM, WM wrote:
Try to choose an eps smaller infinitely many unit fractions. Fail.You just confuse your qualifiers
No, I use quantifiers.
We can choose an eps smaller than ANY GIVEN value.
But you cannot give a value containing only the three smallest unit
fractions although they exist because all exist and are separated.
Regards, WM
Le 17/08/2024 à 16:22, Richard Damon a écrit :
On 8/17/24 9:37 AM, WM wrote:
NUF(x) grows from 0 to more, but at no point it grows by more than 1.
And there is "no point" that is smaller than all unit fractions but
greater than 0, so at that point NUF(x) jumps from 0 to Aleph_0.
That is superstition.
Your problem is NUF(x) may have a clear verbal description, but not a
mathematical one, as it is based on a false assumption that there
exists a smallest unit fraction.
That is not an assumption but it is derived from the mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Regards, WM
On 8/17/2024 9:36 AM, WM wrote:
Le 16/08/2024 à 20:11, Jim Burns a écrit :
On 8/16/2024 12:59 PM, WM wrote:
NUF(x) at no point grows by more than 1.
At unit.fractions. Not at 0.
But NUF(x) grows.
At 0. Not at unit fractions.
Am 17.08.2024 um 23:44 schrieb Jim Burns:
On 8/17/2024 9:36 AM, WM wrote:
NUF(x) at no point grows by more than 1.
NUF does not "grow" anywhere,
It "jumps at 0".
Am Sat, 17 Aug 2024 13:42:03 +0000 schrieb WM:
Try to choose an eps smaller infinitely many unit fractions. Fail.Why should it be possible?
Am Sat, 17 Aug 2024 13:37:51 +0000 schrieb WM:
NUF(x) grows from 0 to more, but at no point it grows by more than 1.How does it even reach infinity then?
On 8/19/24 7:29 AM, WM wrote:
That is not an assumption but it is derived from the mathematics:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
But that doesn't say there is a smallest unit fraction,
Am Sat, 17 Aug 2024 13:39:52 +0000 schrieb WM:
There we see that counting down to 0 requires dark numbers.Counting down from infinity is not possible.
Your "dark" numbers
are a nonstandard extension.
Le 17/08/2024 à 23:44, Jim Burns a écrit :
On 8/17/2024 9:36 AM, WM wrote:
NUF(x) at no point grows by more than 1.
At unit.fractions. Not at 0.
But NUF(x) grows.
At 0. Not at unit fractions.
You are not entitled to change the definition of NUF(x).
NUF(x) grows at every unit fractions by 1.
At 0 there is no unit fractions,
therefore NUF(0) = 0
as for all negative arguments.
The firstincrease happens at the first unit fraction.
Le 18/08/2024 à 11:30, joes a écrit :
Am Sat, 17 Aug 2024 13:37:51 +0000 schrieb WM:
NUF(x) grows from 0 to more,
but at no point it grows by more than 1.
How does it even reach infinity then?
It passes through darkness.
There is no end visible.
That means growth without end.
We call it infinity.
Le 18/08/2024 à 10:16, joes a écrit :
Am Sat, 17 Aug 2024 13:42:03 +0000 schrieb WM:
Try to choose an eps smaller infinitely many unit fractions. Fail.Why should it be possible?
Because when infinitely many exist, then also finitely many must exist - according to logic and mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 . Everything else is the superstition of matheologians.
Regards, WM
On 8/19/2024 7:44 AM, WM wrote:
NUF(x) grows at every unit fractions by 1.
No. ℵ₀+1 = ℵ₀
At 0 there is no unit fractions,
therefore NUF(0) = 0
as for all negative arguments.
The first increase happens at the first unit fraction.
No positive lower.bound, darkᵂᴹ or visibleᵂᴹ, exists of
visibleᵂᴹ unit.fractions.
The unit fractions get arbitrarily close to zero, but no unit
fraction ever equals zero...
Do you even now how to make a Cantor Set Fractal?
Le 21/08/2024 à 12:58, FromTheRafters a écrit :Why should there be an end in the first place? Are there not infinitely
WM wrote :It is an end which proves dark numbers.
Le 20/08/2024 à 23:25, FromTheRafters a écrit :Correct, that is the wrong end.
WM explained :
Le 20/08/2024 à 12:31, FromTheRafters a écrit :
on 8/19/2024, Richard Damon supposed :
You can not derive a first number > 0 in any of the Number System >>>>>>> that we have been talking about, Unit Fractions, Rationals or
Reals, so you can't claim it to exist.
Not in their natural ordering.
Dark numbers have no discernible order. It is impossible to find the >>>>> smallest unit fraction or the next one or the next one. It is only
possible to prove that NUF(x) grows by 1 at every unit fraction. It
starts from 0.
Normally, the unit fractions are listed in the sequence one over one,
one over two, one over three etcetera. There is a first but no last.
Now you have started from the wrong 'end'
No, I have started from the other end.
Since no unit fraction is below or at zero, the end is before. WhatIt exists at x > 0 because NUF(0) = 0.
else?
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:
Since no unit fraction is below or at zero, the end is before.
Am 21.08.2024 um 15:13 schrieb joes:
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:
Since no unit fraction is below or at zero, the end is before.
Since there is no smallest unit fraction, there is no end.
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:
It is an end which proves dark numbers.Why should there be an end in the first place?
Are there not infinitely
many unit fractions?
Le 21/08/2024 à 15:13, joes a écrit :The unit fractions get denser toward 0. They are all positive.
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:
It is an end which proves dark numbers.Why should there be an end in the first place?
There is indisputably an interval without unit fractions, namely (-oo,
0].
That indicates an end before. It proves an end before. Only cranks can
deny that.
I was talking about the regular ones.Are there not infinitely many unit fractions?Yes, the dark unit fractions cannot be counted through. There are more
than can be counted.
Am Wed, 21 Aug 2024 14:53:27 +0000 schrieb WM:
Le 21/08/2024 à 15:51, Moebius a écrit :
Am 21.08.2024 um 15:13 schrieb joes:The other way round. There is an end, because at and below zero there is
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:Since there is no smallest unit fraction, there is no end.
Since no unit fraction is below or at zero, the end is before.
no unit fraction. Therefore there is an end before. Simple as that.
Stupid as this: „There is a last negative power of 2.”
Le 21/08/2024 à 15:51, Moebius a écrit :Stupid as this: „There is a last negative power of 2.”
Am 21.08.2024 um 15:13 schrieb joes:The other way round. There is an end, because at and below zero there is
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:Since there is no smallest unit fraction, there is no end.
Since no unit fraction is below or at zero, the end is before.
no unit fraction. Therefore there is an end before. Simple as that.
Le 17/08/2024 à 16:29, Richard Damon a écrit :Neither is darkness.
On 8/17/24 9:28 AM, WM wrote:You can't see it and you are unable to derive it from mathematics. But blindness is not an argument.
Le 16/08/2024 à 19:39, Jim Burns a écrit :But there is no point (>0) where it has a finite value,
no element of ℕᵈᵉᶠ is its upper.end, because for each diminishable kSBZ(x) starts with 0 at 0 and increases, but at no point x it
diminishable k+1 disproves by counter.example that k is the upper.end
of ℕᵈᵉᶠ
increases by more than 1 because of ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
Therefore there is a smallest unit fractions and vice versa a greatest
natnumber. What can't you understand?
WM used his keyboard to write :
The other way round. There is an end, because at and below zero there is no >> unit fraction. Therefore there is an end before. Simple as that.
The "sequence" doesn't end.
Am Wed, 21 Aug 2024 14:58:50 +0000 schrieb WM:
Le 21/08/2024 à 15:13, joes a écrit :The unit fractions get denser toward 0. They are all positive.
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:
It is an end which proves dark numbers.Why should there be an end in the first place?
There is indisputably an interval without unit fractions, namely (-oo,
0].
That indicates an end before. It proves an end before. Only cranks can
deny that.
I was talking about the regular ones.Are there not infinitely many unit fractions?Yes, the dark unit fractions cannot be counted through. There are more
than can be counted.
Am Wed, 21 Aug 2024 14:53:27 +0000 schrieb WM:
Le 21/08/2024 à 15:51, Moebius a écrit :Stupid as this: „There is a last negative power of 2.”
Am 21.08.2024 um 15:13 schrieb joes:The other way round. There is an end, because at and below zero there is
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:Since there is no smallest unit fraction, there is no end.
Since no unit fraction is below or at zero, the end is before.
no unit fraction. Therefore there is an end before. Simple as that.
Am Mon, 19 Aug 2024 11:32:44 +0000 schrieb WM:
ButNeither is darkness.
blindness is not an argument.
On 8/21/24 8:32 AM, WM wrote:
No, it is a finite number. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 holds for all and >> only reciprocals of natural numbers.
Can't be, because if it WAS 1/n, then 1/(n+1) would be before it,
Yeah, it's like claiming: "There is an end (to the natural numbers),
because at and above omega there is no natural number.
Le 21/08/2024 à 18:20, FromTheRafters a écrit :Why is no end impossible?
WM presented the following explanation :
In a linear order of elements which all have distances from each other,Since no unit fraction is below or at zero, the end is before. WhatNo end, Duh!!
else?
there is necessarily a last one (if nothing follows) because the only alternative would be more than one. Matheologial conjuring trick is
outside of mathematics.
Le 21/08/2024 à 18:21, FromTheRafters a écrit :It continues in ever smaller steps.
WM used his keyboard to write :
Then it must go on below zero. But it does not.The other way round. There is an end, because at and below zero thereThe "sequence" doesn't end.
is no unit fraction. Therefore there is an end before. Simple as that.
Le 21/08/2024 à 20:34, Moebius a écrit :The reals do, but not the unit fractions. 1/0 is not a unit fraction.
Yeah, it's like claiming: "There is an end (to the natural numbers),Of course, but omega is somewhat ghostly. Do the natnumbers reach till
because at and above omega there is no natural number.
omega? Is there a gap? Zero is fixed and firm, existing with certainty.
The positive reals reach till zero with no doubt. There NUF = 0. And
because of ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 it cannot increase by more than 1 per point x.
Le 21/08/2024 à 20:15, joes a écrit :What do you mean by „complete”?
Am Wed, 21 Aug 2024 14:53:27 +0000 schrieb WM:*If there is a complete chain*, then it has a last member.
Le 21/08/2024 à 15:51, Moebius a écrit :Stupid as this: „There is a last negative power of 2.”
Am 21.08.2024 um 15:13 schrieb joes:The other way round. There is an end, because at and below zero there
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:Since there is no smallest unit fraction, there is no end.
Since no unit fraction is below or at zero, the end is before.
is no unit fraction. Therefore there is an end before. Simple as that.
Am Thu, 22 Aug 2024 11:09:13 +0000 schrieb WM:
*If there is a complete chain*, then it has a last member.
What do you mean by „complete”?
Am Thu, 22 Aug 2024 10:54:46 +0000 schrieb WM:
Le 21/08/2024 à 18:20, FromTheRafters a écrit :
WM presented the following explanation :
Since no unit fraction is below or at zero, the end is before. WhatNo end, Duh!!
else?
In a linear order of elements which all have distances from each other,
there is necessarily a last one
Why is no end impossible?
Le 21/08/2024 à 20:14, joes a écrit :How so?
Am Wed, 21 Aug 2024 14:58:50 +0000 schrieb WM:They all have gaps. This implies a last one if all are existing.
Le 21/08/2024 à 15:13, joes a écrit :The unit fractions get denser toward 0. They are all positive.
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:There is indisputably an interval without unit fractions,
It is an end which proves dark numbers.Why should there be an end in the first place?
namely (-oo, 0].
That indicates an end before. It proves an end before. Only cranks can
deny that.
What then?They are potentially infinite: After each one many more appear. They doI was talking about the regular ones.Are there not infinitely many unit fractions?Yes, the dark unit fractions cannot be counted through. There are more
than can be counted.
never end. But that's not my topic.
Le 21/08/2024 à 20:14, joes a écrit :
Am Wed, 21 Aug 2024 14:58:50 +0000 schrieb WM:
Le 21/08/2024 à 15:13, joes a écrit :The unit fractions get denser toward 0. They are all positive.
Am Wed, 21 Aug 2024 12:24:14 +0000 schrieb WM:
It is an end which proves dark numbers.Why should there be an end in the first place?
There is indisputably an interval without unit fractions, namely (-oo,
0].
That indicates an end before. It proves an end before. Only cranks can
deny that.
They all have gaps. This implies a last one if all are existing.
I was talking about the regular ones.Are there not infinitely many unit fractions?Yes, the dark unit fractions cannot be counted through. There are more
than can be counted.
They are potentially infinite: After each one many more appear. They do
never end. But that's not my topic.
Regards, WM
Am Thu, 22 Aug 2024 11:07:42 +0000 schrieb WM:
How so?The unit fractions get denser toward 0. They are all positive.They all have gaps. This implies a last one if all are existing.
Why is no end impossible?
WM formulated the question :
*If there is a complete chain*, then it has a last member.
Just as the last saucer completes the tea set?
Le 22/08/2024 à 19:24, joes a écrit :Why? One can always divide the space remaining from a supposed
Am Thu, 22 Aug 2024 11:07:42 +0000 schrieb WM:
Every point is either in the empty domain or in the populated domain.How so?The unit fractions get denser toward 0. They are all positive.They all have gaps. This implies a last one if all are existing.
One point is the last one in the populated domain.
Le 22/08/2024 à 16:19, FromTheRafters a écrit :The unit fractions don’t reach 0.
WM formulated the question :
No, just as there is a border between populated and empty domains.*If there is a complete chain*, then it has a last member.Just as the last saucer completes the tea set?
Le 22/08/2024 à 19:24, joes a écrit :
Am Thu, 22 Aug 2024 11:07:42 +0000 schrieb WM:
Le 21/08/2024 à 20:14, joes a écrit :
The unit fractions get denser toward 0.
They are all positive.
They all have gaps.
This implies a last one if all are existing.
How so?
Every point is either in the empty domain
or in the populated domain.
One point is the last one in the populated domain.
On 8/20/2024 10:15 PM, Moebius wrote:
Am 20.08.2024 um 21:35 schrieb Chris M. Thomasson:
Do you even now how to make a Cantor Set Fractal?
Talking about fractals, there's an extremely simple "fractal".
Just consider the real line from 0 to 1 and the points 1/2, 1/4,
1/8, ... (ad infinitum).
If you "zoom in" at 0 bei the factors 2, 4, 8, etc. and just "focus"
on the (new) "0 ... 1 part" of the "magnified" line, you will always
"see" the same pattern.
Noes?
Right. That is a very simple fractal that shows self similarity all the
way down for sure.
Le 21/08/2024 à 20:34, Moebius a écrit :
Yeah, it's like claiming:
"There is an end
(to the natural numbers),
because at and above omega
there is no natural number.
Of course, but
omega is somewhat ghostly.
Do the natnumbers reach till omega?
Do the natnumbers reach till omega?
Is there a gap?
Zero is fixed and firm, existing
with certainty.
The positive reals reach till zero
with no doubt.
There NUF = 0.
And because of
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
it cannot increase by more than
1 per point x.
Am 24.08.2024 um 00:04 schrieb Jim Burns:
x = 1/0 :⇔ 0⋅x = 1
¬∃x ∈ R: x = 1/0
Nonsense.
x = 1/0 :⇔ 0⋅x = 1
¬∃x ∈ R: x = 1/0
Am 24.08.2024 um 00:04 schrieb Jim Burns:
x = 1/0 :⇔ 0⋅x = 1
¬∃x ∈ R: x = 1/0
Nonsense.
On 8/22/2024 8:16 AM, WM wrote:
Le 21/08/2024 à 20:34, Moebius a écrit :
Yeah, it's like claiming:
"There is an end
(to the natural numbers),
because at and above omega
there is no natural number.
Of course, but
omega is somewhat ghostly.
Do the natnumbers reach till omega?
ω is an upper.bound of ℕᵈᵉᶠ.
Of all upper.bounds of ℕᵈᵉᶠ, the lowest is ω.
Each element of ℕᵈᵉᶠ is not upper.bound of ℕᵈᵉᶠ.
No upper.bound of ℕᵈᵉᶠ is in ℕᵈᵉᶠ
Do the natnumbers reach till omega?
Define
the natnumbers reach k ⇔
(∀ᵒʳᵈj≤k:(∃ᵒʳᵈi:j=i∪{i} ⇐ j≠0) ∧ 0<k) ∨ 0=k
k ∈ ω ⇔ the natnumbers reach k
The natnumbers only reach elements of ℕᵈᵉᶠ.
ω, an upper.bound of ℕᵈᵉᶠ, is not.in ℕᵈᵉᶠ.
The natnumbers do not reach ω
On 8/23/2024 10:32 AM, WM wrote:
Le 22/08/2024 à 19:24, joes a écrit :
Am Thu, 22 Aug 2024 11:07:42 +0000 schrieb WM:
Le 21/08/2024 à 20:14, joes a écrit :
The unit fractions get denser toward 0.
They are all positive.
They all have gaps.
This implies a last one if all are existing.
How so?
Every point is either in the empty domain
or in the populated domain.
One point is the last one in the populated domain.
Only in some cases.
Split ℚ⁺ the positive rationals into {p²≤2}ꟴ, {2≤p²}ꟴ
No rational √2 exists.
Even so,
each two rationals are separated by a distance > 0.
Le 23/08/2024 à 20:13, Jim Burns a écrit :Except if you include the Häufungspunkt, of course.
All finiteⁿᵒᵗᐧᵂᴹ ordered sets _and their subsets_Every set of unit fractions extended over an interval has two ends in
are 2.ended, except for {}.
that interval.
Am Fri, 23 Aug 2024 14:27:11 +0000 schrieb WM:
Le 22/08/2024 à 16:19, FromTheRafters a écrit :The unit fractions don’t reach 0.
WM formulated the question :No, just as there is a border between populated and empty domains.
*If there is a complete chain*, then it has a last member.Just as the last saucer completes the tea set?
Am Fri, 23 Aug 2024 14:32:34 +0000 schrieb WM:
Le 22/08/2024 à 19:24, joes a écrit :Why? One can always divide the space remaining from a supposed
Am Thu, 22 Aug 2024 11:07:42 +0000 schrieb WM:Every point is either in the empty domain or in the populated domain.
How so?The unit fractions get denser toward 0. They are all positive.They all have gaps. This implies a last one if all are existing.
One point is the last one in the populated domain.
least unit fraction to zero.
Le 23/08/2024 à 20:06, joes a écrit :lolwtfbbq no there is infinite space before
Am Fri, 23 Aug 2024 14:27:11 +0000 schrieb WM:Of course not. Therefore they must cease before.
Le 22/08/2024 à 16:19, FromTheRafters a écrit :The unit fractions don’t reach 0.
WM formulated the question :No, just as there is a border between populated and empty domains.
*If there is a complete chain*, then it has a last member.Just as the last saucer completes the tea set?
Am Sat, 24 Aug 2024 14:17:58 +0000 schrieb WM:
Le 23/08/2024 à 20:13, Jim Burns a écrit :Except if you include the Häufungspunkt, of course.
All finiteⁿᵒᵗᐧᵂᴹ ordered sets _and their subsets_Every set of unit fractions extended over an interval has two ends in
are 2.ended, except for {}.
that interval.
Le 23/08/2024 à 20:42, Jim Burns a écrit :
On 8/23/2024 10:32 AM, WM wrote:
Le 22/08/2024 à 19:24, joes a écrit :
Am Thu, 22 Aug 2024 11:07:42 +0000 schrieb WM:
Le 21/08/2024 à 20:14, joes a écrit :
The unit fractions get denser toward 0.
They are all positive.
They all have gaps.
This implies a last one if all are existing.
How so?
Every point is either in the empty domain
or in the populated domain.
One point is the last one in the populated domain.
Only in some cases.
In cawsews with distances.
Split ℚ⁺ the positive rationals into {p²≤2}ꟴ, {2≤p²}ꟴ
No rational √2 exists.
But the last rational approximation exists.
It is dark.
But the last rational approximation exists.
It is dark.
Even so,
each two rationals are separated by a distance > 0.
Of course.
But the necklace analogon is easier to comprehend.
Le 24/08/2024 à 00:04, Jim Burns a écrit :
On 8/22/2024 8:16 AM, WM wrote:
Le 21/08/2024 à 20:34, Moebius a écrit :
Yeah, it's like claiming:
"There is an end
(to the natural numbers),
because at and above omega
there is no natural number.
Of course, but
omega is somewhat ghostly.
Do the natnumbers reach till omega?
ω is an upper.bound of ℕᵈᵉᶠ.
Of all upper.bounds of ℕᵈᵉᶠ, the lowest is ω.
Yes.
Nevertheless
almost all natural numbers are bewteen ℕᵈᵉᶠ and ω.
Each element of ℕᵈᵉᶠ is not upper.bound of ℕᵈᵉᶠ.
No upper.bound of ℕᵈᵉᶠ is in ℕᵈᵉᶠ
Correct.
Do the natnumbers reach till omega?
Define
the natnumbers reach k ⇔
(∀ᵒʳᵈj≤k:(∃ᵒʳᵈi:j=i∪{i} ⇐ j≠0) ∧ 0<k) ∨ 0=k
k ∈ ω ⇔ the natnumbers reach k
The natnumbers only reach elements of ℕᵈᵉᶠ.
No.
Do the natnumbers reach till omega?
ω, an upper.bound of ℕᵈᵉᶠ, is not.in ℕᵈᵉᶠ.
The natnumbers do not reach ω
ω - 1 is the greates naturak number.
Easy to understand by the smallest unit fraction.
WM has brought this to us :
The unit fractions don’t reach 0.
Of course not. Therefore they must cease before.
Why must they cease at all?
On 8/24/2024 10:11 AM, WM wrote:
ω is an upper.bound of ℕᵈᵉᶠ.
Of all upper.bounds of ℕᵈᵉᶠ, the lowest is ω.
Yes.
That's a definition of ω
Nevertheless
almost all natural numbers are bewteen ℕᵈᵉᶠ and ω.
Something between ℕᵈᵉᶠ and ω is
a lower.than.lowest upper.bound of ℕᵈᵉᶠ
Something between ℕᵈᵉᶠ and ω
does not exist.
Each non.empty set of ordinals holds a first.
Do the natnumbers reach till omega?
ω, an upper.bound of ℕᵈᵉᶠ, is not.in ℕᵈᵉᶠ.
The natnumbers do not reach ω
ω - 1 is the greates natural number.
if
ω-1 and each j before ω-1 can be stepped.uo.to,
then
ω and each j before ω can be stepped.up.to.
Le 25/08/2024 à 12:21, Jim Burns a écrit :
On 8/24/2024 10:11 AM, WM wrote:
On 8/24/2024 10:11 AM, WM wrote:
Le 24/08/2024 à 00:04, Jim Burns a écrit :
ω is an upper.bound of ℕᵈᵉᶠ.
Of all upper.bounds of ℕᵈᵉᶠ, the lowest is ω.
Yes.
That's a definition of ω
Nevertheless
almost all natural numbers are bewteen ℕᵈᵉᶠ and ω.
Something between ℕᵈᵉᶠ and ω is
a lower.than.lowest upper.bound of ℕᵈᵉᶠ
No.
Dark numbers are not any bounds.
Bounds are definable.
Something between ℕᵈᵉᶠ and ω
does not exist.
So the blind man argues.
Each non.empty set of ordinals holds a first.
Like each set of unit fractions.
Do the natnumbers reach till omega?
ω, an upper.bound of ℕᵈᵉᶠ, is not.in ℕᵈᵉᶠ.
The natnumbers do not reach ω
ω - 1 is the greates natural number.
if
ω-1 and each j before ω-1 can be stepped.uo.to,
then
ω and each j before ω can be stepped.up.to.
Dark numbers cannot be stepped to.
Le 24/08/2024 à 22:16, Jim Burns a écrit :
Many of us are more familiar with necklaces which
have some minimum size of bead (or of whatever).
The unit fractions have no minimum distance but more than nothing.
For rationals,
there is a greatest lower bound of distances
between different rationals,
but,
for a minimum distance to exist,
the greatest lower bound needs to be a distance.
For a distance to exist in every case mathematics is sufficicient:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
The greatest lower bound of distances is 0,
The distances between all unit fractions are positive. The GLB is to
comfort blind persons who cannot think.
Regards, WM
Le 24/08/2024 à 22:16, Jim Burns a écrit :
Many of us are more familiar with necklaces which
have some minimum size of bead (or of whatever).
The unit fractions have no minimum distance
but more than nothing.
For rationals,
there is a greatest lower bound of distances
between different rationals,
but,
for a minimum distance to exist,
the greatest lower bound needs to be a distance.
For a distance to exist in every case
mathematics is sufficicient:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
The greatest lower bound of distances is 0,
The distances between all unit fractions are positive.
The GLB is to comfort blind persons who cannot think.
On 8/23/2024 8:05 PM, Moebius wrote:
Am 24.08.2024 um 00:04 schrieb Jim Burns:
x = 1/0 :⇔ 0⋅x = 1
¬∃x ∈ R: x = 1/0
Nonsense.
| 1.3 Stipulative definitions
|
| A stipulative definition imparts a meaning to the defined term,
| and involves no commitment that
| the assigned meaning agrees with prior uses (if any) of the term.
| Stipulative definitions are epistemologically special.
| They yield judgments with epistemological characteristics that
| are puzzling elsewhere.
| If one stipulatively defines a “raimex” as, say,
| a rational, imaginative, experiencing being
| then the judgment “raimexes are rational” is assured of
| being necessary, certain, and a priori.
https://plato.stanford.edu/entries/definitions/
x = 1/0 :⇔ 0⋅x = 1
Lemma.
On 8/25/2024 3:41 PM, WM wrote:
For a distance to exist in every case
mathematics is sufficicient:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
For each n ∈ ℕ:
1/(n+1) is a counter.example disproving the claim
that 1/n is the smallest unit.fraction:
1/n - 1/(n+1) > 0
On 8/25/2024 3:35 PM, WM wrote:
Dark numbers are not any bounds.
Each non.empty set of ordinals holds a first.
Like each set of unit fractions.
Ordered largest.first, yes.
Each non.empty set of unit.fractions holds a largest.
For each unit.fraction u
the claim that u is smallest is proved false
by counter.example ⅟(1+⅟u)
Not all sets of unit.fractions hold a smallest.
For example, the set of all unit.fractions doesn't.
ω - 1 is the greates natural number.
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Each non.empty set of ordinals holds a first.
Like each set of unit fractions.
Fascinating.
On the other hand, if u is a unit fraction, then 1/(1/s + 1) is a
smaller one.
Ordered largest.first, yes.
Each non.empty set of unit.fractions holds a largest.
For each unit.fraction u
the claim that u is smallest is proved false
by counter.example ⅟(1+⅟u)
Indeed! :-)
Am 26.08.2024 um 00:04 schrieb Moebius:
Am 24.08.2024 um 06:30 schrieb Jim Burns:
Seems you really don't get it. (*sigh*)
Without proof of
1. Ex(raimex(x))
and
2. AxAy(raimex(x) & raimex(y) -> x = y)
we may not use the "definition"
x = the_raimex :<-> rational(x) & imaginative(x) &
experiencing(x) & a_being(x)
In other words, we are not allowed to talk about _the_ raimex (not
having a proof that there is exactly one such entity, i.e. exactly one
x such that x is a raimex).
If we would, say, allow for the "definition"
x = 1/0 :<-> 0 * x = 1
and hence the "lemma"
~Ex(x = 1/0) ,
we would get:
Ax(x =/= 1/0) .
Then by specification (AE) we would get:
1/0 =/= 1/0
from this, and then by (EI):
Ex(x =/= x) .
Not a desirable result.
(That's why _one_ of the rules for proper definitions of constants
requires an existence proof first.)
Am 24.08.2024 um 06:30 schrieb Jim Burns:
Seems you really don't get it. (*sigh*)
Without proof of
1. Ex(raimex(x))
and
2. AxAy(raimex(x) & raimex(y) -> x = y)
we may not use the "definition"
x = the_raimex :<-> rational(x) & imaginative(x) & experiencing(x) & a_being(x)
In other words, we are not allowed to talk about _the_ raimex (not
having a proof that there is exactly one such entity, i.e. exactly
one x such that x is a raimex).
Am 26.08.2024 um 02:15 schrieb Moebius:
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Each non.empty set of ordinals holds a first.
Like each set of unit fractions.
Fascinating.
On the other hand, if u is a unit fraction, then 1/(1/s + 1) is a
smaller one.
For each unit.fraction u
the claim that u is smallest is proved false
by counter.example ⅟(1+⅟u)
Indeed! :-)
Btw. To show this we [usually] don't refer to "the smallest unit fraction"
WM (<- a constant denoting "the smallest unit fraction").
I mean, we don't introduce (define) WM ("the smallest unit fraction") by
a "definition" just to show (afterwards) that it does not exist: ~Ex e
SB: s = WM.
Of course, in a proof by contradiction we may proceed the following way:
Assume that there IS _a_ smallest unit fractions. i.e.
Es e SB: As' e SB\{s}: s < s' .
Let wm (<- an arbitrary name or parameter or ...) be such a unit
fraction.
With other words:
wm e SB & As' e SB\{wm}: wm < s'
and hence
As' e SB\{wm}: wm < s'.
Now from wm e SB we get that 1/(1/wm + 1) e SB and 1/(1/wm + 1) < wm, contradicting
As' e SB\{wm}: wm < s' .
qed.
x = 1/0 :⇔ 0⋅x = 1
Again, that's why we have to prove an existence claim
Ex(Px)
before stating/using the defintion
x = c :<-> Px .
Actually, in addition to this proof we also need a proof of "uniquines" before ...
Le 25/08/2024 à 12:21, Jim Burns a écrit :That’s the problem.
On 8/24/2024 10:11 AM, WM wrote:
No. Dark numbers are not any bounds. Bounds are definable.Something between ℕᵈᵉᶠ and ω is a lower.than.lowest upper.bound of ℕᵈᵉᶠω is an upper.bound of ℕᵈᵉᶠ.Nevertheless almost all natural numbers are bewteen ℕᵈᵉᶠ and ω.
Of all upper.bounds of ℕᵈᵉᶠ, the lowest is ω.
Proof?Something between ℕᵈᵉᶠ and ω does not exist.So the blind man argues.
Each non.empty set of ordinals holds a first.Like each set of unit fractions.
Too bad.Dark numbers cannot be stepped to.if ω-1 and each j before ω-1 can be stepped.uo.to,ω - 1 is the greates natural number.Do the natnumbers reach till omega?ω, an upper.bound of ℕᵈᵉᶠ, is not.in ℕᵈᵉᶠ.
The natnumbers do not reach ω
then ω and each j before ω can be stepped.up.to.
Am Sun, 25 Aug 2024 19:35:09 +0000 schrieb WM:
Dark numbers cannot be stepped to.
Too bad.
Le 24/08/2024 à 19:06, Richard Damon a écrit :How is this accomplished? How would this be prevented?
I guess by your finite logic, Achilles can't pass the tortoise,
Achilles and the tortoise run a race. The tortoise gets a start and the
race begins (state 0). When Achilles reaches this point, the tortoise
has advanced further already (state 1). When Achilles reaches that
point, the tortoise has advanced again (state 2). And so on (states 3,
4, 5, ...). Since Achilles runs much faster than the tortoise, he will overtake (state ), but only after infinitely many finitely indexed
states of the described kind. Their number must be completed.
Otherwisemust = can?
Achilles will not overtake. But there must not be a last visible
finitely indexed state.
(The last 1000 states Achilles remembers have indices much smaller than[Assuming at the point of overtaking]
.) This can only be realized by means of dark states.Why? What do you think of as (in)complete?
According to set theory, all states can be put in bijection with all
natural numbers. This is impossible as completeness and well-order
require a last mark.
The three notions "all" and "infinite" and--
"well-ordered" do not match. This dilemma can only be solved by
refraining from well-order of the set Y of dark-numbered states.
On 8/26/2024 10:08 AM, Moebius wrote:
Am 24.08.2024 um 00:04 schrieb Jim Burns:
x = 1/0 :⇔ 0⋅x = 1
Using a slight variant of your "definition"
The issue is that
Am 24.08.2024 um 00:04 schrieb Jim Burns:
x = 1/0 :⇔ 0⋅x = 1
Using a slight variant of your "definition"
Using a slight variant of your "definition"
it's easy to show why your approach
(concerning definitions)
is not tenable in math:
Def.: x = 1/0 :⇔ x ∈ IR & 0⋅x = 1
Now from identity theory
(say in the context of FOPL=) we get
Ax(x = x)
and hence (by specialisiation, AE):
1/0 = 1/0 .
[We may do this, because "1/0" is now -
after your definition -
a term in our system.]
Am 26.08.2024 um 20:17 schrieb Jim Burns:
On 8/26/2024 10:08 AM, Moebius wrote:
Am 24.08.2024 um 00:04 schrieb Jim Burns:
x = 1/0 :⇔ 0⋅x = 1
Using a slight variant of your "definition"
The issue is that
you are an ignorant idiot.
bye
*plonk*
Therefore,
there is no ω-1,
Something between ℕᵈᵉᶠ and ω
does not exist.
So the blind man argues.
But the sum of the infinite series adding up the terms of
1/n - 1/(n+1)
never gets to 1,
You logic just condemns Achilles to never be able to pass the tortoise
as every time he reaches where the tortoise was, the tortoise has moved
some more so Achilles needs to move again.
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Dark numbers are not any bounds.
Fascinating: So we have that there are "dark numbers" which are smaller (larger) than each and every s e SB (n e IN), but they are no bounds?
On 8/27/2024 12:20 PM, WM wrote:
Le 25/08/2024 à 23:18, Richard Damon a écrit :
But the sum [...] adding up the terms of
1/n - 1/(n+1)
never gets to 1,
1/1 - 1/2 = .5
1/2 - 1/3 = .1(6)
1/3 - 1/4 = .08(3)
It does, but only in the darkness.
Are you a full blown moron or just _mostly_ stupid?
On 8/27/2024 12:36 PM, WM wrote:
Le 26/08/2024 à 02:15, Moebius a écrit :
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Dark numbers are not any bounds.
Fascinating: So we have that there are "dark numbers" which are
smaller (larger) than each and every s e SB (n e IN), but they are no
bounds?
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but
larger than their bound 0.
Define visible? lol. ;^) Give me any natural number x. 1/x. Okay, what
about (1/(x+1))? There is no smallest unit fraction just like there is
no largest natural number... See?
Am 27.08.2024 um 22:07 schrieb Chris M. Thomasson:
On 8/27/2024 12:36 PM, WM wrote:
Le 26/08/2024 à 02:15, Moebius a écrit :
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Dark numbers are not any bounds.
Fascinating: So we have that there are "dark numbers" which are
smaller (larger) than each and every s e SB (n e IN), but they are
no bounds?
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but
larger than their bound 0.
Define visible? lol. ;^) Give me any natural number x. 1/x. Okay, what
about (1/(x+1))? There is no smallest unit fraction just like there is
no largest natural number... See?
No. He is blind. :-)
On 8/27/2024 1:01 PM, Moebius wrote:
Am 27.08.2024 um 21:34 schrieb Chris M. Thomasson:
On 8/27/2024 12:20 PM, WM wrote:
Le 25/08/2024 à 23:18, Richard Damon a écrit :
But the sum [...] adding up the terms of
1/n - 1/(n+1)
never gets to 1,
1/1 - 1/2 = .5
1/2 - 1/3 = .1(6)
1/3 - 1/4 = .08(3)
Don't forget to sum up the terms:
(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) = ?
3/4
Yup. Never equals one... However it "tends" to it...
Le 25/08/2024 à 23:18, Jim Burns a écrit :
Therefore,
there is no ω-1,
If the set of ordinal numbers is complete, then ω-1 precedes ω - by definition.
Something between ℕᵈᵉᶠ and ω
does not exist.
So the blind man argues.
Or actual infinity does not exist. But then there is no ω, and ℕ = ℕᵈᵉᶠ
is potentially infinite only.
Regards, WM
Le 26/08/2024 à 02:15, Moebius a écrit :
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Dark numbers are not any bounds.
Fascinating: So we have that there are "dark numbers" which are
smaller (larger) than each and every s e SB (n e IN), but they are no
bounds?
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but
larger than their bound 0.
Regards, WM
Le 25/08/2024 à 23:18, Richard Damon a écrit :
But the sum of the infinite series adding up the terms of
1/n - 1/(n+1)
never gets to 1,
It does, but only in the darkness.
You logic just condemns Achilles to never be able to pass the tortoise
as every time he reaches where the tortoise was, the tortoise has
moved some more so Achilles needs to move again.
Overtaking is possible only in the darkness.
Regards, WM
Le 27/08/2024 à 21:31, Moebius a écrit :„Sufficient” meaning „not infinitesimal”.
On 8/25/2024 3:45 PM, WM wrote:
This function exists because nothing contradicts its existence.
Right. But the values of the function NUF are:
0 for all real arguments <= 0 and aleph_0 for all real arguments > 0.
Correction: For all ends of real intervals containing ℵo unit fractions. The interval (0, x) cannot contain ℵo unit fractions unless x is sufficiently large.
Le 25/08/2024 à 23:18, Jim Burns a écrit :How does it go again?
Therefore, there is no ω-1,If the set of ordinal numbers is complete, then ω-1 precedes ω - by definition.
Le 25/08/2024 à 23:18, Richard Damon a écrit :Better known as the limit.
It does, but only in the darkness.
But the sum of the infinite series adding up the terms of
1/n - 1/(n+1)
never gets to 1,
--You logic just condemns Achilles to never be able to pass the tortoiseOvertaking is possible only in the darkness.
as every time he reaches where the tortoise was, the tortoise has moved
some more so Achilles needs to move again.
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:
Le 27/08/2024 à 21:31, Moebius a écrit :
On 8/25/2024 3:45 PM, WM wrote:
This function exists because nothing contradicts its existence.
Right. But the values of the function NUF are:
0 for all real arguments <= 0 and aleph_0 for all real arguments > 0.
[Fun fact]: For all ends of real intervals containing ℵo unit fractions. >> The interval (0, x) cannot contain ℵo unit fractions unless x is
sufficiently large.
„Sufficient” meaning „not infinitesimal”.
Hence we write "SUM_(k=1..oo) 1/k - 1/(k+1) = 1" and speak of an
"infinite series" with sum 1 (thouh we do not actually "sum up"
infinitely many terms).
WM wrote :
Le 25/08/2024 à 23:18, Jim Burns a écrit :
Therefore,
there is no ω-1,
If the set of ordinal numbers is complete, then ω-1 precedes ω - by
definition.
What is the definition of subtraction here?
On 8/27/2024 12:36 PM, WM wrote:
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but
larger than their bound 0.
Define visible?
Okay, what
about (1/(x+1))?
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:
The interval (0, x) cannot contain ℵo unit fractions unless x is„Sufficient” meaning „not infinitesimal”.
sufficiently large.
So, we can't SEE Achilles pass the tortoise, so how can we know that he did?
There is no maximum "visible" natural number
Am Tue, 27 Aug 2024 19:36:08 +0000 schrieb WM:
Le 26/08/2024 à 02:15, Moebius a écrit :
Am 25.08.2024 um 23:18 schrieb Jim Burns:
On 8/25/2024 3:35 PM, WM wrote:
Dark numbers are not any bounds.
Fascinating: So we have that there are "dark numbers" which are smaller
(larger) than each and every s e SB (n e IN), but they are no bounds?
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but
larger than their bound 0.
This sounds an awful lot like the hyperreal numbers.
On 8/27/2024 12:20 PM, WM wrote:
Le 25/08/2024 à 23:18, Richard Damon a écrit :
You logic just condemns Achilles to never be able to pass the tortoise
as every time he reaches where the tortoise was, the tortoise has
moved some more so Achilles needs to move again.
Overtaking is possible only in the darkness.
Blah! What a joke.
Am Tue, 27 Aug 2024 19:26:25 +0000 schrieb WM:
Le 25/08/2024 à 23:18, Jim Burns a écrit :
Therefore, there is no ω-1,
If the set of ordinal numbers is complete, then ω-1 precedes ω - by
definition.
How does it go again?
Am 28.08.2024 um 06:20 schrieb joes:
„Sufficient” meaning „not infinitesimal”.
Right. After all, each and every _real_ number x > 0 is "sufficiently
large".
Am Tue, 27 Aug 2024 19:20:26 +0000 schrieb WM:
Le 25/08/2024 à 23:18, Richard Damon a écrit :Better known as the limit.
It does, but only in the darkness.
But the sum of the infinite series adding up the terms of
1/n - 1/(n+1)
never gets to 1,
We'd better define:
A set of ordinals is /Mückenheim complete/ iff each and every
ordinal in the set (except 0)
has an immediate predecessor
(which precedes it).
Theorem: {0, 1, 2, 3, ... ω} is not Mückenheim complete. :-)
Though it's not clear what's missing here. :-)
Am 28.08.2024 um 06:23 schrieb joes:
Am Tue, 27 Aug 2024 19:26:25 +0000 schrieb WM:
Le 25/08/2024 à 23:18, Jim Burns a écrit :
Therefore, there is no ω-1,
*sigh*
Therefore there is no ordinal number o such that o + 1 = ω.
If the set of ordinal numbers is complete, then ω-1 precedes ω - by
definition.
How does it go again?
WM defines "complete" comcerning (sets of) ordinal numbers the following
way:
A set of ordinals is /complete/ iff each and every ordinal
in the set (except 0) has an immediate predecessor (which
precedes it).
Simple as that.
In this sense, {0, 1, 2, 3, ... ω} is not (Mückenheim) complete.
Though clearly no ordinal between 0 and ω is "missing" - LOL. :-)
Hint: ~Eo e ORD: An e IN: n < o < ω.
Hence the term "complete" as defined by Mückenheim is quite
"misleading" (to say the least).
Le 28/08/2024 à 06:20, joes a écrit :So… you agree.
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:
No. "Sufficient" means that ℵo non-empty finite gaps between unitThe interval (0, x) cannot contain ℵo unit fractions unless x is„Sufficient” meaning „not infinitesimal”.
sufficiently large.
fractions fit into the interval. That is much larger than infinitesimal.
On 8/28/2024 5:51 AM, WM wrote:
Le 27/08/2024 à 21:34, "Chris M. Thomasson" a écrit :
On 8/27/2024 12:20 PM, WM wrote:
Le 25/08/2024 à 23:18, Richard Damon a écrit :
You logic just condemns Achilles to never be able to pass the
tortoise as every time he reaches where the tortoise was, the
tortoise has moved some more so Achilles needs to move again.
Overtaking is possible only in the darkness.
Blah! What a joke.
Try to enlighten the step where overtaking happens.
Le 28/08/2024 à 05:24, Richard Damon a écrit :I am confused wrt. to my signature:
So, we can't SEE Achilles pass the tortoise, so how can we know that heBy the same mathematics which proves the existence of dark states.
did?
There is no maximum "visible" natural numberCorrect. With n also n+1, 2n and n^n^n are visible. Potential infinity!
Le 28/08/2024 à 14:47, Moebius a écrit :I think you overlooked a „not”.
Am 28.08.2024 um 06:20 schrieb joes:
Wrong. "Sufficient" means that ℵo non-empty finite gaps between unit fractions fit into the interval. One gap already is larger some x > 0.„Sufficient” meaning „not infinitesimal”.Right. After all, each and every _real_ number x > 0 is "sufficiently
large".
Why do you think that ℵo non-empty finite gaps are smaller than everyI don’t. Get your quantifiers in order.
x > 0?
Am Wed, 28 Aug 2024 13:11:10 +0000 schrieb WM:
Why do you think that ℵo non-empty finite gaps are smaller than everyI don’t.
x > 0?
Get your quantifiers in order.
He seems to think that n being some undisclosed (dark) natural number
means that n+1 is merely another undisclosed (dark) natural number and
you cannot pair (thinking of matching,
Try to enlighten the step where overtaking happens.
Quite simple: The "step" (rather /stage/) can be labeled with "omega".
Am Wed, 28 Aug 2024 13:01:46 +0000 schrieb WM:
I am confused wrt. to my signature:There is no maximum "visible" natural numberCorrect. With n also n+1, 2n and n^n^n are visible. Potential infinity!
Le 29/08/2024 à 12:48, joes a écrit :
Am Wed, 28 Aug 2024 13:11:10 +0000 schrieb WM:
Why do you think that ℵo non-empty finite gaps are smaller than everyI don’t.
x > 0?
How many would you accept?
Get your quantifiers in order.
This order is just that in question: Can two or more unit fractions fit into every interval (0, x)?
Regards, WM
WM presented the following explanation :
Le 27/08/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 8/27/2024 12:36 PM, WM wrote:
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but larger >>>> than their bound 0.
Define visible?
The simplest definition is this: A visible number can be expressed in
decimals or binaries.
Can be? Or is? Or has previously been?
on 8/28/2024, WM supposed :
Le 28/08/2024 à 14:47, Moebius a écrit :
Am 28.08.2024 um 06:20 schrieb joes:
„Sufficient” meaning „not infinitesimal”.
Right. After all, each and every _real_ number x > 0 is "sufficiently
large".
Wrong. "Sufficient" means that ℵo non-empty finite gaps between unit
fractions fit into the interval. One gap already is larger some x > 0. Why do
you think that ℵo non-empty finite gaps are smaller than every x > 0?
Because of the way the reals are constructed.
Am Wed, 28 Aug 2024 13:04:57 +0000 schrieb WM:
Le 28/08/2024 à 06:20, joes a écrit :So… you agree.
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:No. "Sufficient" means that ℵo non-empty finite gaps between unit
The interval (0, x) cannot contain ℵo unit fractions unless x is„Sufficient” meaning „not infinitesimal”.
sufficiently large.
fractions fit into the interval. That is much larger than infinitesimal.
Le 29/08/2024 à 12:48, joes a écrit :Nothing is smaller than all x.
Am Wed, 28 Aug 2024 13:11:10 +0000 schrieb WM:
How many would you accept?Why do you think that ℵo non-empty finite gaps are smaller than everyI don’t.
x > 0?
However, the number of unit fractions less than any given x is infinite.Get your quantifiers in order.
This order is just that in question: Can two or more unit fractions fit--
into every interval (0, x)?
Le 29/08/2024 à 13:00, FromTheRafters a écrit :By induction, they all are.
He seems to think that n being some undisclosed (dark) natural numberNo. If n is a visible nabtuarl number, then also n^n^n is also a visible natural number. That is potential infinity.
means that n+1 is merely another undisclosed (dark) natural number and
you cannot pair (thinking of matching,
On 8/29/2024 6:15 AM, WM wrote:
Le 29/08/2024 à 13:00, FromTheRafters a écrit :
He seems to think that n being some undisclosed (dark) natural number
means that n+1 is merely another undisclosed (dark) natural number
and you cannot pair (thinking of matching,
No. If n is a visible nabtuarl number, then also n^n^n is also a
visible natural number. That is potential infinity.
if n is a natural number then n + 1 is a natural number.
Le 28/08/2024 à 16:00, FromTheRafters a écrit :
WM presented the following explanation :
The simplest definition is this:
A visible number can be expressed in decimals or binaries.
Can be? Or is? Or has previously been?
Is or has previously been.
If not yet expressed in the system,
it is dark in the system.
But small dark numbers can become visisble.
(They have been called grey numbers.)
Le 28/08/2024 à 16:00, FromTheRafters a écrit :By induction that goes for every number.
WM presented the following explanation :
Le 27/08/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 8/27/2024 12:36 PM, WM wrote:
Dark natural numbers are larger than any visible natural number but
smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but
larger than their bound 0.
Define visible?
The simplest definition is this: A visible number can be expressed in
decimals or binaries.
Can be? Or is? Or has previously been?
Is or has previously been. If not yet expressed in the system, it is
dark in the system. But small dark numbers can become visisble. (They
have been called grey numbers.)
Am Thu, 29 Aug 2024 13:37:03 +0000 schrieb WM:
Le 28/08/2024 à 17:06, joes a écrit :
Am Wed, 28 Aug 2024 13:04:57 +0000 schrieb WM:
Le 28/08/2024 à 06:20, joes a écrit :
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:
The interval (0, x) cannot contain ℵo unit fractions unless x is >>>>>> sufficiently large.
„Sufficient” meaning „not infinitesimal”.
No. "Sufficient" means that ℵo non-empty finite gaps between unit
fractions fit into the interval. That is much larger than
infinitesimal.
So… you agree.
Yes, I misread.
Thus: The interval (0, x) contains finitely many unit fractions only
for infinitesimal x.
on 8/29/2024, WM supposed :
Le 29/08/2024 à 13:00, FromTheRafters a écrit :
He seems to think that n being some undisclosed (dark) natural number means >>> that n+1 is merely another undisclosed (dark) natural number and you cannot >>> pair (thinking of matching,
No. If n is a visible nabtuarl number, then also n^n^n is also a visible
natural number. That is potential infinity.
What about n+n if n is visible? or n+1?
Am Thu, 29 Aug 2024 13:13:20 +0000 schrieb WM:
Le 29/08/2024 à 12:48, joes a écrit :Nothing is smaller than all x.
Am Wed, 28 Aug 2024 13:11:10 +0000 schrieb WM:How many would you accept?
Why do you think that ℵo non-empty finite gaps are smaller than every >>>> x > 0?I don’t.
However, the number of unit fractions less than any given x is infinite.Get your quantifiers in order.
Am 30.08.2024 um 09:21 schrieb joes:
Thus: The interval (0, x) contains finitely many unit fractions only
for infinitesimal x.
Actually, it either contains NO unit fractions (if x = 0 or x is infinitesimal but > 0) or it contains infinitely many (ℵo) unit fractions.
So no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.
WM formulated the question :
Le 28/08/2024 à 16:00, FromTheRafters a écrit :
WM presented the following explanation :
Le 27/08/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 8/27/2024 12:36 PM, WM wrote:
Dark natural numbers are larger than any visible natural number but >>>>>> smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but >>>>>> larger than their bound 0.
Define visible?
The simplest definition is this: A visible number can be expressed in
decimals or binaries.
Can be? Or is? Or has previously been?
Is or has previously been. If not yet expressed in the system, it is dark in >> the system. But small dark numbers can become visisble. (They have been
called grey numbers.)
Oh no! Grey numbers? You're pulling my leg now huh? How would you
distinguish a small enough dark or grey number from a fully dark larger number?
Am Thu, 29 Aug 2024 13:34:14 +0000 schrieb WM:
Le 28/08/2024 à 16:00, FromTheRafters a écrit :By induction that goes for every number.
WM presented the following explanation :
Le 27/08/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 8/27/2024 12:36 PM, WM wrote:
Dark natural numbers are larger than any visible natural number but >>>>>> smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but >>>>>> larger than their bound 0.
Define visible?
The simplest definition is this: A visible number can be expressed in
decimals or binaries.
Can be? Or is? Or has previously been?
Is or has previously been. If not yet expressed in the system, it is
dark in the system. But small dark numbers can become visisble. (They
have been called grey numbers.)
Am Thu, 29 Aug 2024 13:37:03 +0000 schrieb WM:
Le 28/08/2024 à 17:06, joes a écrit :Thus: The interval (0, x) contains finitely many unit fractions only
Am Wed, 28 Aug 2024 13:04:57 +0000 schrieb WM:Yes, I misread.
Le 28/08/2024 à 06:20, joes a écrit :So… you agree.
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:No. "Sufficient" means that ℵo non-empty finite gaps between unit
The interval (0, x) cannot contain ℵo unit fractions unless x is >>>>>> sufficiently large.„Sufficient” meaning „not infinitesimal”.
fractions fit into the interval. That is much larger than
infinitesimal.
for infinitesimal x.
Le 30/08/2024 à 09:21, joes a écrit :
Am Thu, 29 Aug 2024 13:34:14 +0000 schrieb WM:
Le 28/08/2024 à 16:00, FromTheRafters a écrit :By induction that goes for every number.
WM presented the following explanation :
Le 27/08/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 8/27/2024 12:36 PM, WM wrote:
Dark natural numbers are larger than any visible natural number but >>>>>>> smaller than their bound omega.
Dark unit fractions are smaller than any visible unit fractions but >>>>>>> larger than their bound 0.
Define visible?
The simplest definition is this: A visible number can be expressed in >>>>> decimals or binaries.
Can be? Or is? Or has previously been?
Is or has previously been. If not yet expressed in the system, it is
dark in the system. But small dark numbers can become visisble. (They
have been called grey numbers.)
*If actual infinity exists*, then induction fails to reach most numbers.
Regards, WM
Le 30/08/2024 à 12:13, Moebius a écrit :
Am 30.08.2024 um 09:21 schrieb joes:
Thus: The interval (0, x) contains finitely many unit fractions only
for infinitesimal x.
Actually, it either contains NO unit fractions (if x = 0 or x is
infinitesimal but > 0) or it contains infinitely many (ℵo) unit
fractions.
That contradicts the fact that ℵo unit fractions occupy an interval that can be reduced.
So no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.
Are there two unit fractions possible lessorequal than all unit fractions?
Regards, WM
Le 30/08/2024 à 09:21, joes a écrit :
Am Thu, 29 Aug 2024 13:37:03 +0000 schrieb WM:
Le 28/08/2024 à 17:06, joes a écrit :Thus: The interval (0, x) contains finitely many unit fractions only
Am Wed, 28 Aug 2024 13:04:57 +0000 schrieb WM:Yes, I misread.
Le 28/08/2024 à 06:20, joes a écrit :So… you agree.
Am Tue, 27 Aug 2024 19:57:39 +0000 schrieb WM:No. "Sufficient" means that ℵo non-empty finite gaps between unit
The interval (0, x) cannot contain ℵo unit fractions unless x is >>>>>>> sufficiently large.„Sufficient” meaning „not infinitesimal”.
fractions fit into the interval. That is much larger than
infinitesimal.
for infinitesimal x.
Every gap between two unit fractions is finite, not infinitesimal.
Regards, WM
Le 30/08/2024 à 12:13, Moebius a écrit :What do you mean by „reducing”? That the number of unit fractions
Am 30.08.2024 um 09:21 schrieb joes:
That contradicts the fact that ℵo unit fractions occupy an interval that can be reduced.Thus: The interval (0, x) contains finitely many unit fractions onlyActually, it either contains NO unit fractions (if x = 0 or x is
for infinitesimal x.
infinitesimal but > 0) or it contains infinitely many (ℵo) unit
fractions.
--So no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.Are there two unit fractions possible lessorequal than all unit
fractions?
Le 30/08/2024 à 03:09, Richard Damon a écrit :Does not follow.
On 8/29/24 9:27 AM, WM wrote:0 is below the end. Hence there is an end.
Le 29/08/2024 à 01:48, Richard Damon a écrit :Because you can't start at an end that isn't there.
On 8/28/24 8:57 AM, WM wrote:
Le 28/08/2024 à 04:13, Richard Damon a écrit :There number exists, it is aleph_0.
Well NUF(x) does not exist, but that doesn't say that infinity isSo the unit fractions are actually existing but their number isn't?
not actual,
You just can't count them from the "end" that doesn't have an end.
That is not a unit fraction.Try to tell me the actual number you are going to start at.I start with NUF(0) = 0.
--Logic that depends on the existance of something that doesn't exist is
just broken.
On 8/30/24 8:59 AM, WM wrote:
Le 30/08/2024 à 12:13, Moebius a écrit :
Am 30.08.2024 um 09:21 schrieb joes:
No, because nothing is smaller than itself, and two unequal order thingsSo no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.
Are there two unit fractions possible lessorequal than all unit fractions? >>
must have an order.
But there ARE two unit fractions less than ANY given unit fraction.
You just apply the wrong qualifiers because you logic system is broken.
Le 30/08/2024 à 15:23, Richard Damon a écrit :
On 8/30/24 8:59 AM, WM wrote:
Le 30/08/2024 à 12:13, Moebius a écrit :
Am 30.08.2024 um 09:21 schrieb joes:
No, because nothing is smaller than itself, and two unequal orderSo no matter if x > 0 is infinitesimal or not: NUF(x) =/= 1.
Are there two unit fractions possible lessorequal than all unit
fractions?
things must have an order.
Not both can be the smallest. Hence only one is the first.
But there ARE two unit fractions less than ANY given unit fraction.
Of course. Any given unit fraction can be given and therefore is visible.
You just apply the wrong qualifiers because you logic system is broken.
No. I apply this way in oder to show the more intelligent readers among
you, than there is NUF(x) = 1.
Regards, WM
Am Fri, 30 Aug 2024 12:59:47 +0000 schrieb WM:
Le 30/08/2024 à 12:13, Moebius a écrit :
Am 30.08.2024 um 09:21 schrieb joes:
Thus: The interval (0, x) contains finitely many unit fractions only
for infinitesimal x.
Actually, it either contains NO unit fractions (if x = 0 or x is
infinitesimal but > 0) or it contains infinitely many (ℵo) unit
fractions.
That contradicts [bla bla bla]
Am Fri, 30 Aug 2024 13:08:14 +0000 schrieb WM:
0 is below the end. Hence there is an end.
Does not follow.
Logic that depends on the existance of something that doesn't exist is
just broken.
Le 30/08/2024 à 09:21, joes a écrit :
[...]
Every gap between two unit fractions is finite,
not infinitesimal.
Le 29/08/2024 à 18:15, FromTheRafters a écrit :
[...]
The collection of visible numbers is potentially infinite.
It reaches as far as induction reaches.
But *if actual infinity is existing*, then
almost all natural numbers are not reached by induction.
But *if actual infinity is existing*, then
almost all natural numbers are not reached by induction.
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
How does that fit with WM who thinks there is a smallest unit fraction
to start counting from?
On 01.09.2024 21:09, Chris M. Thomasson wrote:
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
Don't get confused by that nonsense. Everybody knows that unit fractions
are different from each other. Therefore they cannot be counted at the
same x, let alone at less than all positive x, i.e., at zero.
How does that fit with WM who thinks there is a smallest unit fraction
to start counting from?
I do not think that but I prove that by the simple fact that not more
than one unit fraction can be lessequal than all unit fractions.
Regards, WM
On 02.09.2024 19:13, Richard Damon wrote:
On 9/2/24 12:56 PM, WM wrote:
On 01.09.2024 21:09, Chris M. Thomasson wrote:Except that it starts with the incorrect assumption that such a unit fraction exists.
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
Don't get confused by that nonsense. Everybody knows that unit
fractions are different from each other. Therefore they cannot be
counted at the same x, let alone at less than all positive x, i.e., at
zero.
How does that fit with WM who thinks there is a smallest unit
fraction to start counting from?
I do not think that but I prove that by the simple fact that not more
than one unit fraction can be lessequal than all unit fractions.
No, it shows that no unit fractions exist unless a first one exists.
Regards, WM
On 9/2/24 12:56 PM, WM wrote:
On 01.09.2024 21:09, Chris M. Thomasson wrote:Except that it starts with the incorrect assumption that such a unit fraction exists.
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
Don't get confused by that nonsense. Everybody knows that unit
fractions are different from each other. Therefore they cannot be
counted at the same x, let alone at less than all positive x, i.e., at
zero.
How does that fit with WM who thinks there is a smallest unit
fraction to start counting from?
I do not think that but I prove that by the simple fact that not more
than one unit fraction can be lessequal than all unit fractions.
On 9/3/24 7:19 AM, WM wrote:
No, it shows that no unit fractions exist unless a first one exists.
Sol you think that 1/1 doesn't exist?
On 9/2/2024 9:56 AM, WM wrote:
On 01.09.2024 21:09, Chris M. Thomasson wrote:
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
Don't get confused by that nonsense. Everybody knows that unit
fractions are different from each other. Therefore they cannot be
counted at the same x, let alone at less than all positive x, i.e., at
zero.
I suppose you think you can prove there is a largest natural number as
well?
On 02.09.2024 19:13, Richard Damon wrote:
On 9/2/24 12:56 PM, WM wrote:
On 01.09.2024 21:09, Chris M. Thomasson wrote:Except that it starts with the incorrect assumption that such a unit
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
Don't get confused by that nonsense. Everybody knows that unit
fractions are different from each other. Therefore they cannot be
counted at the same x, let alone at less than all positive x, i.e.,
at zero.
How does that fit with WM who thinks there is a smallest unit
fraction to start counting from?
I do not think that but I prove that by the simple fact that not more
than one unit fraction can be lessequal than all unit fractions.
fraction exists.
No, it shows that no unit fractions exist unless a first one exists.
Regards, WM
On 02.09.2024 20:47, Chris M. Thomasson wrote:
On 9/2/2024 9:56 AM, WM wrote:
On 01.09.2024 21:09, Chris M. Thomasson wrote:
On 8/31/2024 8:27 PM, Moebius wrote:
In general, for all x e IR, x > 0: NUF(x) = aleph_0.
Don't get confused by that nonsense. Everybody knows that unit
fractions are different from each other. Therefore they cannot be
counted at the same x, let alone at less than all positive x, i.e.,
at zero.
I suppose you think you can prove there is a largest natural number as
well?
The proof with unit fractions is more convincing because 0 has been
better accepted than ω. And it is really simple: At 0 there are no unit fractions. Then their set increases but all have different positions.
Regards, WM
On 03.09.2024 13:26, Richard Damon wrote:
On 9/3/24 7:19 AM, WM wrote:
No, it shows that no unit fractions exist unless a first one exists.
Sol you think that 1/1 doesn't exist?
No, I know that 1/1 does not exist unless the smallest unit fraction
exists.
Regards, WM
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