[email protected] schrieb am Samstag, 6. Januar 2024 um 03:17:23 UTC+1:exists) or nothing (if infinity is potential only).
zero is not a quantity so it would be absolute
But above zero is first quantity or infinitesimal
of the calculus. That is sub finite.
The first real numbers above zero are dark in that they can only be used collectively. (But some of them must be there below any defined point.) Moreover, between any two defined points on the real line there are ℵ dark points (if actual infinity
Regards, WM
On 1/7/2024 2:34 PM, WM wrote:
[email protected] schrieb am Samstag, 6. Januar 2024 um 03:17:23 UTC+1: >>> zero is not a quantity so it would be absolute
But above zero is first quantity or infinitesimal
of the calculus. That is sub finite.
The first real numbers above zero are dark in that they can only be used
collectively. (But some of them must be there below any defined point.) Moreover,
between any two defined points on the real line there are ℵ dark points (if
actual infinity exists) or nothing (if infinity is potential only).
There is no _first_ real number above zero.
On 1/7/24 5:34 PM, WM wrote:
[email protected] schrieb am Samstag, 6. Januar 2024 um 03:17:23 UTC+1: >>> zero is not a quantity so it would be absolute
But above zero is first quantity or infinitesimal
of the calculus. That is sub finite.
The first real numbers above zero are dark in that they can only be used
collectively. (But some of them must be there below any defined point.) Moreover,
between any two defined points on the real line there are ℵ dark points (if
actual infinity exists) or nothing (if infinity is potential only).
Nope, your reasoning is "Dark", as it just doesn't understand Unbouded sets.
Can you name a set that only has "Dark" numbers?
You say they can be used collectively, so give a collection that only
has dark numbers!
If not, then are your dark numbers actually in existance?
From your previous comments, the "darkness" isn't actually a property
of the "Number" but of the observes knowledge.
It seems you just don't know how to keep the numbers you are talking
about in the set that they need to be in.
Le 08/01/2024 à 00:26, Richard Damon a écrit :
On 1/7/24 5:34 PM, WM wrote:
[email protected] schrieb am Samstag, 6. Januar 2024 um 03:17:23
UTC+1:
zero is not a quantity so it would be absolute
But above zero is first quantity or infinitesimal
of the calculus. That is sub finite.
The first real numbers above zero are dark in that they can only be
used collectively. (But some of them must be there below any defined
point.) Moreover, between any two defined points on the real line
there are ℵ dark points (if actual infinity exists) or nothing (if
infinity is potential only).
Nope, your reasoning is "Dark", as it just doesn't understand Unbouded
sets.
I understand that every real x > 0 that you can name has ℵ smaller
positive reals ℵ of which you cannot name.
Can you name a set that only has "Dark" numbers?
If existing {ω, ω+1, ω+2, ...} because there is no finite initial
segment for any of its elements.
You say they can be used collectively, so give a collection that only
has dark numbers!
That is not a precondition for the existence of dark numbers.
If not, then are your dark numbers actually in existance?
I don't know ehtehre dar numbers are existing. I have proved only that,
if actual infinity exists, then dark numbers exist too.
From your previous comments, the "darkness" isn't actually a property
of the "Number" but of the observes knowledge.
It seems you just don't know how to keep the numbers you are talking
about in the set that they need to be in.
Try to understand: Every x > 0 is not the smallest positive number.
Either there are no (not yet) numbers below the smallest defined x > 0,
or they are dark.
Regards, WM
On 1/8/24 5:56 AM, WM wrote:
I understand that every real x > 0 that you can name has ℵ smaller
positive reals ℵ of which you cannot name.
why do you think I can not name any of them?
On Saturday 6 January 2024 at 03:17:23 UTC+1, [email protected] wrote:
zero is not a quantity so it would be absolutedoesn't exist you nimrod
But above zero is first quantity or infinitesimal
of the calculus. That is sub finite.
[email protected] schrieb am Montag, 8. Januar 2024 um 07:37:56 UTC+1:
On Saturday 6 January 2024 at 03:17:23 UTC+1, [email protected]
wrote: > zero is not a quantity so it would be absolute > But above
zero is first quantity or infinitesimal > of the calculus. That is sub
finite.
doesn't exist you nimrod
There are ℵ real numbers in every interval (0, eps) for every eps > 0
that you can choose. That proves that ℵ real numbers cannot be chosen individually but only collectively as I just described them. They are dark.
Regards, WM
Le 08/01/2024 à 13:21, Richard Damon a écrit :
On 1/8/24 5:56 AM, WM wrote:
I understand that every real x > 0 that you can name has ℵ smaller
positive reals ℵ of which you cannot name.
why do you think I can not name any of them?
You can name many of them, but every attempt fails to reduce the
remainder below ℵ. It can be reduced however to the empty set collectively.
Regards, WM
On 1/8/24 7:29 AM, WM wrote:
Le 08/01/2024 à 13:21, Richard Damon a écrit :
On 1/8/24 5:56 AM, WM wrote:
I understand that every real x > 0 that you can name has ℵ smaller
positive reals ℵ of which you cannot name.
why do you think I can not name any of them?
You can name many of them, but every attempt fails to reduce the
remainder below ℵ. It can be reduced however to the empty set collectively.
Why do you expect otherwise?
We have an infinite set.
Individually naming is a finite operation, so can only name a finite
number of them at a time.
Thus there will alway be an infinite number that we haven't produced the name.
That doesn't mean the rest can't be named.
"naming" is the opposite of your "dark", you can only do it
individually, but not to an infinite group.
It doesn't mean that any particular ones left can't be named.
It seems your problem is you can't keep the right domain in focus.
Your "dark" numbers have some of the properties of the transfinite
numbers,
Le 09/01/2024 à 03:31, Richard Damon a écrit :
On 1/8/24 7:29 AM, WM wrote:
Le 08/01/2024 à 13:21, Richard Damon a écrit :
On 1/8/24 5:56 AM, WM wrote:
I understand that every real x > 0 that you can name has ℵ smaller >>>>> positive reals ℵ of which you cannot name.
why do you think I can not name any of them?
You can name many of them, but every attempt fails to reduce the
remainder below ℵ. It can be reduced however to the empty set
collectively.
Why do you expect otherwise?
I don't. I only note that fact.
We have an infinite set.
Individually naming is a finite operation, so can only name a finite
number of them at a time.
But if you could name every number, then you could name one with lesser successors.
Thus there will alway be an infinite number that we haven't produced
the name.
That doesn't mean the rest can't be named.
ℵ will always be without name. But they can be removed such that none remains, collectively.
"naming" is the opposite of your "dark", you can only do it
individually, but not to an infinite group.
So it is. And not for the numbers belonging to the last ℵ.
It doesn't mean that any particular ones left can't be named.
It does. The last ℵ numbers will remain.
It seems your problem is you can't keep the right domain in focus.
Your "dark" numbers have some of the properties of the transfinite
numbers,
Yes, they are also dark because they have no finite initial segment of natural numbers.
Regards, WM
On 1/9/24 12:23 PM, WM wrote:
Le 09/01/2024 à 03:31, Richard Damon a écrit :
On 1/8/24 7:29 AM, WM wrote:
Le 08/01/2024 à 13:21, Richard Damon a écrit :
On 1/8/24 5:56 AM, WM wrote:
I understand that every real x > 0 that you can name has ℵ smaller >>>>>> positive reals ℵ of which you cannot name.
why do you think I can not name any of them?
You can name many of them, but every attempt fails to reduce the
remainder below ℵ. It can be reduced however to the empty set
collectively.
Why do you expect otherwise?
I don't. I only note that fact.
We have an infinite set.
Individually naming is a finite operation, so can only name a finite
number of them at a time.
But if you could name every number, then you could name one with lesser
successors.
Why?
If you could name a number with a finite number of points after it, you
could find the last number, but since no number is the last, you can't
do that.
There is no problem with having infinite sets that don't have an end,
that is just the nature of unbounded numbers.
ℵ will always be without name. But they can be removed such that none
remains, collectively.
Nope, they all have a name, we just can't express them all at once.
"naming" is the opposite of your "dark", you can only do it
individually, but not to an infinite group.
So it is. And not for the numbers belonging to the last ℵ.
No, we didn't "use up" the names, so we still have names for them all.
It doesn't mean that any particular ones left can't be named.
It does. The last ℵ numbers will remain.
Nope. I gave the formula to name any of them individually.
WM used his keyboard to write :
That proves that ℵ real numbers cannot be chosen individually
but only collectively as I just described them. They are dark.
No it doesn't.
Above zero is first non zero fundamental infinitesimal.
Le 10/01/2024 à 03:58, Richard Damon a écrit :
On 1/9/24 12:23 PM, WM wrote:
Le 09/01/2024 à 03:31, Richard Damon a écrit :
On 1/8/24 7:29 AM, WM wrote:
Le 08/01/2024 à 13:21, Richard Damon a écrit :
On 1/8/24 5:56 AM, WM wrote:
I understand that every real x > 0 that you can name has ℵ
smaller positive reals ℵ of which you cannot name.
why do you think I can not name any of them?
You can name many of them, but every attempt fails to reduce the
remainder below ℵ. It can be reduced however to the empty set
collectively.
Why do you expect otherwise?
I don't. I only note that fact.
We have an infinite set.
Individually naming is a finite operation, so can only name a finite
number of them at a time.
But if you could name every number, then you could name one with
lesser successors.
Why?
Because they are existing.
If you could name a number with a finite number of points after it,
you could find the last number, but since no number is the last, you
can't do that.
There is a smallest unit fraction with no dount because
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
is valid for all unit fractions.
But you cannot name it.
There is no problem with having infinite sets that don't have an end,
that is just the nature of unbounded numbers.
It is the nature of all visible numbers.
ℵ will always be without name. But they can be removed such that none
remains, collectively.
Nope, they all have a name, we just can't express them all at once.
You can remove all. But you cannot name those with less than ℵo
successors. So you cannot name all.
"naming" is the opposite of your "dark", you can only do it
individually, but not to an infinite group.
So it is. And not for the numbers belonging to the last ℵ.
No, we didn't "use up" the names, so we still have names for them all.
Apply them. Nevertheless ℵo will remain without name.
It doesn't mean that any particular ones left can't be named.
It does. The last ℵ numbers will remain.
Nope. I gave the formula to name any of them individually.
But you cannot give the names.
Regards, WM
What Natural number exists with only a finite number of numbers above it?
Finite sets are always namable, so you should be able to name it.
Le 11/01/2024 à 01:49, Richard Damon a écrit :
What Natural number exists with only a finite number of numbers above it?
Finite sets are always namable, so you should be able to name it.
Not within the dark domain.
Regards, WM
On 1/11/24 4:33 PM, WM wrote:
Le 11/01/2024 à 01:49, Richard Damon a écrit :
What Natural number exists with only a finite number of numbers above it? >>>
Finite sets are always namable, so you should be able to name it.
Not within the dark domain.
Why?
Le 12/01/2024 à 04:11, Richard Damon a écrit :
On 1/11/24 4:33 PM, WM wrote:You must try it. The last ℵo natural numbers and the first ℵo unit fractions cannot be subdivided.
Le 11/01/2024 à 01:49, Richard Damon a écrit :
What Natural number exists with only a finite number of numbers
above it?
Finite sets are always namable, so you should be able to name it.
Not within the dark domain.
Why?
Regards, WM
On 1/12/24 9:17 AM, WM wrote:
Le 12/01/2024 à 04:11, Richard Damon a écrit :
On 1/11/24 4:33 PM, WM wrote:You must try it. The last ℵo natural numbers and the first ℵo unit
Le 11/01/2024 à 01:49, Richard Damon a écrit :
What Natural number exists with only a finite number of numbers
above it?
Finite sets are always namable, so you should be able to name it.
Not within the dark domain.
Why?
fractions cannot be subdivided.
Why not?
WHere is the actual property derivable from the actual definition of
Natural Numbers that creates these non-subdividable numbers?
I think it is all just in your head, caused by using faulty logic.
Le 13/01/2024 à 00:46, Richard Damon a écrit :
On 1/12/24 9:17 AM, WM wrote:
Le 12/01/2024 à 04:11, Richard Damon a écrit :
On 1/11/24 4:33 PM, WM wrote:You must try it. The last ℵo natural numbers and the first ℵo unit
Le 11/01/2024 à 01:49, Richard Damon a écrit :
What Natural number exists with only a finite number of numbers
above it?
Finite sets are always namable, so you should be able to name it.
Not within the dark domain.
Why?
fractions cannot be subdivided.
Why not?
Because otherwise the last one could be identified. But it proves
impossible like the splitting of a quark. There remain always ℵ
elements. Therefore they are called dark.
WHere is the actual property derivable from the actual definition of
Natural Numbers that creates these non-subdividable numbers?
The axioms only create the definable natnumbers.
I think it is all just in your head, caused by using faulty logic.
It is caused by your inability to reduce the amounts to less than ℵ.
Regards, WM
On 1/13/24 5:23 AM, WM wrote:
The axioms only create the definable natnumbers.
But since the axioms create ALL the Natural Numbers, you are just
admitting that your "Dark Numbers" are not part of them.
It is caused by your inability to reduce the amounts to less than ℵ.
And why do I need to?
That seems to be your problem, you are assuming you can do what isn't allowed,
Le 13/01/2024 à 14:22, Richard Damon a écrit :
On 1/13/24 5:23 AM, WM wrote:
The axioms only create the definable natnumbers.
But since the axioms create ALL the Natural Numbers, you are just
admitting that your "Dark Numbers" are not part of them.
No, the matter is not so easy. Peano's axioms create the natural
numbers, a potentially infinite sequence or collection, not a set.Only Zermelo's axioms, which are adopted from Peano or Dedekind, create all natural numbers because in set theory there is the complete set.
It is caused by your inability to reduce the amounts to less than ℵ.
And why do I need to?
In order to prove that you could.
That seems to be your problem, you are assuming you can do what isn't
allowed,
It is not only disallowed (a revolutionary spirit would violate this prohibition), but it is impossible. Why? Because it is impossible. I
call that dark. But the notion is irrelevant. The fact is what counts
(and hinders the counting).
Regards, WM
On 1/14/24 7:35 AM, WM wrote:
That seems to be your problem, you are assuming you can do what isn't
allowed,
It is not only disallowed (a revolutionary spirit would violate this
prohibition), but it is impossible. Why? Because it is impossible. I
call that dark. But the notion is irrelevant. The fact is what counts
(and hinders the counting).
So, your logic is based on the concept that since we can't divide an
infinite set into two finite sets, there must be elements that we can
not name in the set?
Le 14/01/2024 à 19:42, Richard Damon a écrit :
On 1/14/24 7:35 AM, WM wrote:
That seems to be your problem, you are assuming you can do what
isn't allowed,
It is not only disallowed (a revolutionary spirit would violate this
prohibition), but it is impossible. Why? Because it is impossible. I
call that dark. But the notion is irrelevant. The fact is what counts
(and hinders the counting).
So, your logic is based on the concept that since we can't divide an
infinite set into two finite sets, there must be elements that we can
not name in the set?
There *are* elements we cannot name, even almost all, namely ℵ, whereas
we can name only finitely many.
Regards, WM
On 1/15/24 3:18 AM, WM wrote:
There *are* elements we cannot name, even almost all, namely ℵ, whereas
we can name only finitely many.
But ℵ isn't a number in the domain of regard.
Note, we CAN name any of a countable infinite number of them, as we have
a countable infinite number of names to use.
We can only name a finitely many at a time, but a countable infinite
number have names, and that number matches the ℵ0 of them that are.
Le 15/01/2024 à 13:52, Richard Damon a écrit :
On 1/15/24 3:18 AM, WM wrote:
There *are* elements we cannot name, even almost all, namely ℵ,
whereas we can name only finitely many.
But ℵ isn't a number in the domain of regard.
But the ℵ elements are in the domain.
Note, we CAN name any of a countable infinite number of them, as we
have a countable infinite number of names to use.
Why do always infinitely many unit fractions remain between the smallest named one and zero?
We can only name a finitely many at a time, but a countable infinite
number have names, and that number matches the ℵ0 of them that are.
But you cannot use a name of a unit fraction having less than ℵ smaller ones.
Regards, WM
On 1/17/24 6:36 AM, WM wrote:
Le 15/01/2024 à 13:52, Richard Damon a écrit :
On 1/15/24 3:18 AM, WM wrote:
There *are* elements we cannot name, even almost all, namely ℵ,
whereas we can name only finitely many.
But ℵ isn't a number in the domain of regard.
But the ℵ elements are in the domain.
Yes, but the value ℵ isn't
Note, we CAN name any of a countable infinite number of them, as we
have a countable infinite number of names to use.
Why do always infinitely many unit fractions remain between the smallest
named one and zero?
Because the set is unbounded.
We can only name a finitely many at a time, but a countable infinite
number have names, and that number matches the ℵ0 of them that are.
But you cannot use a name of a unit fraction having less than ℵ smaller
ones.
because such a number doesn't exist. You don't need to name the
non-existant.
The set is bounded by its smallest upper bound 0.
Le 15/01/2024 à 13:52, Richard Damon a écrit :
Note, we CAN name any of
a countable infinite number of them, as
we have a countable infinite number of
names to use.
Why do always
infinitely many unit fractions remain
between the smallest named one and zero?
Le 17/01/2024 à 13:56, Richard Damon a écrit :
On 1/17/24 6:36 AM, WM wrote:
Le 15/01/2024 à 13:52, Richard Damon a écrit :
On 1/15/24 3:18 AM, WM wrote:
There *are* elements we cannot name, even almost all, namely ℵ,
whereas we can name only finitely many.
But ℵ isn't a number in the domain of regard.
But the ℵ elements are in the domain.
Yes, but the value ℵ isn't
Note, we CAN name any of a countable infinite number of them, as we
have a countable infinite number of names to use.
Why do always infinitely many unit fractions remain between the
smallest named one and zero?
Because the set is unbounded.
Wrong. The set is bounded by its smallest upper bound 0.
We can only name a finitely many at a time, but a countable infinite
number have names, and that number matches the ℵ0 of them that are.
But you cannot use a name of a unit fraction having less than ℵ
smaller ones.
because such a number doesn't exist. You don't need to name the
non-existant.
NUF(0) = 0
NUF(x>0) = ℵo
This means an increase. But that increase cannot happen between 0 and
(0, 1] because
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
So you are wrong, if mathematics is right.
Regards, WM
On 1/17/24 9:08 AM, WM wrote:
Note, we CAN name any of a countable infinite number of them, as we
have a countable infinite number of names to use.
Why do always infinitely many unit fractions remain between the
smallest named one and zero?
Because the set is unbounded.
Wrong. The set is bounded by its largest lower bound 0
But 0 isn't member of the set, so the set itself is unbounded.
Thus NUF(x) with x finite, jumps, perhaps in domain of the
infinitesimals between 0 and the bottom of (0,1]
On 1/17/2024 6:36 AM, WM wrote:
Le 15/01/2024 à 13:52, Richard Damon a écrit :
Note, we CAN name any of
a countable infinite number of them, as
we have a countable infinite number of
names to use.
Why do always
infinitely many unit fractions remain
between the smallest named one and zero?
Some cardinalities can change by 1.
They are the cardinalities of
flocks of sheep and of pockets of pebbles.
They are finite.
The cardinality of
cardinalities which can [not] change by 1
is larger.than any cardinality which can change by 1,
is not any cardinality which can change by 1.
That cardinality cannot change by 1.
|⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...
Le 18/01/2024 à 02:36, Richard Damon a écrit :
On 1/17/24 9:08 AM, WM wrote:
Note, we CAN name any of a countable infinite number of them, as
we have a countable infinite number of names to use.
Why do always infinitely many unit fractions remain between the
smallest named one and zero?
Because the set is unbounded.
Wrong. The set is bounded by its largest lower bound 0
But 0 isn't member of the set, so the set itself is unbounded.
No. The bound exists. All unit fractions must fit into the positive
interval. They have internal distances. Hence there is a linear chain
having a first element. The only alternative, namely infinitely many
between 0 and (0, 1], can be excluded by mathematics.
Thus NUF(x) with x finite, jumps, perhaps in domain of the
infinitesimals between 0 and the bottom of (0,1]
That is merely another name for dark real numbers. We cannot address
them as individuals.
Regards, WM
On 1/18/24 3:54 AM, WM wrote:
There is not need for a first element, since the chain is unbounded in length.
That is merely another name for dark real numbers. We cannot address
them as individuals.
But they are not elements of the finite numbers, and thus not Real,
Rational, or Unit Fractions.
Le 17/01/2024 à 21:04, Jim Burns a écrit :
The cardinality of
cardinalities which can [not] change by 1
is larger.than
any cardinality which can change by 1,
is not
any cardinality which can change by 1.
That cardinality cannot change by 1.
|⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...
That statement is correct for
nameable unit fractions, but in general
it is in contradiction with
mathematics because of
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
|⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...
That statement is correct for
nameable unit fractions, but in general
it is in contradiction with
mathematics because of
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
Le 18/01/2024 à 13:46, Richard Damon a écrit :
On 1/18/24 3:54 AM, WM wrote:
There is not need for a first element, since the chain is unbounded in
length.
No, it is bounded in length by the origin 0.
That is merely another name for dark real numbers. We cannot address
them as individuals.
But they are not elements of the finite numbers, and thus not Real,
Rational, or Unit Fractions.
All unit fractions ate unit fractions and are elements of the finite
numbers.
Regards, WM
On 1/18/24 12:34 PM, WM wrote:
Le 18/01/2024 à 13:46, Richard Damon a écrit :
On 1/18/24 3:54 AM, WM wrote:
There is not need for a first element, since the chain is unbounded in
length.
No, it is bounded in length by the origin 0.
Which isn't an element of the set,
All unit fractions are unit fractions and are elements of the finite
numbers.
yes, and all unit fractions, being one divided by a Natural Number, are individually definable by that Natural Number, and thus not "dark"
I will give
cardinal which can change by 1
a shorter name: final ordinal.
0 = {} is first with cardinality 0
5 = {0,1,2,3,4} is first with cardinality 5
ω = {0,1,2,…} is first with cardinality ℵ₀
0, 5, and ω are initial ordinals.
0 is last with cardinality 0
5 is last with cardinality 5
ω _is not_ last with cardinality ℵ₀
ω+1 = {0,1,…;ω} is also with cardinality ℵ₀
|ω| = |ω+1|
Le 19/01/2024 à 02:26, Richard Damon a écrit :
On 1/18/24 12:34 PM, WM wrote:
Le 18/01/2024 à 13:46, Richard Damon a écrit :
On 1/18/24 3:54 AM, WM wrote:
There is not need for a first element, since the chain is unbounded
in length.
No, it is bounded in length by the origin 0.
Which isn't an element of the set,
But NUF(x) is well defined.
All unit fractions are unit fractions and are elements of the finite
numbers.
yes, and all unit fractions, being one divided by a Natural Number,
are individually definable by that Natural Number, and thus not "dark"
Most unit fractions are dark because most natural numbers are dark.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
Regards, WM
On 1/19/24 3:55 AM, WM wrote:
No, it is bounded in length by the origin 0.
Which isn't an element of the set,
But NUF(x) is well defined.
No, it isn't.
Most unit fractions are dark because most natural numbers are dark.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
But that doesn't define a set of just dark number.
Your problem is you don't have a definiton of what "N_def" actually is,
You are implicitly assuming that there is a "highest" defined number,
Le 19/01/2024 à 16:54, Richard Damon a écrit :
On 1/19/24 3:55 AM, WM wrote:
No, it is bounded in length by the origin 0.
Which isn't an element of the set,
But NUF(x) is well defined.
No, it isn't.
NUF(x) is the number of unit fractions between 0 and x.
Most unit fractions are dark because most natural numbers are dark.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
But that doesn't define a set of just dark number.
There is no set of only dark numbers. There is an infinite set ℕ, a
finite part of which is visible, the complement is dark.
Your problem is you don't have a definiton of what "N_def" actually is,
Every number that is defined individually is visible.
You are implicitly assuming that there is a "highest" defined number,
Not a constant number but only temporarily.
Regards, WM
On 1/19/24 12:07 PM, WM wrote:
Le 19/01/2024 à 16:54, Richard Damon a écrit :
On 1/19/24 3:55 AM, WM wrote:
No, it is bounded in length by the origin 0.
Which isn't an element of the set,
But NUF(x) is well defined.
No, it isn't.
NUF(x) is the number of unit fractions between 0 and x.
Which is infinite for all x > 0.
And thus NUF(x) doesn't have a finite value answer for any x > 0, and if
you are saying you are working in the domain of finite values, NUF(x)
isn't actually defined for any x > 0, since its value isn't defined to something in the domain of regard.
Nothing in that definition allows it to have values other than 0 or
infinity, so claiming it has other values is just an error.
To presume it has the value 1 somewhere, presumes that there exists a smallest unit fraction, which is a false assumption.
Most unit fractions are dark because most natural numbers are dark.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
But that doesn't define a set of just dark number.
There is no set of only dark numbers. There is an infinite set ℕ, a
finite part of which is visible, the complement is dark.
But why is it only a finite part that is visible?
Your problem is you don't have a definiton of what "N_def" actually is,
Every number that is defined individually is visible.
So, all Natural Numbers are visible, as we can find the individual
definition of any of them.
You are implicitly assuming that there is a "highest" defined number,
Not a constant number but only temporarily.
So not actually existing.
All you are doing is showing that there exist number bigger than HAVE
BEEN named, not bigger than CAN be named.
This becomes a problem of what we have seen, not of what is, so not a property of the numbers themselves, but of the observer.
Le 18/01/2024 à 19:20, Jim Burns a écrit :
[...]
In order to excorzise the bijective meaning
I will henceforth use only ℵ,
meaning "infinitely many".
|ℕ| = ℵ, |ℚ| = ℵ, |ℝ| = ℵ,
Le 19/01/2024 à 19:09, Richard Damon a écrit :
On 1/19/24 12:07 PM, WM wrote:
Le 19/01/2024 à 16:54, Richard Damon a écrit :
On 1/19/24 3:55 AM, WM wrote:
No, it is bounded in length by the origin 0.
Which isn't an element of the set,
But NUF(x) is well defined.
No, it isn't.
NUF(x) is the number of unit fractions between 0 and x.
Which is infinite for all x > 0.
Not when mathematics is applied.
And thus NUF(x) doesn't have a finite value answer for any x > 0, and
if you are saying you are working in the domain of finite values,
NUF(x) isn't actually defined for any x > 0, since its value isn't
defined to something in the domain of regard.
Nothing in that definition allows it to have values other than 0 or
infinity, so claiming it has other values is just an error.
To presume it has the value 1 somewhere, presumes that there exists a
smallest unit fraction, which is a false assumption.
Not when mathematics is applied.
Most unit fractions are dark because most natural numbers are dark.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
But that doesn't define a set of just dark number.
There is no set of only dark numbers. There is an infinite set ℕ, a
finite part of which is visible, the complement is dark.
But why is it only a finite part that is visible?
Try to make more visible. Fail. Then you know it.
Every number that is defined individually is visible.
Your problem is you don't have a definiton of what "N_def" actually is, >>>
So, all Natural Numbers are visible, as we can find the individual
definition of any of them.
Try it.
You are implicitly assuming that there is a "highest" defined number,
Not a constant number but only temporarily.
So not actually existing.
The number is actually existing like the biggest known prime number.
All you are doing is showing that there exist number bigger than HAVE
BEEN named, not bigger than CAN be named.
Try to name all. Fail
This becomes a problem of what we have seen, not of what is, so not a
property of the numbers themselves, but of the observer.
Exactly. Nevertheless no observer can see all numbers.
Regards, WM
On 1/19/2024 5:19 AM, WM wrote:
Le 18/01/2024 à 19:20, Jim Burns a écrit :
[...]
In order to excorzise the bijective meaning
I will henceforth use only ℵ,
meaning "infinitely many".
¬(𝒫(ℕ) ⇉ ℕ)
| Assume otherwise.
| Assume 𝒫(ℕ) ⇉ ℕ
| Exists f: 𝒫(ℕ) ⇉ ℕ
| ∀S ∈ 𝒫(ℕ): ∃k ∈ ℕ: k = f(S)
|
| However,
| consider D = {f(S) ∈ ℕ| S ∈ 𝒫(ℕ) ∧ f(S) ∉ S }
| f(D) ∈ D ∨ f(D) ∉ D
|ℝ| > |ℚ|
On 1/19/24 2:18 PM, WM wrote:
To presume it has the value 1 somewhere, presumes that there exists a
smallest unit fraction, which is a false assumption.
Not when mathematics is applied.
What mathematics?
BY YOUR DEFINITON, since there is an infinte number of unit fractions
below any finite x,
NUF(x) has an infinite value for all finite x
greater than 0.
There is no point where NF(x) can be 1, as that implies that there is a smallest unit fraction and thus a highest Natural Number, but ALL
Natural numbers, by definition, have a successor,
But why is it only a finite part that is visible?
Try to make more visible. Fail. Then you know it.
So, all Natural Numbers are visible, as we can find the individual
definition of any of them.
Try it.
I did. You just don't understand it.
You are implicitly assuming that there is a "highest" defined number, >>>>Not a constant number but only temporarily.
So not actually existing.
The number is actually existing like the biggest known prime number.
Nope, "Known" is different than "Existing".
All you are doing is showing that there exist number bigger than HAVE
BEEN named, not bigger than CAN be named.
Try to name all. Fail
Didn't say I could name ALL, I said I could name ANY.
Of course we can't name ALL of an unbounded set, as NAMING is a finite function.
After all, we already have a theory of (finite) and infinite cardinals in set theory.
Le 19/01/2024 à 20:53, Jim Burns a écrit :
On 1/19/2024 5:19 AM, WM wrote:
Le 18/01/2024 à 19:20, Jim Burns a écrit :
[...]
In order to excorzise the bijective meaning
I will henceforth use only ℵ,
meaning "infinitely many".
¬(𝒫(ℕ) ⇉ ℕ)
| Assume otherwise.
| Assume 𝒫(ℕ) ⇉ ℕ
| Exists f: 𝒫(ℕ) ⇉ ℕ
| ∀S ∈ 𝒫(ℕ): ∃k ∈ ℕ: k = f(S)
|
| However,
| consider D = {f(S) ∈ ℕ| S ∈ 𝒫(ℕ) ∧ f(S) ∉ S }
| f(D) ∈ D ∨ f(D) ∉ D
This proof presumes that infinite bijections exist and that all subsets
of ℕ are definable. Both is wrong. Same with Cantor's diagonal. It
assumes that all natnumbers were definable. That is wrong. Therefore
that is wrong:
|ℝ| > |ℚ|
Regards, WM
Le 19/01/2024 à 22:48, Richard Damon a écrit :
On 1/19/24 2:18 PM, WM wrote:
To presume it has the value 1 somewhere, presumes that there exists
a smallest unit fraction, which is a false assumption.
Not when mathematics is applied.
What mathematics?
∀n ∈ ℕ: 1/n - 1/(n+1) > 0
BY YOUR DEFINITON, since there is an infinte number of unit fractions
below any finite x,
Wrong premise.
NUF(x) has an infinite value for all finite x greater than 0.
Wrong result.
There is no point where NF(x) can be 1, as that implies that there is
a smallest unit fraction and thus a highest Natural Number, but ALL
Natural numbers, by definition, have a successor,
This definition is obtained from definble numbers and erroneously generatlized to all numbers.
But why is it only a finite part that is visible?
Try to make more visible. Fail. Then you know it.
So, all Natural Numbers are visible, as we can find the individual
definition of any of them.
Try it.
I did. You just don't understand it.
I don't accept your lies.
You are implicitly assuming that there is a "highest" defined number, >>>>>Not a constant number but only temporarily.
So not actually existing.
The number is actually existing like the biggest known prime number.
Nope, "Known" is different than "Existing".
All numbers are existing. Only few are visible.
All you are doing is showing that there exist number bigger than
HAVE BEEN named, not bigger than CAN be named.
Try to name all. Fail
Didn't say I could name ALL, I said I could name ANY.
Wrong. You can name only any which has ℵ successors. That is a big difference.
Of course we can't name ALL of an unbounded set, as NAMING is a finite
function.
You must let almost all remain unnamed.
Regards, WM
On 1/19/24 5:41 PM, WM wrote:
But what Natural Number or Unit Fraction isn't definable?
Wrong. You can name only any which has ℵ successors. That is a big
difference.
Which is all of them.
Of course we can't name ALL of an unbounded set, as NAMING is a finite
function.
You must let almost all remain unnamed.
Nope.
On 1/19/24 5:46 PM, WM wrote:
This proof presumes that infinite bijections exist and that all subsetsSo you think, but can not prove.
of ℕ are definable. Both is wrong. Same with Cantor's diagonal. It
assumes that all natnumbers were definable. That is wrong. Therefore
that is wrong:
Le 20/01/2024 à 00:18, Richard Damon a écrit :
On 1/19/24 5:46 PM, WM wrote:
This proof presumes that infinite bijections exist and that allSo you think, but can not prove.
subsets of ℕ are definable. Both is wrong. Same with Cantor's
diagonal. It assumes that all natnumbers were definable. That is
wrong. Therefore that is wrong:
Simple. The diagonal number has no last digit. Therefore it is undefined.
Regards, WM
On 1/20/24 6:08 AM, WM wrote:
Are you jumping to the Reals now?So you think, but can not prove.
Simple. The diagonal number has no last digit. Therefore it is undefined.
On 1/20/24 6:05 AM, WM wrote:
Le 20/01/2024 à 00:18, Richard Damon a écrit :
On 1/19/24 5:41 PM, WM wrote:The smallest ones. Those which, according to you, sit at infinitesimals
But what Natural Number or Unit Fraction isn't definable?
between 0 and (0, 1].
Wrong. You can name only any which has ℵ successors. That is a big
difference.
Which is all of them.
No, it is a small always finite part.
Nope.
Which ones, NAME THEM OR THEIR SET, don't
You must let almost all remain unnamed.
Nope.
Le 20/01/2024 à 13:29, Richard Damon a écrit :
On 1/20/24 6:05 AM, WM wrote:
Le 20/01/2024 à 00:18, Richard Damon a écrit :
On 1/19/24 5:41 PM, WM wrote:The smallest ones. Those which, according to you, sit at
But what Natural Number or Unit Fraction isn't definable?
infinitesimals between 0 and (0, 1].
Do you agree?
Wrong. You can name only any which has ℵ successors. That is a big >>>>> difference.
Which is all of them.
No, it is a small always finite part.
Nope.
Which ones, NAME THEM OR THEIR SET, don't
Those which, according to you, sit at infinitesimals between 0 and (0, 1].
Since you can't define what set you consider Visible, I can't tell whatYou must let almost all remain unnamed.
Nope.
Name one of them.
Regards, WM
Le 20/01/2024 à 13:29, Richard Damon a écrit :
On 1/20/24 6:08 AM, WM wrote:
Are you jumping to the Reals now?So you think, but can not prove.
Simple. The diagonal number has no last digit. Therefore it is
undefined.
I prove that the diagonal number is not defined.
Regards, WM
On 1/21/24 2:58 AM, WM wrote:
Yes, it seems that what you think of as your dark numbers (of unit
fractions) are actually the infinitesimals, but they don't have the non-definable property of your dark numbers.
They are dark to you only
Those which, according to you, sit at infinitesimals between 0 and (0, 1].
In other words, you are admitting that you "dark numbers" are just the transfinite numbers, the infintesimals between 0 and (0, and the
infinite number beyond the finites.
Le 21/01/2024 à 14:01, Richard Damon a écrit :
On 1/21/24 2:58 AM, WM wrote:
Yes, it seems that what you think of as your dark numbers (of unit
fractions) are actually the infinitesimals, but they don't have the
non-definable property of your dark numbers.
They cannot be distinguished.
They are dark to you only
Can you distinguish them?
Those which, according to you, sit at infinitesimals between 0 and
(0, 1].
In other words, you are admitting that you "dark numbers" are just the
transfinite numbers, the infintesimals between 0 and (0, and the
infinite number beyond the finites.
No. Unit fractions are not infinitesimals. But you raised this topic
because you cannot distinguish the smallest ℵ unit fractions.
Regards, WM
On 1/21/2024 5:03 AM, Richard Damon wrote:
On 1/21/24 2:59 AM, WM wrote:
Le 20/01/2024 à 13:29, Richard Damon a écrit :
On 1/20/24 6:08 AM, WM wrote:
Are you jumping to the Reals now?So you think, but can not prove.
Simple. The diagonal number has no last digit. Therefore it is
undefined.
I prove that the diagonal number is not defined.
Regards, WM
So, you think that 0.333.... isn't defined?
.(3) is how base 10 handles 1 divided by 3.
You need to get your definitions straight.
And the Reals need better logic than you have, since you can't even
handle the Natural Numbers.
:^)
On 1/21/24 2:59 AM, WM wrote:
I prove that the diagonal number is not defined.
So, you think that 0.333.... isn't defined?
There are various notations used, but it still has no last digit, as
that is just a notation for repeating.
On 1/21/24 11:52 AM, WM wrote:
Can you distinguish them?
I just did.
Admittedly, you need to adjust your definition of NUF(x) to allow
NUF(delta) == 1, but if you don't, then you just have to accept that
NUF(x), not taking into account the infintesimals, just jumps from 0 to
ℵ0 as it passes over the GAP caused by them.
Le 21/01/2024 à 18:52, Richard Damon a écrit :
On 1/21/24 11:52 AM, WM wrote:
Can you distinguish them?
I just did.
Admittedly, you need to adjust your definition of NUF(x) to allow
NUF(delta) == 1, but if you don't, then you just have to accept that
NUF(x), not taking into account the infintesimals, just jumps from 0
to ℵ0 as it passes over the GAP caused by them.
Fine. Say NUF(delta) == 1. But you cannot express delta in real numbers.
Regards, WM
Le 21/01/2024 à 21:45, Richard Damon a écrit :
There are various notations used, but it still has no last digit, as
that is just a notation for repeating.
If all digits are determined by a formula, there is no need for a last
digit. But if the digits follow upon each other by chance, the knowledge
of all of them would require the knowledge of the last one. But that is impossible because either the last one is not existing or it is dark.
Regards
On 1/22/24 3:05 AM, WM wrote:
Le 21/01/2024 à 18:52, Richard Damon a écrit :
On 1/21/24 11:52 AM, WM wrote:
Can you distinguish them?
I just did.
Admittedly, you need to adjust your definition of NUF(x) to allow
NUF(delta) == 1, but if you don't, then you just have to accept that
NUF(x), not taking into account the infintesimals, just jumps from 0
to ℵ0 as it passes over the GAP caused by them.
Fine. Say NUF(delta) == 1. But you cannot express delta in real numbers.
Right, because delta is transfinite.
The infinitesimals allow you to define your NUF(x) to be step-wise continuous, if that is what you need.
It still doesn't make any of the finite number "dark"
Again, your reasoning just doesn't take into account infinitte unbounded sets.
Your need for a smallest (last) unit fraction just is not supported by
the math.
To get that, you will need transfinite numbers, which are NOT part of
the finite number, so are not just "dark" finite numbers.
On 1/22/24 3:10 AM, WM wrote:
Le 21/01/2024 à 21:45, Richard Damon a écrit :
There are various notations used, but it still has no last digit, as
that is just a notation for repeating.
If all digits are determined by a formula, there is no need for a last
digit. But if the digits follow upon each other by chance, the knowledge
of all of them would require the knowledge of the last one. But that is
impossible because either the last one is not existing or it is dark.
but the diagonal arguement IS a "formula",
guess you agree that it defines a number.
Le 22/01/2024 à 13:59, Richard Damon a écrit :
On 1/22/24 3:05 AM, WM wrote:
Le 21/01/2024 à 18:52, Richard Damon a écrit :
On 1/21/24 11:52 AM, WM wrote:
Can you distinguish them?
I just did.
Admittedly, you need to adjust your definition of NUF(x) to allow
NUF(delta) == 1, but if you don't, then you just have to accept that
NUF(x), not taking into account the infintesimals, just jumps from 0
to ℵ0 as it passes over the GAP caused by them.
Fine. Say NUF(delta) == 1. But you cannot express delta in real numbers.
Right, because delta is transfinite.
The infinitesimals allow you to define your NUF(x) to be step-wise
continuous, if that is what you need.
It still doesn't make any of the finite number "dark"
Where NUF increases, there lies a unit fraction. All unit fractions are finite real numbers.
Again, your reasoning just doesn't take into account infinitte
unbounded sets.
Your need for a smallest (last) unit fraction just is not supported by
the math.
To get that, you will need transfinite numbers, which are NOT part of
the finite number, so are not just "dark" finite numbers.
Unit fractions are finite numbers. The first is not visible. It is dark.
Regards, WM
On 1/22/24 5:35 PM, WM wrote:
Where NUF increases, there lies a unit fraction. All unit fractions are
finite real numbers.
And with that definition, NUF just jumps from 0 to Aleph0 in moving from
0 to (0,1]
NUF(x) just jumps from 0 (at x = 0) to aleph_0 for each and every x > 0
It happens that WM formulated :
But we can construct a list which defines a number as its limit.
0.0
0.10
0.110
..
Replacing 0 by 1 we get the diagonal number 0.111... . In decimals it has the
limit 1/9. This limit is not in the list and not in the diagonal. But all
terms of the sequence approaching 1/9 are there - in the list and in the
diagonal. Cantor's diagonal argument fails.
Define failure. It looks to me like no such list can exist which covers
all real numbers without missing any.
Le 23/01/2024 à 05:00, Richard Damon a écrit :
On 1/22/24 5:35 PM, WM wrote:
Where NUF increases, there lies a unit fraction. All unit fractions
are finite real numbers.
And with that definition, NUF just jumps from 0 to Aleph0 in moving
from 0 to (0,1]
Impossible, because all unit fractions have distances from each other.
∀n ∈ ℕ: 1/n - 1/(n+1) is greater than the space between 0 and (0, 1].
Regards, WM
Which means there is no value that NUF(x) == 1.
On 1/23/24 6:36 AM, WM wrote:
Le 23/01/2024 à 05:00, Richard Damon a écrit :
On 1/22/24 5:35 PM, WM wrote:
Where NUF increases, there lies a unit fraction. All unit fractions
are finite real numbers.
And with that definition, NUF just jumps from 0 to Aleph0 in moving
from 0 to (0,1]
But that says nothing between 0 and (0,1]
Impossible, because all unit fractions have distances from each other.
∀n ∈ ℕ: 1/n - 1/(n+1) is greater than the space between 0 and (0, 1]. >>
Le 23/01/2024 à 14:12, Richard Damon a écrit :
On 1/23/24 6:36 AM, WM wrote:
Le 23/01/2024 à 05:00, Richard Damon a écrit :
On 1/22/24 5:35 PM, WM wrote:
Where NUF increases, there lies a unit fraction. All unit fractions
are finite real numbers.
And with that definition, NUF just jumps from 0 to Aleph0 in moving
from 0 to (0,1]
If so, you could not distinguish these unit fractions by their position
or number. But that is
But that says nothing between 0 and (0,1]
Impossible, because all unit fractions have distances from each other.
∀n ∈ ℕ: 1/n - 1/(n+1) is greater than the space between 0 and (0, 1]. >>>
There is nothing between 0 and (0, 1]. All unit fractions are elements
of (0, 1].
Regards, WM
On 1/23/24 11:50 AM, WM wrote:
Le 23/01/2024 à 14:12, Richard Damon a écrit :Nope. They just closer and closer and never end
Which means there is no value that NUF(x) == 1.Never two unit fractions sit at the same x. Therefore there is one first
unit fraction.
Let me riddle you a few questions,
Do you accept the principle of Induction?
If so, do you accept that 0 is definable/visible?
Do you accept that if the number n is definable/visible, then so will be
n+1?
If you accept these, then you have to accept that ALL Natural Numbers
are definable/visible by the necessary consequence of these properties.
If you don't accept the principle of Induction, you can't be using ZFC,
as the principle of Induction is provable from the axioms of ZFC,
∀ x ∈ (0, 1]: ∃^oo y ∈ {1/n : n e IN}: 0 < y < x (true)That implies that by increasing the interval (0, 1] by one point to [0, 1] leads to an infinity of points no longer fitting into the interval. This
and
∀ x ∈ [0, 1]: ∃^oo y ∈ {1/n : n e IN}: 0 < y < x (false).
On Tuesday, January 23, 2024 at 5:50:11 PM UTC+1, WM wrote:
there is one first [smallest] unit fraction.
Nein, there is NO first/smallest unit fraction,
On 1/23/24 11:53 AM, WM wrote:
Le 23/01/2024 à 14:12, Richard Damon a écrit :
On 1/23/24 6:36 AM, WM wrote:
Le 23/01/2024 à 05:00, Richard Damon a écrit :
On 1/22/24 5:35 PM, WM wrote:
Where NUF increases, there lies a unit fraction. All unit fractions >>>>>> are finite real numbers.
And with that definition, NUF just jumps from 0 to Aleph0 in moving
from 0 to (0,1]
If so, you could not distinguish these unit fractions by their position
or number. But that is
But you can. by counting from 1/1, the direction they are indexed.
But that says nothing between 0 and (0,1]
Impossible, because all unit fractions have distances from each other. >>>> ∀n ∈ ℕ: 1/n - 1/(n+1) is greater than the space between 0 and (0, 1].
There is nothing between 0 and (0, 1]. All unit fractions are elements
of (0, 1].
There are no unit fractions, but that doesn't mean there is nothing.
And your NUF(x) isn't defined there,
so nothing prevents it from jumping.
On Tuesday, January 23, 2024 at 12:36:15 PM UTC+1, WM wrote:
the space between 0 and (0, 1].There is no "space between 0 and (0, 1]".
Fritz Feldhase schrieb am Dienstag,
23. Januar 2024 um 18:50:09 UTC+1:
On Tuesday, January 23, 2024
at 12:36:15 PM UTC+1, WM wrote:
the space between 0 and (0, 1].
There is no "space between 0 and (0, 1]".
There is the empty set of real numbers,
and the empty set of transcendental numbers,
and the empty set of algebraic numbers,
and the empty set of rational numbers,
and the empty set of integers,
and the empty set of natnumbers,
and the empty set of even numbers,
and the empty set of prime numbers ---
quite a lot of ZF-objects.
Le 24/01/2024 à 04:02, Richard Damon a écrit :
On 1/23/24 11:53 AM, WM wrote:
Le 23/01/2024 à 14:12, Richard Damon a écrit :
On 1/23/24 6:36 AM, WM wrote:
Le 23/01/2024 à 05:00, Richard Damon a écrit :
On 1/22/24 5:35 PM, WM wrote:
Where NUF increases, there lies a unit fraction. All unit
fractions are finite real numbers.
And with that definition, NUF just jumps from 0 to Aleph0 in
moving from 0 to (0,1]
If so, you could not distinguish these unit fractions by their
position or number. But that is
But you can. by counting from 1/1, the direction they are indexed.
Not the smallest ℵ which you attributed to infinitesimals.
But that says nothing between 0 and (0,1]
Impossible, because all unit fractions have distances from each other. >>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) is greater than the space between 0 and (0, 1].
There is nothing between 0 and (0, 1]. All unit fractions are
elements of (0, 1].
There are no unit fractions, but that doesn't mean there is nothing.
Just this is meant. The set of elements between them is empty.
And your NUF(x) isn't defined there,
There is nothing. NUF is defined at every real point.
so nothing prevents it from jumping.
It cannot jump without meeting unit fractions.
Regards, WM
On Wednesday, January 24, 2024 at 5:05:36 PM UTC-8, Richard Damon wrote:
On 1/24/24 11:04 AM, WM wrote:
Fritz Feldhase schrieb am Mittwoch, 24. Januar 2024 um 11:00:51 UTC+1:Nope, only if you can count from infinity, which you can't.
On Tuesday, January 23, 2024 at 5:50:11 PM UTC+1, WM wrote:
there is one first [smallest] unit fraction.Nein, there is NO first/smallest unit fraction,
NUF(0) = 0 and NUF(1) = greater. Hence there is a point where the first
unit fraction must sit or where more than one must sit. That kind of logic >>> cannot be defeated by the fools of matheology.
Regards, WM
You need a "first" from that end, which just doesn't exist in the
formulation of Natural Numbers.
... "in the _usual_ formulation", ....
... Speaking for those for whom it's perfectly natural,
that one or the other and both of zero and infinity,
book-end the numbers, like those book-ends that are books.
... And that it's entirely understood that arithmetic in your theory,
is incomplete, at best.
See, this is totally easy, because it's of quite a better theory.
Or, you know, "INFINITY".
On 1/24/24 10:51 AM, WM wrote:
It [NUF] cannot jump without meeting unit fractions.
That is an incorrect assumption.
Can you try to actually PROVE it from a set of consistent axioms?
On 1/24/24 10:46 AM, WM wrote:
They end at zero. Therefore they begin above zero.
But there is no "begin" as the values are dense there.
Let me riddle you a few questions,
Do you accept the principle of Induction?
Yes, but it holds for the potentially infinite collection of visible
numbers.
It hold for ALL Natural Numbers.
Le 25/01/2024 à 02:05, Richard Damon a écrit :
On 1/24/24 10:51 AM, WM wrote:
It [NUF] cannot jump without meeting unit fractions.
That is an incorrect assumption.
Can you try to actually PROVE it from a set of consistent axioms?
It is so by definition. NUF counts unit fractions and cannot count where
none are present. And noweher more than one can be present.
Regards, WM
But we can't count from an unbounded end,
Just because you put words to something, doesn't mean you have a definition.
On 1/25/24 6:15 AM, WM wrote:
Le 25/01/2024 à 02:05, Richard Damon a écrit :
On 1/24/24 10:46 AM, WM wrote:
They end at zero. Therefore they begin above zero.
But there is no "begin" as the values are dense there.
They are not dense but all have distances.
Not the Reals or the Rationals.
On Wednesday, January 24, 2024
at 5:05:36 PM UTC-8, Richard Damon wrote:
You need a "first" from that end,
which just doesn't exist in
the formulation of Natural Numbers.
... "in the _usual_ formulation", ....
See, this is totally easy,
because it's of quite a better theory.
Or, you know, "INFINITY".
Le 25/01/2024 à 13:22, Richard Damon a écrit :
But we can't count from an unbounded end,
True, therefore this elements there are dark.
Just because you put words to something, doesn't mean you have a
definition.
Simple: Uncountable elements are dark.
Regards, WM
Le 25/01/2024 à 13:22, Richard Damon a écrit :
On 1/25/24 6:15 AM, WM wrote:
Le 25/01/2024 à 02:05, Richard Damon a écrit :
On 1/24/24 10:46 AM, WM wrote:
They end at zero. Therefore they begin above zero.
But there is no "begin" as the values are dense there.
They are not dense but all have distances.
Not the Reals or the Rationals.
We talk about unit fractions.
Regards, WM
WM formulated the question :
Le 25/01/2024 à 13:22, Richard Damon a écrit :
On 1/25/24 6:15 AM, WM wrote:
Le 25/01/2024 à 02:05, Richard Damon a écrit :
On 1/24/24 10:46 AM, WM wrote:
They end at zero. Therefore they begin above zero.
But there is no "begin" as the values are dense there.
They are not dense but all have distances.
Not the Reals or the Rationals.
We talk about unit fractions.
As embedded in the reals. You know what a real interval is?
WM formulated the question :
Le 25/01/2024 à 13:22, Richard Damon a écrit :
On 1/25/24 6:15 AM, WM wrote:
Le 25/01/2024 à 02:05, Richard Damon a écrit :
On 1/24/24 10:46 AM, WM wrote:
They end at zero. Therefore they begin above zero.
But there is no "begin" as the values are dense there.
They are not dense but all have distances.
Not the Reals or the Rationals.
We talk about unit fractions.
As embedded in the reals.
On 1/25/24 7:42 AM, WM wrote:
Le 25/01/2024 à 13:22, Richard Damon a écrit :
But we can't count from an unbounded end,
True, therefore this elements there are dark.
But the NUMBERS are all finite and thus defined.
It is the SET that is unbounded.
So, your "dark" numbers are numbers that can't be described as a set,
but only individually.
On 1/25/24 7:43 AM, WM wrote:
We talk about unit fractions.
And they are dense approaching 0.
Le 26/01/2024 à 00:49, Richard Damon a écrit :
On 1/25/24 7:43 AM, WM wrote:
We talk about unit fractions.
And they are dense approaching 0.
No. All have distances: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Regards, WM
Le 26/01/2024 à 00:48, Richard Damon a écrit :
On 1/25/24 7:42 AM, WM wrote:
Le 25/01/2024 à 13:22, Richard Damon a écrit :
But we can't count from an unbounded end,
True, therefore this elements there are dark.
But the NUMBERS are all finite and thus defined.
You just said the contrary. And even earlier: Thus NUF(x) with x finite, jumps, perhaps in domain of the infinitesimals between 0 and the bottom
of (0,1].
It is the SET that is unbounded.
NUF counts the unit fractions.
So, your "dark" numbers are numbers that can't be described as a set,
but only individually.
The other way round.
Regards, WM
Le 25/01/2024 à 14:06, FromTheRafters a écrit :
WM formulated the question :
Le 25/01/2024 à 13:22, Richard Damon a écrit :
On 1/25/24 6:15 AM, WM wrote:
Le 25/01/2024 à 02:05, Richard Damon a écrit :
On 1/24/24 10:46 AM, WM wrote:
They end at zero. Therefore they begin above zero.
But there is no "begin" as the values are dense there.
They are not dense but all have distances.
Not the Reals or the Rationals.
We talk about unit fractions.
As embedded in the reals.
Irrelevant for the statement to be wrong.
Regards, WM
On Friday, January 26, 2024 at 10:52:07 AM UTC+1, WM wrote:
Le 25/01/2024 à 15:57, Jim Burns a écrit :
∀x ∈ (0,1]: NUF(x) = ℵ₀
ℵ₀ unit fractions [...], do not fit between 0 and every x > 0.
Hint:
∀ x > 0: E^ℵ₀ u ∈ UF: 0 < u < x (true).
E^ℵ₀ u ∈ UF: ∀ x > 0: 0 < u < x (false).
The correct order of quantifiers is essential.
(Hint: We are dealing with two completely different claims here.)
On 1/26/24 5:07 AM, WM wrote:
I never said the distance goes becomes 0,
Below ANY unit fraction are more, so there is no smallest unit fraction.
And all those smaller are definable, so don't need to be "dark"
Point of note, you claim that there is only "One Mathematics", because
you seem to be a "Naturalist" for mathematics, using the properties
observed in ordinary math.
The problem with this method is that it can only define the mathematics
for the numbers we "see", and thus can't HAVE "dark" numbers.
Your "dark" numbers are just numbers that you haven't discovered the
actual properties that make them do what you see happening.
And the problem is that YOU apparently can't see the nature of unbounded sets, and thus you can't use them in this mathematics.
On 1/26/24 5:05 AM, WM wrote:
NUF counts the unit fractions.
And thus jumps to aleph0 for all x > 0 since there is no smallest unit fraction.
So, your "dark" numbers are numbers that can't be described as a set,
but only individually.
The other way round.
So you say, except that you have shown that you can't actually define
them as a set, but at least some of them have individual descriptions,
they are the points x where NUF(x) has the various finite values.
On 1/26/24 4:48 AM, WM wrote:
But you still don't understand that there is no smallest unit fraction,
so you can't start counting at it.
You talk about all have distance BETWEEN them, but "first" isn't between
two of them.
I have shown that for ANY value 1/n, there is room for at least n more
unit fractions below it, so your "finite gap" idea doesn't show there
must be an smallest, but that there CAN'T be a smallest.
You are just using bad logic, because you logic doesn't actually have anything to reason with.
Le 26/01/2024 à 13:32, Richard Damon a écrit :
On 1/26/24 5:05 AM, WM wrote:
NUF counts the unit fractions.
And thus jumps to aleph0 for all x > 0 since there is no smallest unit
fraction.
Wrong by mathematics. It cannot jump before (0, 1] because there are no
unit fractions and it can only jump by 1 because all unit fractions have distances.
So, your "dark" numbers are numbers that can't be described as a
set, but only individually.
The other way round.
So you say, except that you have shown that you can't actually define
them as a set, but at least some of them have individual descriptions,
they are the points x where NUF(x) has the various finite values.
These points have no descriptions and can never be described.
Regards, WM
On 1/26/2024 4:32 AM, Richard Damon wrote:
[...]
I hope you are taking the proper brain coolants when conversing with
WM... A circular firing squad comes to mind... ;^)
On 1/26/24 1:03 PM, WM wrote:
Le 26/01/2024 à 13:32, Richard Damon a écrit :
On 1/26/24 5:05 AM, WM wrote:
NUF counts the unit fractions.
And thus jumps to aleph0 for all x > 0 since there is no smallest unit
fraction.
Wrong by mathematics. It cannot jump before (0, 1] because there are no
unit fractions and it can only jump by 1 because all unit fractions have
distances.
WHy?
You can't actually get to infinity, as that doesn't exist in the
physical observable universe.
You can't get to "Dark" numbers, as those are just things beyond the
border of what you have explored. You can't get them by "Subtracting" a finite collection from the infinite set, as you can't actually define
that infinite set, it just is more than you can do.
Mathematics understands that there is no "biggest" numbers, or "Smallest Positive Number" as its math shows that given any number you can make a bigger one, or one closer to zero.
So, if you REALLY want to claim you are just using "Mathematics: and not
an axiometic Mathology, you need to stop making claims that it just
doesn't handle.
Now, if you want to work with Aleph or Omega, you need to accept that
you ARE using Axioms, and figure out which ones you are using, and stick
with it.
On 1/26/24 1:00 PM, WM wrote:
Le 26/01/2024 à 13:31, Richard Damon a écrit :
On 1/26/24 5:07 AM, WM wrote:
I never said the distance goes becomes 0,
Then NUF cannot increase by more than 1 at a point.
But where it increase isn't AT a point.
Below ANY unit fraction are more, so there is no smallest unit fraction.
If the function can't start, it cannot count.
Right, since ther is no spot to start, you can't actually count from
that end.
On 1/26/24 1:09 PM, WM wrote:
Le 26/01/2024 à 13:35, Richard Damon a écrit :
On 1/26/24 4:48 AM, WM wrote:
But you still don't understand that there is no smallest unit
fraction, so you can't start counting at it.
Real points on the real line must be existing.
Yes. But there is no "Smallest Positive"
Your N_def isn't defined universally.
Yes, If N_def is defined as the set of the initial sequence below a
GIVEN natural number, you get an infinite set above it, but that isn't
all the "definable" numbers, as n+1 is defin
You have never exhausted the set. But that is possible, collectively:
|ℕ \ {1, 2, 3, ...}| = 0
So.
You are just using bad logic, because you logic doesn't actually have
anything to reason with.
I am avoiding matheology.
No, you are not, you are USING it to talk about ℵo.
On 1/26/2024 5:22 PM, Richard Damon wrote:
On 1/26/24 8:09 PM, Chris M. Thomasson wrote:
On 1/26/2024 4:32 AM, Richard Damon wrote:
[...]
I hope you are taking the proper brain coolants when conversing with
WM... A circular firing squad comes to mind... ;^)
It provides useful practice for handling non-logical idiots.
Hummm... Touche! :^)
Le 27/01/2024 à 01:50, Richard Damon a écrit :
On 1/26/24 1:00 PM, WM wrote:
Le 26/01/2024 à 13:31, Richard Damon a écrit :
On 1/26/24 5:07 AM, WM wrote:
I never said the distance goes becomes 0,
Then NUF cannot increase by more than 1 at a point.
But where it increase isn't AT a point.
It can only increase at points 1/n.
Below ANY unit fraction are more, so there is no smallest unit
fraction.
If the function can't start, it cannot count.
Right, since ther is no spot to start, you can't actually count from
that end.
I can't but the points 1/n must be there.
Regards, WM
Le 27/01/2024 à 01:50, Richard Damon a écrit :
On 1/26/24 1:03 PM, WM wrote:
Le 26/01/2024 à 13:32, Richard Damon a écrit :
On 1/26/24 5:05 AM, WM wrote:
NUF counts the unit fractions.
And thus jumps to aleph0 for all x > 0 since there is no smallest
unit fraction.
Wrong by mathematics. It cannot jump before (0, 1] because there are
no unit fractions and it can only jump by 1 because all unit
fractions have distances.
WHy?
It cannot jump before (0, 1] because there are no unit fractions and it
can only jump by 1 because all unit fractions have distances.
You can't actually get to infinity, as that doesn't exist in the
physical observable universe.
But it can be assumed between every pair of definable real numbers.
You can't get to "Dark" numbers, as those are just things beyond the
border of what you have explored. You can't get them by "Subtracting"
a finite collection from the infinite set, as you can't actually
define that infinite set, it just is more than you can do.
Dark numbers can be manipulated collectively.
Mathematics understands that there is no "biggest" numbers, or
"Smallest Positive Number" as its math shows that given any number you
can make a bigger one, or one closer to zero.
So, if you REALLY want to claim you are just using "Mathematics: and
not an axiometic Mathology, you need to stop making claims that it
just doesn't handle.
Now, if you want to work with Aleph or Omega, you need to accept that
you ARE using Axioms, and figure out which ones you are using, and
stick with it.
I use only the numbers existing under the premise of actul infinity.
Regards, WM
On 1/27/24 6:07 AM, WM wrote:
It cannot jump before (0, 1] because there are no unit fractions and it
can only jump by 1 because all unit fractions have distances.
WHy can't it "Jump" since there is a boundry of domain that it crossed.
You can't actually get to infinity, as that doesn't exist in the
physical observable universe.
But it can be assumed between every pair of definable real numbers.
What "Infinity" can be assumed?
On 1/27/24 6:17 AM, WM wrote:
Yes, If N_def is defined as the set of the initial sequence below a
GIVEN natural number, you get an infinite set above it, but that isn't
all the "definable" numbers, as n+1 is defin
Hard to accept, but fact.
So you are stating it is a FACT that there is no actual finite set of "Defined" Natural Numbers, as any such proposed set leaves out a defined Natural number.
Good that you accept the "hard" fact that your ideas are based on
falsehoods.
I use ℵo or better ℵ only as an expression for infinitely many.
Then you should use the right symbol: ∞
You can't define a count from a point that doesn't eist.
Le 27/01/2024 à 16:25, Richard Damon a écrit :
On 1/27/24 6:07 AM, WM wrote:
It cannot jump before (0, 1] because there are no unit fractions and
it can only jump by 1 because all unit fractions have distances.
WHy can't it "Jump" since there is a boundry of domain that it crossed.
It cannot jump by more than 1 because all unit fraction have distances.
Only the unit fractions are counted.
You can't actually get to infinity, as that doesn't exist in the
physical observable universe.
But it can be assumed between every pair of definable real numbers.
What "Infinity" can be assumed?
Infinitely many dark points on the real axis can be assumed to exist.
Regards, WM
On Sunday, January 28, 2024 at 12:02:20 PM UTC+1, WM wrote:
The first count happens at the first unit fraction.
Hence a "first count" never happens.
Hint: There is no "first unit fraction".
On 1/28/24 6:12 AM, WM wrote:
But you don't have a starting point. You need a starting point to start counting.
We already agreed that the (definable) Real Numbers had no gaps.
Where are these "Dark" points?
On 1/28/24 6:06 AM, WM wrote:
That assumes that there IS a "First Unit Fraction" and that implies that there is a Highest Natural Number, which we know BY DEFINITION doesn't
exist. (for any number n, we can make a number n+1).
I use ℵo or better ℵ only as an expression for infinitely many.
Then you should use the right symbol: ∞
That expresses potential infinity.
What "logic" is telling you that?
∞ is the only "infinity" of natural mathematics, and it isn't even a number, just a concept.
On 1/28/24 6:02 AM, WM wrote:
Le 27/01/2024 à 16:26, Richard Damon a écrit :
You can't define a count from a point that doesn't eist.
The point 0 exists. There NUF(x) = 0. The first count happens at the
first unit fraction. It can only be one because of ∀n ∈ ℕ: 1/n - 1/(n+1)
0.
Which means your NUF(x) is only defined if there *IS* a first unit
fraction.
But any unit fraction that you might want to claim to be first, as
aleph0 unit fractions below it, so it wasn't the first.
Therefore, your function NUF(x) isn't actually definable by "counting".
Annd thus NUF(x) == 1 doesn't exist.
Le 28/01/2024 à 13:50, Richard Damon a écrit :
On 1/28/24 6:12 AM, WM wrote:
But you don't have a starting point. You need a starting point toStarting point is 0 with NUF(0) = 0.
start counting.
We already agreed that the (definable) Real Numbers had no gaps.
Therefore they have dark points between them.
Where are these "Dark" points?
The dark unit fractions are at 1/n, n ∈ ℕ.
The dark points between visible real numbers are the space between them.
Regards, WM
On 1/30/24 3:46 AM, WM wrote:
Le 28/01/2024 à 13:50, Richard Damon a écrit :
On 1/28/24 6:06 AM, WM wrote:
That assumes that there IS a "First Unit Fraction" and that implies
that there is a Highest Natural Number, which we know BY DEFINITION
doesn't exist. (for any number n, we can make a number n+1).
Therefore either this definition is wrong or actual infinity does not
exist (and with it the completeness of the set ℕ).
Why do you say actual infinity does not exist? And what do you mean by that?
On 1/30/24 3:41 AM, WM wrote:
Le 28/01/2024 à 13:50, Richard Damon a écrit :
On 1/28/24 6:12 AM, WM wrote:Starting point is 0 with NUF(0) = 0.
But you don't have a starting point. You need a starting point to
start counting.
But what is the first unit fraction?
The dark points between visible real numbers are the space between them.
So, they aren't real numbers,
On 1/30/24 3:49 AM, WM wrote:
Annd thus NUF(x) == 1 doesn't exist.
Then NUF(x) = 2 and higher numbers do not exist. Since NUF cannot jump
from 0 to more than 1 because all unit fractions are separate points,
the non-existence of actual infinity is proved.
Right, NUF(x) for any finite value of NUF, just doesn't exist.
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:41 AM, WM wrote:
Le 28/01/2024 à 13:50, Richard Damon a écrit :
On 1/28/24 6:12 AM, WM wrote:Starting point is 0 with NUF(0) = 0.
But you don't have a starting point. You need a starting point to
start counting.
But what is the first unit fraction?
It is dark but provably existing.
The dark points between visible real numbers are the space between them.
So, they aren't real numbers,
Many can be made visible. Then they are real real numbers.
Regards, WM
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:46 AM, WM wrote:
Le 28/01/2024 à 13:50, Richard Damon a écrit :
On 1/28/24 6:06 AM, WM wrote:
That assumes that there IS a "First Unit Fraction" and that implies
that there is a Highest Natural Number, which we know BY DEFINITION
doesn't exist. (for any number n, we can make a number n+1).
Therefore either this definition is wrong or actual infinity does not
exist (and with it the completeness of the set ℕ).
Why do you say actual infinity does not exist? And what do you mean by
that?
Between the smallest visible positive point x and 0 we can create x/2.
Either this was dark before, or there was nothing before.
Beyond the largest visible natural number n we can create 2n.
Either this was dark before, or there was nothing before.
Regards, WM
On 1/31/24 3:08 AM, WM wrote:
Why do you say actual infinity does not exist? And what do you mean by
that?
Between the smallest visible positive point x and 0 we can create x/2.
Either this was dark before, or there was nothing before.
But since it DID EXIST as a visible number
That or your "visible" isn't a property of the numbers, but of the
knowledge of the observer, and thus not a part of "math".
On 1/31/24 3:04 AM, WM wrote:
But what is the first unit fraction?
It is dark but provably existing.
Then prove it actually exists.
Note, to PROVE something, you are going to need to admit to the axioms
you are going to use as a basis of proof.
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
If they can be made visible, they weren't dark, or darkness isn't a
property of the actual number.
On 1/31/24 3:12 AM, WM wrote:
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:49 AM, WM wrote:
Annd thus NUF(x) == 1 doesn't exist.
Then NUF(x) = 2 and higher numbers do not exist. Since NUF cannot
jump from 0 to more than 1 because all unit fractions are separate
points, the non-existence of actual infinity is proved.
Right, NUF(x) for any finite value of NUF, just doesn't exist.
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
Then there are no gaps between the unit fractions, and ∀n ∈ ℕ: 1/n - >> 1/(n+1) > 0 is wrong. That nowever is impossible. Therefore only nothing
can exist between the smallest known unit fraction and zero. Then there
is no aleph.
You can't count what doesn't exist, and thus all you are concluding is
that there is no gaps between the unit fractions that don't exist.
Thats fine, they doen't exist.
Le 31/01/2024 à 13:42, Richard Damon a écrit :
On 1/31/24 3:04 AM, WM wrote:
But what is the first unit fraction?
It is dark but provably existing.
Then prove it actually exists.
Note, to PROVE something, you are going to need to admit to the axioms
you are going to use as a basis of proof.
This is the axiom: If some discrete points are on a line, then at both
sides there is a first one.
In other words, you are saying that there exist an n that there does notNUF(0) = 0,
NUF (x>0) > 0
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
Every axiom violating this conclusion is nonsense.
If they can be made visible, they weren't dark, or darkness isn't a
property of the actual number.
So it is.
Regards, WM
Le 31/01/2024 à 13:42, Richard Damon a écrit :
On 1/31/24 3:12 AM, WM wrote:
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:49 AM, WM wrote:
Annd thus NUF(x) == 1 doesn't exist.
Then NUF(x) = 2 and higher numbers do not exist. Since NUF cannot
jump from 0 to more than 1 because all unit fractions are separate
points, the non-existence of actual infinity is proved.
Right, NUF(x) for any finite value of NUF, just doesn't exist.
NUF(0) = 0,
NUF (x>0) > 0
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
Simplest logic. Too hard for you? Never mind. There are many who can't comprehend it.
Then there are no gaps between the unit fractions, and ∀n ∈ ℕ: 1/n - >>> 1/(n+1) > 0 is wrong. That nowever is impossible. Therefore only
nothing can exist between the smallest known unit fraction and zero.
Then there is no aleph.
You can't count what doesn't exist, and thus all you are concluding is
that there is no gaps between the unit fractions that don't exist.
Thats fine, they doen't exist.
Then they do not make up a complete set and cannot be counted.
Regards, WM
By your axiom since there are discrete point built by the Natural
Numbers, there must be an end on both sides, and thus there is a Highest Natural Number,
On 2/1/24 4:58 AM, WM wrote:
Le 31/01/2024 à 13:42, Richard Damon a écrit :
On 1/31/24 3:08 AM, WM wrote:
Why do you say actual infinity does not exist? And what do you mean
by that?
Between the smallest visible positive point x and 0 we can create x/2. >>>> Either this was dark before, or there was nothing before.
But since it DID EXIST as a visible number
The smallest one was x by assumption. To find it is already a hard task.
Then it is not hard to find x/2. But you cannot exhaust the remaining
space by halving in eternity. After potential infinity there is actual
darkness - or nothing.
And you are are just admitting that you logic is based on making
baseless assumptions.
The assumption of things that are not to prove that which is not true.
That or your "visible" isn't a property of the numbers, but of the
knowledge of the observer, and thus not a part of "math".
The knowlege of the observer is math. Why would people try to expand the
knowledge?
Nope. The rules of math exist outside of knowledge, and part of our goal
is to discover it.
On 2/1/24 5:02 AM, WM wrote:
Le 31/01/2024 à 13:42, Richard Damon a écrit :
On 1/31/24 3:12 AM, WM wrote:
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:49 AM, WM wrote:
Annd thus NUF(x) == 1 doesn't exist.
Then NUF(x) = 2 and higher numbers do not exist. Since NUF cannot
jump from 0 to more than 1 because all unit fractions are separate >>>>>> points, the non-existence of actual infinity is proved.
Right, NUF(x) for any finite value of NUF, just doesn't exist.
NUF(0) = 0,
NUF (x>0) > 0
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
Simplest logic. Too hard for you? Never mind. There are many who can't
comprehend it.
You don't get the third line from the previous two.
It isn't that they can't be counted, they don't exist because you logic
has axioms that prohibit unbounded sets
Le 01/02/2024 à 13:36, Richard Damon a écrit :
By your axiom since there are discrete point built by the Natural
Numbers, there must be an end on both sides, and thus there is a
Highest Natural Number,
By inverting them we can prove it:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 and ∀n ∈ ℕ: 1/n > 0.
NUF(0) = 0, NUF(x > 0) > 0.
Regards, WM
On 2/2/24 7:40 AM, WM wrote:
Le 01/02/2024 à 13:37, Richard Damon a écrit :
On 2/1/24 5:02 AM, WM wrote:
Le 31/01/2024 à 13:42, Richard Damon a écrit :
On 1/31/24 3:12 AM, WM wrote:
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:49 AM, WM wrote:
Right, NUF(x) for any finite value of NUF, just doesn't exist.Annd thus NUF(x) == 1 doesn't exist.
Then NUF(x) = 2 and higher numbers do not exist. Since NUF cannot >>>>>>>> jump from 0 to more than 1 because all unit fractions are
separate points, the non-existence of actual infinity is proved. >>>>>>
NUF(0) = 0,
NUF (x>0) > 0
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
Simplest logic. Too hard for you? Never mind. There are many who
can't comprehend it.
You don't get the third line from the previous two.
This was assumed too: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Which says that there is no smallest unit fraction as there is always
room for 1/(n+1) below 1/n.
It isn't that they can't be counted, they don't exist because you
logic has axioms that prohibit unbounded sets
Infinitely many exist. They don't sit at one point.
Right.
But there also isn't a smallest, which you are trying to start
your count at.
On 2/2/24 7:34 AM, WM wrote:
Le 01/02/2024 à 13:36, Richard Damon a écrit :
By your axiom since there are discrete point built by the Natural
Numbers, there must be an end on both sides, and thus there is a
Highest Natural Number,
By inverting them we can prove it:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 and ∀n ∈ ℕ: 1/n > 0.
NUF(0) = 0, NUF(x > 0) > 0.
It doesn't prove that NUF(x) needs to have the value of 1 anywhere.
Le 02/02/2024 à 15:33, Richard Damon a écrit :
On 2/2/24 7:34 AM, WM wrote:
Le 01/02/2024 à 13:36, Richard Damon a écrit :
By your axiom since there are discrete point built by the Natural
Numbers, there must be an end on both sides, and thus there is a
Highest Natural Number,
By inverting them we can prove it:
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 and ∀n ∈ ℕ: 1/n > 0.
NUF(0) = 0, NUF(x > 0) > 0.
It doesn't prove that NUF(x) needs to have the value of 1 anywhere.
What is the alternative?
More than one is impossible.
Regards, WM
Le 02/02/2024 à 15:33, Richard Damon a écrit :
On 2/2/24 7:40 AM, WM wrote:
Le 01/02/2024 à 13:37, Richard Damon a écrit :
On 2/1/24 5:02 AM, WM wrote:
Le 31/01/2024 à 13:42, Richard Damon a écrit :
On 1/31/24 3:12 AM, WM wrote:
Le 30/01/2024 à 13:50, Richard Damon a écrit :
On 1/30/24 3:49 AM, WM wrote:
Annd thus NUF(x) == 1 doesn't exist.
Then NUF(x) = 2 and higher numbers do not exist. Since NUF
cannot jump from 0 to more than 1 because all unit fractions >>>>>>>>> are separate points, the non-existence of actual infinity is >>>>>>>>> proved.
Right, NUF(x) for any finite value of NUF, just doesn't exist.
NUF(0) = 0,
NUF (x>0) > 0
∃ 1/n (0, 1]: ~∃ 1/m < 1/n.
Simplest logic. Too hard for you? Never mind. There are many who
can't comprehend it.
You don't get the third line from the previous two.
This was assumed too: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0
Which says that there is no smallest unit fraction as there is always
room for 1/(n+1) below 1/n.
No, it does not say that there is always an n+1.
It isn't that they can't be counted, they don't exist because you
logic has axioms that prohibit unbounded sets
Infinitely many exist. They don't sit at one point.
Right.
Then there is a first one.
But there also isn't a smallest, which you are trying to start your
count at.
Then there is no actual infinity.
Regards, WM
It just jumps to infinity.
Why is "More than one" impossible?
On 2/3/24 6:12 AM, WM wrote:
That is a problem with your NUF(x), to define it, you must get to trans-finite mathematics,
On 2/2/2024 6:33 AM, Richard Damon wrote:
Your logic claims there must be a Highest Natural Number and smallest
Unit Fraction.
The Definition of Natural Numbers say there is NOT a Highest Natural
Number, and thus not a smallest Unit Fraction.
So, your darkness is just you putting blinders on yourself.
If WM _really_ thinks there is a highest natural
number, and WM is actually a teacher, it should be fired right now!
Le 03/02/2024 à 12:52, Richard Damon a écrit :
It just jumps to infinity.
Why is "More than one" impossible?
Because 1/n - 1/(n+1) > 0.
Regards, WM
On 2/4/24 3:38 AM, WM wrote:
Le 03/02/2024 à 12:59, Richard Damon a écrit :
On 2/3/24 6:12 AM, WM wrote:
That is a problem with your NUF(x), to define it, you must get to
trans-finite mathematics,
No, every counted unit fraction is finite.
Yes, every counted unit fraction is finite, but you can't start counted
from the unbounded end with finite numbers.
On 2/4/24 3:37 AM, WM wrote:
Le 03/02/2024 à 12:52, Richard Damon a écrit :
It just jumps to infinity.
How does that follow?
Why is "More than one" impossible?
Because 1/n - 1/(n+1) > 0.
since 1/n - 1/(n+1) > 0,
So, for any point 1/n, there is a smaller point 1/(n+1) below it,
So there is no point in the Finite Numbers for NUF(x) to be 1, so it has
to jump if we are only looking at Finite Numbers.
You just don't understand how logic works, and are getting "darkness" as
a result of not being able to look at thing that are clearly there.
Le 04/02/2024 à 13:48, Richard Damon a écrit :
On 2/4/24 3:38 AM, WM wrote:
Le 03/02/2024 à 12:59, Richard Damon a écrit :
On 2/3/24 6:12 AM, WM wrote:
That is a problem with your NUF(x), to define it, you must get to
trans-finite mathematics,
No, every counted unit fraction is finite.
Yes, every counted unit fraction is finite, but you can't start
counted from the unbounded end with finite numbers.
So they are dark.
Regards, WM
Le 04/02/2024 à 13:48, Richard Damon a écrit :
On 2/4/24 3:37 AM, WM wrote:
Le 03/02/2024 à 12:52, Richard Damon a écrit :
It just jumps to infinity.
That is true for visible unit fractions. But it means that ℵ unit
fractions of this jump cannot be distinguished.
How does that follow?
Why is "More than one" impossible?
Because 1/n - 1/(n+1) > 0.
Don't be so stupid.
since 1/n - 1/(n+1) > 0,
So, for any point 1/n, there is a smaller point 1/(n+1) below it,
For any chosen point 1/n are ℵ smaller unit fractions below it. That
means any attempt to distinguish them must fail.
So there is no point in the Finite Numbers for NUF(x) to be 1, so it
has to jump if we are only looking at Finite Numbers.
This junmp cannot be further analyzed.
You just don't understand how logic works, and are getting "darkness"
as a result of not being able to look at thing that are clearly there.
If you could look at the unit fractions in the jump, then there was no
jump. But you can't. Every attempt leaves ℵ unit fractions not distingusihed. Yor claim that all could be distinguished or chosen fails.
Regards, WM
On 2/5/2024 4:51 AM, Richard Damon wrote:
WM is trying to tell us that
there is a smallest unit fraction...
wow.
Trying to do logic on things that don't exist
can create all sorts of problems.
On Monday, February 5, 2024 at 11:57:00 AM UTC-8, Jim Burns wrote:
WM's proof that the smallest unit.fraction exists is that it must
exist.
That it has contradictory properties is proof that it is a
non.unit.fraction unit.fraction.
Inverse infinity is a smallest fraction.
Calculus believes in that infinitesimal math.
Math is before man. That makes it God.. Not the mathematician playingMathematics actually _owes_ physics the mathematics of infinity.Math had infinity before human physics... ;^)
Which is very large in numbers, ....
God.
On 2/5/24 4:07 AM, WM wrote:
Le 04/02/2024 à 13:48, Richard Damon a écrit :
On 2/4/24 3:37 AM, WM wrote:
Le 03/02/2024 à 12:52, Richard Damon a écrit :
It just jumps to infinity.
That is true for visible unit fractions. But it means that ℵ unit
fractions of this jump cannot be distinguished.
Nope, NUF(x) just jumps because it is ill-defined, to count from an end
that doesn't have an ed.
How does that follow?Why is "More than one" impossible?
Because 1/n - 1/(n+1) > 0.
Don't be so stupid.
Really? You can't actually SHOW how you get there?
Your problem is that for you to actually SHOW how to prove your claim,
you will need to make explicit the assumption (aka axioms) that you are using, and those appear to be so obviously flawed (at least to someone
who understands what infinity actually is) so you need to hide them.
For any chosen point 1/n are ℵ smaller unit fractions below it. That
means any attempt to distinguish them must fail.
WHy?
You don't think we can distinguish an infinite number of points?
We have an infinite number of names to give them.
So there is no point in the Finite Numbers for NUF(x) to be 1, so it
has to jump if we are only looking at Finite Numbers.
This junmp cannot be further analyzed.
Maybe, with your logic, but doesn't mean it can't happen.
Nope. You are trying to distingush something that doesn't exist, that
first unit fraction.
On 2/5/24 4:08 AM, WM wrote:
Yes, every counted unit fraction is finite, but you can't start
counted from the unbounded end with finite numbers.
So they are dark.
So the first from the "end" without and "end" doesn't exsit.
Le 05/02/2024 à 13:50, Richard Damon a écrit :
On 2/5/24 4:08 AM, WM wrote:
Yes, every counted unit fraction is finite, but you can't start
counted from the unbounded end with finite numbers.
So they are dark.
So the first from the "end" without and "end" doesn't exsit.
For every eps > 0 there exist ℵ unit fractions in (0, eps), don't they?
But ℵ of which cannot be distinguished by any means, can they.
Regards, WM
Le 05/02/2024 à 13:51, Richard Damon a écrit :
On 2/5/24 4:07 AM, WM wrote:
Le 04/02/2024 à 13:48, Richard Damon a écrit :
On 2/4/24 3:37 AM, WM wrote:
Le 03/02/2024 à 12:52, Richard Damon a écrit :
It just jumps to infinity.
That is true for visible unit fractions. But it means that ℵ unit
fractions of this jump cannot be distinguished.
Nope, NUF(x) just jumps because it is ill-defined, to count from an
end that doesn't have an ed.
Since 0 is below the end, there is an end.
Where do the ℵ unit fractions sit in your opinion?
How does that follow?Why is "More than one" impossible?
Because 1/n - 1/(n+1) > 0.
Don't be so stupid.
Really? You can't actually SHOW how you get there?
Don't be so stupid.
Your problem is that for you to actually SHOW how to prove your claim,
you will need to make explicit the assumption (aka axioms) that you
are using, and those appear to be so obviously flawed (at least to
someone who understands what infinity actually is) so you need to hide
them.
This is obviously not flawed and shown explicitly: 1/n - 1/(n+1) > 0
For any chosen point 1/n are ℵ smaller unit fractions below it. That
means any attempt to distinguish them must fail.
WHy?
You don't think we can distinguish an infinite number of points?
We have an infinite number of names to give them.
Do it, such thzat none remains without name.
So there is no point in the Finite Numbers for NUF(x) to be 1, so it
has to jump if we are only looking at Finite Numbers.
This junmp cannot be further analyzed.
Maybe, with your logic, but doesn't mean it can't happen.
Then do it with your logic. How are the ℵ unit fractuions distributed?
Nope. You are trying to distingush something that doesn't exist, that
first unit fraction.
ℵ unit fractions exist in (0, eps). They cannot be distinguished by any eps. How else could it be done?
Regards, WM
On 2/7/24 2:15 AM, WM wrote:
Le 05/02/2024 à 13:50, Richard Damon a écrit :
On 2/5/24 4:08 AM, WM wrote:
Yes, every counted unit fraction is finite, but you can't start
counted from the unbounded end with finite numbers.
So they are dark.
So the first from the "end" without and "end" doesn't exsit.
For every eps > 0 there exist ℵ unit fractions in (0, eps), don't they?
But ℵ of which cannot be distinguished by any means, can they.
YOU may not be able to distinguish them, because you close your eyes to
those definition, but, as I have shown, I can distinguish them.
On Wednesday, February 7, 2024 at 8:21:56 AM UTC+1, WM wrote:
Since 0 is below the end, there is an end.Ein wunderbares Argument, Mückenheim!
Vielleicht ein klein wenig Zirkulär, aber was macht das schon?
| Weil Batman schwächer ist als Superman, gibt es Superman.
Ja, schon klar, Mückenheim. So ist es!
You don't think we can distinguish an infinite number of points? [RD]
We have an infinite number of names to give them. [RD]
Do it, such that none remains without name.
Nun ja, man macht das natürlich nicht VON HAND,
Hier eine kleine Hilfestellung: Sei WM ein BELIEBIGER Stammbruch.
Le 07/02/2024 à 13:29, Richard Damon a écrit :
On 2/7/24 2:15 AM, WM wrote:
Le 05/02/2024 à 13:50, Richard Damon a écrit :
On 2/5/24 4:08 AM, WM wrote:
Yes, every counted unit fraction is finite, but you can't start
counted from the unbounded end with finite numbers.
So they are dark.
So the first from the "end" without and "end" doesn't exsit.
For every eps > 0 there exist ℵ unit fractions in (0, eps), don't
they? But ℵ of which cannot be distinguished by any means, can they.
YOU may not be able to distinguish them, because you close your eyes
to those definition, but, as I have shown, I can distinguish them.
By what eps have you distinguished all?
Rehards, WM
On 2/8/24 6:32 AM, WM wrote:
Le 07/02/2024 à 13:29, Richard Damon a écrit :
On 2/7/24 2:15 AM, WM wrote:
Le 05/02/2024 à 13:50, Richard Damon a écrit :
On 2/5/24 4:08 AM, WM wrote:
Yes, every counted unit fraction is finite, but you can't start
counted from the unbounded end with finite numbers.
So they are dark.
So the first from the "end" without and "end" doesn't exsit.
For every eps > 0 there exist ℵ unit fractions in (0, eps), don't
they? But ℵ of which cannot be distinguished by any means, can they.
YOU may not be able to distinguish them, because you close your eyes
to those definition, but, as I have shown, I can distinguish them.
By what eps have you distinguished all?
That's a non-sense question,
as eps doesn't "distinguish" points at all.
And there doesn't need to be a finite eps that is below all the unit fractions, in fact, there CAN'T Be.
Le 08/02/2024 à 13:35, Richard Damon a écrit :
On 2/8/24 6:32 AM, WM wrote:
Le 07/02/2024 à 13:29, Richard Damon a écrit :
On 2/7/24 2:15 AM, WM wrote:
Le 05/02/2024 à 13:50, Richard Damon a écrit :YOU may not be able to distinguish them, because you close your eyes
On 2/5/24 4:08 AM, WM wrote:
Yes, every counted unit fraction is finite, but you can't start >>>>>>>> counted from the unbounded end with finite numbers.
So they are dark.
So the first from the "end" without and "end" doesn't exsit.
For every eps > 0 there exist ℵ unit fractions in (0, eps), don't
they? But ℵ of which cannot be distinguished by any means, can they. >>>
to those definition, but, as I have shown, I can distinguish them.
By what eps have you distinguished all?
That's a non-sense question,
No.
as eps doesn't "distinguish" points at all.
eps = 4/10 distinguishes 1/2 and 1/3 because 1/3 < 4/10 < 1/2.
And there doesn't need to be a finite eps that is below all the unit
fractions, in fact, there CAN'T Be.
So it is. But for every distinguishable unit fraction there is a smaller
eps.
Regards, WM
On 2/8/24 7:50 AM, WM wrote:
YOU may not be able to distinguish them, because you close your eyes >>>>> to those definition, but, as I have shown, I can distinguish them.
By what eps have you distinguished all?
That's a non-sense question,
No.
Then what does it actually MEAN, and why do you think such a thing needs
to exist.
as eps doesn't "distinguish" points at all.
eps = 4/10 distinguishes 1/2 and 1/3 because 1/3 < 4/10 < 1/2.
And why is that useful or important.
for every distinguishable unit fraction there is a smaller
eps.
That just means that for every unit fraction, there is a smaller unit fraction, so there is no lowest unit fractioin.
Le 08/02/2024 à 13:59, Richard Damon a écrit :
On 2/8/24 7:50 AM, WM wrote:
YOU may not be able to distinguish them, because you close your
eyes to those definition, but, as I have shown, I can distinguish
them.
By what eps have you distinguished all?
That's a non-sense question,
No.
Then what does it actually MEAN, and why do you think such a thing
needs to exist.
Read Borel's nice little book: Les nombres inaccessible:
"This little book is the result of half a century of reflections on the principles of mathematical analysis and, in particular, on the definition
of numbers. Some of these reflections have already been sketched
here and there in the works of this Collection, but it seemed to me
that it would be useful to coordinate them in a connected account."
Borel's thesis is that the overwhelming majority of numbers will
always remain inaccessible to the human race as we know it, in the
sense that it will never be possible to define these numbers effectively
in such a manner that any two mathematicians will be certain that
they are speaking about one and the same entity. (F. Bagemihl)
as eps doesn't "distinguish" points at all.
eps = 4/10 distinguishes 1/2 and 1/3 because 1/3 < 4/10 < 1/2.
And why is that useful or important.
First in order to show that you are wrong.
Second because every visible unit fractions, a smaller eps can be
defined, but not for dark unit fractions.
for every distinguishable unit fraction there is a smaller eps.
That just means that for every unit fraction, there is a smaller unit
fraction, so there is no lowest unit fractioin.
But almost all of them cannot be defined in such a manner that any two mathematicians will be certain that
they are speaking about one and the same entity.
Regards, WM
On 2/8/24 9:37 AM, WM wrote:
Borel's thesis is that the overwhelming majority of numbers will
always remain inaccessible to the human race as we know it, in the
sense that it will never be possible to define these numbers effectively
in such a manner that any two mathematicians will be certain that
they are speaking about one and the same entity. (F. Bagemihl)
But that is a limit of PEOPLE, not the numbers.
All the numbers are "Definable" even if no person will ever write its name.
Using these names "any two mathematicians may be certain that they are speaking
about one and the same entity."
Le 09/02/2024 à 04:40, Richard Damon a écrit :
On 2/8/24 9:37 AM, WM wrote:
Borel's thesis is that the overwhelming majority of numbers will
always remain inaccessible to the human race as we know it, in the
sense that it will never be possible to define these numbers effectively >>> in such a manner that any two mathematicians will be certain that
they are speaking about one and the same entity. (F. Bagemihl)
But that is a limit of PEOPLE, not the numbers.
Not only. The ℵ smallest *existing* unit fractions will remain undefined forever.
All the numbers are "Definable" even if no person will ever write its
name.
All means without exception. That is impossible, because whatever will
be defined, it fails to define almost all unit fractions.
Regards, WM
On 2/9/24 4:53 AM, WM wrote:
Le 09/02/2024 à 04:40, Richard Damon a écrit :
On 2/8/24 9:37 AM, WM wrote:
Borel's thesis is that the overwhelming majority of numbers will
always remain inaccessible to the human race as we know it, in the
sense that it will never be possible to define these numbers effectively >>>> in such a manner that any two mathematicians will be certain that
they are speaking about one and the same entity. (F. Bagemihl)
But that is a limit of PEOPLE, not the numbers.
Not only. The ℵ smallest *existing* unit fractions will remain undefined >> forever.
Ah, so you think that the existance and definition of numbers is
dependent on us?
On 02/09/2024 05:52 AM, Fritz Feldhase wrote:
Assume that a mathematician tells me that he's thinking of a certain unit
fraction and that the name of this unit fraction is X. Then all I have to do is to
count the strokes (|) occurring after "|/" in X. Say the number of these strokes is
n. Then I know that this mathematician is thinking of the unit fraction 1/n.
On Friday, February 9, 2024 at 11:10:53 AM UTC+1, WM wrote:
Fritz Feldhase schrieb am Freitag, 9. Februar 2024 um 10:17:40 UTC+1:
Name a unit fraction that has less than infinitely many smaller ones.
I cannot "name" something which does not exist.
On Friday, February 9, 2024 at 10:53:40 AM UTC+1, WM wrote:
The ℵ smallest [...] unit fractions will remain undefined forever.
Since the notion "the ℵ smallest [...] unit fractions" is not "well-defined"
(i.e. not defined at all), no big problem.
Hint: Does 1/1 belong to "the ℵ smallest [...] unit fractions"? 1/2? 1/3? Which is the largest unit fraction which belongs to "the ℵ smallest [...] unit
fractions"?
All the numbers are "Definable" even if no person will ever write its name.
All means without exception.Oh really?! How would you know?
Le 10/02/2024 à 00:36, Richard Damon a écrit :
On 2/9/24 4:53 AM, WM wrote:
Le 09/02/2024 à 04:40, Richard Damon a écrit :
On 2/8/24 9:37 AM, WM wrote:
Borel's thesis is that the overwhelming majority of numbers will
always remain inaccessible to the human race as we know it, in the
sense that it will never be possible to define these numbers
effectively
in such a manner that any two mathematicians will be certain that
they are speaking about one and the same entity. (F. Bagemihl)
But that is a limit of PEOPLE, not the numbers.
Not only. The ℵ smallest *existing* unit fractions will remain
undefined forever.
Ah, so you think that the existance and definition of numbers is
dependent on us?
No, the infinite cannot be traversed. Therefore almost all remains dark.
Regards, WM
On 2/10/24 1:09 PM, WM wrote:
the infinite cannot be traversed. Therefore almost all remains dark.So, you start by assuming there can not be an infinite that we can understand, and them defining that something is infinite.
Why do you say that the infinite cannot be "traversed",
Le 10/02/2024 à 21:34, Richard Damon a écrit :
On 2/10/24 1:09 PM, WM wrote:
the infinite cannot be traversed. Therefore almost all remains dark.So, you start by assuming there can not be an infinite that we can
understand, and them defining that something is infinite.
No, you cannot understand what I said. The infinite can be understood
(by intelligent persons), but cannot be traversed step by step. Step by
step you will always have ℵ steps in front of you:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ
Only collectively, summing up all the dark steps you can get through:
|ℕ \ {1, 2, 3, ...}| = 0
Why do you say that the infinite cannot be "traversed",
Because it is true.
Regards, WM
On 2/11/24 4:44 AM, WM wrote:
Why do you say that the infinite cannot be "traversed",
Because it is true.
And why do you need to travese it "step by step"?
That restriction is the restriction of finiteness, and doesn't allow for
the infinite.
So, your "Darkness", is just the
Le 11/02/2024 à 13:47, Richard Damon a écrit :
On 2/11/24 4:44 AM, WM wrote:
Why do you say that the infinite cannot be "traversed",
Because it is true.
And why do you need to travese it "step by step"?
That would be possible, if every element could be visited.
That restriction is the restriction of finiteness, and doesn't allow
for the infinite.
So, your "Darkness", is just the
recognition that infinite sets consist of dark elements.
Regards, WM
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