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Evidence for dark natural numbers
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WM
Jan 10, 2020, 5:59:04 AM1/10/20
to
1) Every bijection of |N and another countable set M is defined to happen in one single step. It is impossible to count so far that nothing remains. But it is possible for every definable natural number to count to the pair (n, m).
2) Every definable natural number n is the last number of a FISON (1, 2, 3, ..., n). But no FISON contains an infinite set of natnumbers. The sum of a FISON is n(n+1)/2 and as such well defined. But the sum of all natural numbers is much larger and remains undefined.
3) Every definable endsegment is infinite and is the last element of an intersection of endsegments
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo .
The intersection of all endsegments however is empty
∩{E(1), E(2), E(3), ...} = { }
but not any single definable endsegment can be found to be necessary in
∩{E(x), E(x+1), E(x+2), ...} = { } .
Further the basic rule
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
yields the function of cardinalities
f(n+1) = f(n) - 1
and shows that, if all endsegments were definable, then the empty set could not be realized without preceding finite sets. The function f defined for all natural numbers has values larger than 2 and 0 and cannot exist without the values 2 and 1 existing in the codomain.
Finally there are countably many definitions and uncountably many real numbers, such that most real numbers do not have a definition (infinite "definitions" are blatant nonsense) although all real numbers can be well-ordered in a linear ordering (every non-empty set has a least element). Why should all natural numbers be definable?
Regards, WM
2) Every definable natural number n is the last number of a FISON (1, 2, 3, ..., n). But no FISON contains an infinite set of natnumbers. The sum of a FISON is n(n+1)/2 and as such well defined. But the sum of all natural numbers is much larger and remains undefined.
3) Every definable endsegment is infinite and is the last element of an intersection of endsegments
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo .
The intersection of all endsegments however is empty
∩{E(1), E(2), E(3), ...} = { }
but not any single definable endsegment can be found to be necessary in
∩{E(x), E(x+1), E(x+2), ...} = { } .
Further the basic rule
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
yields the function of cardinalities
f(n+1) = f(n) - 1
and shows that, if all endsegments were definable, then the empty set could not be realized without preceding finite sets. The function f defined for all natural numbers has values larger than 2 and 0 and cannot exist without the values 2 and 1 existing in the codomain.
Finally there are countably many definitions and uncountably many real numbers, such that most real numbers do not have a definition (infinite "definitions" are blatant nonsense) although all real numbers can be well-ordered in a linear ordering (every non-empty set has a least element). Why should all natural numbers be definable?
Regards, WM
Mostowski Collapse
Jan 10, 2020, 6:43:29 AM1/10/20
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Ratatosk set theory has clearly dark numbers.
It doesn't have union, only equality. You
cant say:
{1} U {1, 2} = {1, 2}
But you could say:
{1} =/= {1, 2}
It was discovered in 2020 by Rossi Robot
AI. Dark numbers usually appear when you
nevertheless use union,
like {1} U {1, 2}. Researchers are still
not clear why it then can happen that
{1} U {1, 2} = {1, 2},
they assume there is connex between
Ratatosk set theory and Skolem set theory.
A lot of conjectures are
related to the space filling curve. Mostlikely
such a curve only fills 75% of the space,
and 25% is left with dark matter.
It doesn't have union, only equality. You
cant say:
{1} U {1, 2} = {1, 2}
But you could say:
{1} =/= {1, 2}
It was discovered in 2020 by Rossi Robot
AI. Dark numbers usually appear when you
nevertheless use union,
like {1} U {1, 2}. Researchers are still
not clear why it then can happen that
{1} U {1, 2} = {1, 2},
they assume there is connex between
Ratatosk set theory and Skolem set theory.
A lot of conjectures are
related to the space filling curve. Mostlikely
such a curve only fills 75% of the space,
and 25% is left with dark matter.
Zelos Malum
Jan 14, 2020, 2:20:03 AM1/14/20
to
>1) Every bijection of |N and another countable set M is defined to happen in one single step. It is impossible to count so far that nothing remains. But it is possible for every definable natural number to count to the pair (n, m).
We are not counting anything you fuckign imbecile.
>2) Every definable natural number n is the last number of a FISON (1, 2, 3, ..., n). But no FISON contains an infinite set of natnumbers. The sum of a FISON is n(n+1)/2 and as such well defined. But the sum of all natural numbers is much larger and remains undefined.
>and shows that, if all endsegments were definable, then the empty set could not be realized without preceding finite sets. The function f defined for all natural numbers has values larger than 2 and 0 and cannot exist without the values 2 and 1 existing in the codomain.
Ganzhinterseher
Jan 14, 2020, 10:37:07 AM1/14/20
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Am Dienstag, 14. Januar 2020 08:20:03 UTC+1 schrieb Zelos Malum:
> >1) Every bijection of |N and another countable set M is defined to happen in one single step. It is impossible to count so far that nothing remains. But it is possible for every definable natural number to count to the pair (n, m).
>
> We are not counting anything
That is irrelevant. I can count to every definable number - and so can every mathematician.
> >1) Every bijection of |N and another countable set M is defined to happen in one single step. It is impossible to count so far that nothing remains. But it is possible for every definable natural number to count to the pair (n, m).
>
> We are not counting anything
>
> >2) Every definable natural number n is the last number of a FISON (1, 2, 3, ..., n). But no FISON contains an infinite set of natnumbers. The sum of a FISON is n(n+1)/2 and as such well defined. But the sum of all natural numbers is much larger and remains undefined.
>
> The fact that the sum is a natural number does not in anyway shape or form determine if somethign is well-defined.
The question is whether there exist natural numbers which cannot bne summed. This is the case because not all can be summed.
> >2) Every definable natural number n is the last number of a FISON (1, 2, 3, ..., n). But no FISON contains an infinite set of natnumbers. The sum of a FISON is n(n+1)/2 and as such well defined. But the sum of all natural numbers is much larger and remains undefined.
>
> The fact that the sum is a natural number does not in anyway shape or form determine if somethign is well-defined.
>
> >and shows that, if all endsegments were definable, then the empty set could not be realized without preceding finite sets. The function f defined for all natural numbers has values larger than 2 and 0 and cannot exist without the values 2 and 1 existing in the codomain.
>
> Incorrect, as well.
Only in the eyes of unmathematical cranks. A function defined by
> >and shows that, if all endsegments were definable, then the empty set could not be realized without preceding finite sets. The function f defined for all natural numbers has values larger than 2 and 0 and cannot exist without the values 2 and 1 existing in the codomain.
>
> Incorrect, as well.
f(n+1) = f(n) - 1
for all natural numbers cannot reach 0 without passing 1.
Regards, WM
Chris M. Thomasson
Jan 14, 2020, 12:36:48 PM1/14/20
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thought about yet?
Zelos Malum
Jan 15, 2020, 3:35:26 AM1/15/20
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>That is irrelevant. I can count to every definable number - and so can every mathematician.
Counting is irrelevant here.
>The question is whether there exist natural numbers which cannot bne summed. This is the case because not all can be summed.
>Only in the eyes of unmathematical cranks. A function defined by
>f(n+1) = f(n) - 1
>for all natural numbers cannot reach 0 without passing 1.
And in this case, we are dealing with cardinal arithmetic and sets.
There you do not need to "pass by" the finites to reach 0 (the empty set)
Chris M. Thomasson
Jan 15, 2020, 3:51:44 AM1/15/20
to
On 1/15/2020 12:35 AM, Zelos Malum wrote:
>> That is irrelevant. I can count to every definable number - and so can every mathematician.
>
> Counting is irrelevant here.
>
>> The question is whether there exist natural numbers which cannot bne summed. This is the case because not all can be summed.
>
> All natural numbers can be summed, your dark numbers do not exist.
Oh I think they do exist. Its just that, perhaps, he has not thought
>> That is irrelevant. I can count to every definable number - and so can every mathematician.
>
> Counting is irrelevant here.
>
>> The question is whether there exist natural numbers which cannot bne summed. This is the case because not all can be summed.
>
> All natural numbers can be summed, your dark numbers do not exist.
about them yet. Think about it for a moment... A habitually
condescending Person that does not think negative numbers exist just
might think -1 is dark? Or to go to the extreme: cannot even exist in
the first place? Humm...
There is a person on here that thinks we cannot plot the 2d point (0, 1)
on a graph, or i in the form (0 + 1i). I showed him that graph, and the
axes. Well, the x axis went from (-1, 0) to (1, 0). The y from (1, 0) to
(-1, 0) from top to bottom: and he freaked out... Saying something akin
to a negative x cannot exist or some shi% like that. _No_ negatives, or
this person gets really mad!
Chris M. Thomasson
Jan 15, 2020, 3:53:43 AM1/15/20
to
On 1/15/2020 12:51 AM, Chris M. Thomasson wrote:
> On 1/15/2020 12:35 AM, Zelos Malum wrote:
>>> That is irrelevant. I can count to every definable number - and so
>>> can every mathematician.
>>
>> Counting is irrelevant here.
>>
>>> The question is whether there exist natural numbers which cannot bne
>>> summed. This is the case because not all can be summed.
>>
>> All natural numbers can be summed, your dark numbers do not exist.
>
> Oh I think they do exist. Its just that, perhaps, he has not thought
> about them yet. Think about it for a moment... A habitually
> condescending Person that does not think negative numbers exist just
> might think -1 is dark? Or to go to the extreme: cannot even exist in
> the first place? Humm...
>
> There is a person on here that thinks we cannot plot the 2d point (0, 1)
> on a graph, or i in the form (0 + 1i). I showed him that graph, and the
> axes.
> Well, the x axis went from (-1, 0) to (1, 0).
> The y from (1, 0) to
> (-1, 0) from top to bottom: and he freaked out...
^^^^^^^^^^^^^^^^^^^^
> On 1/15/2020 12:35 AM, Zelos Malum wrote:
>>> That is irrelevant. I can count to every definable number - and so
>>> can every mathematician.
>>
>> Counting is irrelevant here.
>>
>>> The question is whether there exist natural numbers which cannot bne
>>> summed. This is the case because not all can be summed.
>>
>> All natural numbers can be summed, your dark numbers do not exist.
>
> Oh I think they do exist. Its just that, perhaps, he has not thought
> about them yet. Think about it for a moment... A habitually
> condescending Person that does not think negative numbers exist just
> might think -1 is dark? Or to go to the extreme: cannot even exist in
> the first place? Humm...
>
> There is a person on here that thinks we cannot plot the 2d point (0, 1)
> on a graph, or i in the form (0 + 1i). I showed him that graph, and the
> axes.
> Well, the x axis went from (-1, 0) to (1, 0).
> The y from (1, 0) to
> (-1, 0) from top to bottom: and he freaked out...
Ummm.... The y goes from (0, 1) to (0, -1) wrt top to bottom! Yikes!
This is a correction. Sorry for the error!
Ganzhinterseher
Jan 16, 2020, 2:09:49 PM1/16/20
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Am Dienstag, 14. Januar 2020 18:36:48 UTC+1 schrieb Chris M. Thomasson:
> I need to clarify, is a "dark" number, any number that you have not
> thought about yet?
You can say so, if you assume to be your own universe. If you take into account the whole (physical) universe, then a dark number is a number which no-one in the universe has ever thought about yet. The border is permeable. But the set of dark numbers will never lose enough numbers to become finite. It will always remain infinite.
> I need to clarify, is a "dark" number, any number that you have not
> thought about yet?
Regards, WM
Ganzhinterseher
Jan 16, 2020, 2:14:39 PM1/16/20
to
Am Mittwoch, 15. Januar 2020 09:35:26 UTC+1 schrieb Zelos Malum:
> >That is irrelevant. I can count to every definable number - and so can every mathematician.
>
> Counting is irrelevant here.
That is true for dark numbers only. Counting is easy for definable or countable numbrs.
> >That is irrelevant. I can count to every definable number - and so can every mathematician.
>
> Counting is irrelevant here.
>
> >The question is whether there exist natural numbers which cannot bne summed. This is the case because not all can be summed.
>
> All natural numbers can be summed, your dark numbers do not exist.
What is the sum of all natural numbers?
> >The question is whether there exist natural numbers which cannot bne summed. This is the case because not all can be summed.
>
> All natural numbers can be summed, your dark numbers do not exist.
>
> >Only in the eyes of unmathematical cranks. A function defined by
> >f(n+1) = f(n) - 1
> >for all natural numbers cannot reach 0 without passing 1.
>
> Only in NATURAL numbers, however when you deal with other number systems, that is not necisserily true anymore!
Here we deal with natural numbers only.
> >Only in the eyes of unmathematical cranks. A function defined by
> >f(n+1) = f(n) - 1
> >for all natural numbers cannot reach 0 without passing 1.
>
> Only in NATURAL numbers, however when you deal with other number systems, that is not necisserily true anymore!
>
> And in this case, we are dealing with cardinal arithmetic and sets.
The elements of endsegments of natural numbers are natural numbers by definition.
> And in this case, we are dealing with cardinal arithmetic and sets.
>
> There you do not need to "pass by" the finites to reach 0 (the empty set)
Natural numbers cannot pass by the infinite in a decreasing sequence.
> There you do not need to "pass by" the finites to reach 0 (the empty set)
Regards, WM
Zelos Malum
Jan 17, 2020, 1:33:59 AM1/17/20
to
>That is true for dark numbers only. Counting is easy for definable or countable numbrs.
It is not relevant here bceause we are not doing counting you imbecile.
>What is the sum of all natural numbers?
>Here we deal with natural numbers only.
>The elements of endsegments of natural numbers are natural numbers by definition.
Message has been deleted
Ganzhinterseher
Jan 17, 2020, 6:24:06 AM1/17/20
to
Am Freitag, 17. Januar 2020 07:33:59 UTC+1 schrieb Zelos Malum:
> >That is true for dark numbers only. Counting is easy for definable or countable numbrs.
>
> It is not relevant here bceause we are not doing counting you imbecile.
>
> >What is the sum of all natural numbers?
>
> You cannot sum the set of all natural numbers,
But we can sum all definable natural numbers.
> >That is true for dark numbers only. Counting is easy for definable or countable numbrs.
>
> It is not relevant here bceause we are not doing counting you imbecile.
>
> >What is the sum of all natural numbers?
>
> You cannot sum the set of all natural numbers,
> but you can sum any of the natural numbers, or sum any finite subset of natural numbers.
>
> >Here we deal with natural numbers only.
>
> We are not counting IN natural numbers
Never they have been counted.
> >Here we deal with natural numbers only.
>
> We are not counting IN natural numbers
> your argument is about CARDINALITY which means we are doing CARDINAL arithmetic, not natural number arithmetic!
>
> >The elements of endsegments of natural numbers are natural numbers by definition.
>
> Yes, but the cardinality, hwich is the one you go on about, is not natural numbers.
That is irrelevant. We know that for every definable endsegment
> >The elements of endsegments of natural numbers are natural numbers by definition.
>
> Yes, but the cardinality, hwich is the one you go on about, is not natural numbers.
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
∩{E(1), E(2), E(3), ...} = { }
|E(k+1)| = |E(k)| - 1
This can never reach 0 without having passed ...3, 2, 1 before. Simplest arithmetic of finite numbers.
Regards, WM
Chris M. Thomasson
Jan 17, 2020, 10:39:46 PM1/17/20
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On 1/16/2020 10:33 PM, Zelos Malum wrote:
>> That is true for dark numbers only. Counting is easy for definable or countable numbrs.
>
> It is not relevant here bceause we are not doing counting you imbecile.
Think about it... A dark number...
>> That is true for dark numbers only. Counting is easy for definable or countable numbrs.
>
> It is not relevant here bceause we are not doing counting you imbecile.
1, 2, 3, 4, 5, 6, yawn... go to sleep
7, 8, 9, ... are all dark numbers for now.
Zelos Malum
Jan 20, 2020, 1:52:33 AM1/20/20
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>But we can sum all definable natural numbers.
Not the set of them, its still infinite.
>And why is that so? Because any of the natural numbers that can be defined is definable. And any finite set must have a last definable natural number. But the set of all natural numbers consists mainly of undefinable natural numbers. Therefore it cannot be summed.
All natural numbers are "definable" as you want them, there are no "dark numbers" or the likes.
>Fo definable natural numbers we have appropriate arithmetic with natural numbers.
>This can never reach 0 without having passed ...3, 2, 1 before. Simplest arithmetic of finite numbers.
Ganzhinterseher
Jan 20, 2020, 7:56:00 AM1/20/20
to
Am Montag, 20. Januar 2020 07:52:33 UTC+1 schrieb Zelos Malum:
> >But we can sum all definable natural numbers.
>
> Not the set of them, its still infinite.
All definable natnumbers are finite:
> >But we can sum all definable natural numbers.
>
> Not the set of them, its still infinite.
>
> >And why is that so? Because any of the natural numbers that can be defined is definable. And any finite set must have a last definable natural number. But the set of all natural numbers consists mainly of undefinable natural numbers. Therefore it cannot be summed.
>
> >And why is that so? Because any of the natural numbers that can be defined is definable. And any finite set must have a last definable natural number. But the set of all natural numbers consists mainly of undefinable natural numbers. Therefore it cannot be summed.
>
> All natural numbers are "definable" as you want them, there are no "dark numbers" or the likes.
>
> >Fo definable natural numbers we have appropriate arithmetic with natural numbers.
>
> We have arithmetic for cardinal numbers as well.
But not for undefinable natnumbers.
>
> >Fo definable natural numbers we have appropriate arithmetic with natural numbers.
>
> We have arithmetic for cardinal numbers as well.
>
> >This can never reach 0 without having passed ...3, 2, 1 before. Simplest arithmetic of finite numbers.
>
> Incorrect, that is an assumption based on intuition.
It is based on the definition of endsegments. Every endegment in
> >This can never reach 0 without having passed ...3, 2, 1 before. Simplest arithmetic of finite numbers.
>
> Incorrect, that is an assumption based on intuition.
∩{E(1), E(2), E(3), ...} = { } (*)
can eject only one natnumber because of
∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> There is nothign that says we much go past finite cardinalities.
f(k+1) = f(k) - 1 and has values larger 2 and 0 must have the value 1 too.
Regards, WM
Zelos Malum
Jan 21, 2020, 2:00:42 AM1/21/20
to
>All definable natnumbers are finite:
But not the set of them, which is the set of all natural numbers.
>But not for undefinable natnumbers.
>It is based on the definition of endsegments. Every endegment in
>can eject only one natnumber because of
>It is the intermediate value theorem: A function that goes in steps of units
Ganzhinterseher
Jan 21, 2020, 11:19:23 AM1/21/20
to
Am Dienstag, 21. Januar 2020 08:00:42 UTC+1 schrieb Zelos Malum:
> >All definable natnumbers are finite:
>
> But not the set of them, which is the set of all natural numbers.
No. Try to find any decreasing sequence of natnumbers. It will be finite and every attempt to increase it will fail to reach an infinite set.
> >All definable natnumbers are finite:
>
> But not the set of them, which is the set of all natural numbers.
>
> >It is based on the definition of endsegments. Every endegment in
>
> Nope, it is based on your intuition.
>
> >can eject only one natnumber because of
>
> Which in noway implies finite cardinalities must be raeched.
It is implied by ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.
> >It is based on the definition of endsegments. Every endegment in
>
> Nope, it is based on your intuition.
>
> >can eject only one natnumber because of
>
> Which in noway implies finite cardinalities must be raeched.
>
> >It is the intermediate value theorem: A function that goes in steps of units
>
> Does not apply to cardinal arithmetic.
Here we have arithmetic of natural numbers, namely a function with domain |N and codomain (0, 1, >=2} defined by f(1) >= 2 and f(x+1) = f(x) - 1.
> >It is the intermediate value theorem: A function that goes in steps of units
>
> Does not apply to cardinal arithmetic.
Regards, WM
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