A dark quantity

Post by ***@gmail.com
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).

Regards, WM

Chris M. Thomasson

2024-01-07 23:19:16 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Ross Finlayson

2024-01-08 00:24:40 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.

integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum

So, in line reals there _is- a first number above zero.

This is called, for each "definition of continuity", a "completenes", all, a "repleteness".

That is all, ....

Chris M. Thomasson

2024-01-08 03:15:03 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?

.001, or .01?

Oh don't forget .00001. Not to mention .00000000001

These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

We can plot a line in n-ary p0, and p1.

dif = p1 - p0;

Now we have our difference. We can define a normal, say:

n = 2048;
normal_base = 1 / n;

Then we can plot the points:

for (i = 0; i < n; ++i)
{
normal = normal_base * i;
point = p0 + dif * normal;
plot_point(point);
}

Notice that as soon as we define n, its no longer infinite...

Ross Finlayson

2024-01-08 03:21:52 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?
.001, or .01?
Oh don't forget .00001. Not to mention .00000000001
These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

We can plot a line in n-ary p0, and p1.
dif = p1 - p0;
n = 2048;
normal_base = 1 / n;
for (i = 0; i < n; ++i)
{
normal = normal_base * i;
point = p0 + dif * normal;
plot_point(point);
}
Notice that as soon as we define n, its no longer infinite...

What you do is you get a compass and you get an edge,
and you throw away the compass, and get some chalk,
then you mark down the edge, the chalk, and,
you just wrote infinity points in a row.

It's sort of similar with fixed point, it's just, you know,
1, with however deep a radix in all the storage there is.

I mean, your real numbers have 1/3, right. They have n/d,
for unsigned n, d. Then just let n go from zero to d,
as sizeof(typeof(d)), or in accords with types, goes to infinity.

It's called the equivalency function, its range has a sigma algebra.

See also Ross Finlayson's 10,000's posts to sci.math,
particularly his, "slates".

...

Chris M. Thomasson

2024-01-08 03:29:11 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?
.001, or .01?
Oh don't forget .00001. Not to mention .00000000001
These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

What you do is you get a compass and you get an edge,
and you throw away the compass, and get some chalk,
then you mark down the edge, the chalk, and,
you just wrote infinity points in a row.

Yes. Think of drawing a line from p0 to p1, we just drew infinite
points. Fine. However, wrt infinity and real numbers there is no first
one greater than zero:

0---->1/2---->1

Well, we have:

0---->1/4---->1/2

Oh what about:

0---->1/8---->1/4

0---->1/16---->1/8

Humm, oh we also have:

0---->1/32---->1/16

This is basically zooming in on line segments. Those are all covered,
however there is no first real number greater than zero... We can zoom
in forever...

Post by Ross Finlayson
It's sort of similar with fixed point, it's just, you know,
1, with however deep a radix in all the storage there is.
I mean, your real numbers have 1/3, right. They have n/d,
for unsigned n, d. Then just let n go from zero to d,
as sizeof(typeof(d)), or in accords with types, goes to infinity.
It's called the equivalency function, its range has a sigma algebra.
See also Ross Finlayson's 10,000's posts to sci.math,
particularly his, "slates".
...

Chris M. Thomasson

2024-01-08 03:31:58 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?
.001, or .01?
Oh don't forget .00001. Not to mention .00000000001
These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

What you do is you get a compass and you get an edge,
and you throw away the compass, and get some chalk,
then you mark down the edge, the chalk, and,
you just wrote infinity points in a row.
It's sort of similar with fixed point, it's just, you know,
1, with however deep a radix in all the storage there is.
I mean, your real numbers have 1/3, right. They have n/d,
for unsigned n, d. Then just let n go from zero to d,
as sizeof(typeof(d)), or in accords with types, goes to infinity.
It's called the equivalency function, its range has a sigma algebra.
See also Ross Finlayson's 10,000's posts to sci.math,
particularly his, "slates".

Just keep in mind that the concept of a "first" real number greater than
zero doesn't exist within the framework of real numbers...

Ross Finlayson

2024-01-08 05:15:22 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?
.001, or .01?
Oh don't forget .00001. Not to mention .00000000001
These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

What you do is you get a compass and you get an edge,
and you throw away the compass, and get some chalk,
then you mark down the edge, the chalk, and,
you just wrote infinity points in a row.
It's sort of similar with fixed point, it's just, you know,
1, with however deep a radix in all the storage there is.
I mean, your real numbers have 1/3, right. They have n/d,
for unsigned n, d. Then just let n go from zero to d,
as sizeof(typeof(d)), or in accords with types, goes to infinity.
It's called the equivalency function, its range has a sigma algebra.
See also Ross Finlayson's 10,000's posts to sci.math,
particularly his, "slates".

Just keep in mind that the concept of a "first" real number greater than
zero doesn't exist within the framework of real numbers...

... "Standard", "standard" real numbers, indeed, the entire idea of
the "non-standard", real numbers, is these sorts things.

Then it's what's called "continuous domains", and that on the "linear continuum",
which are all, only, and everywhere real numbers, are these standard and non-standard,
real numbers.

For example if you look into Jordan measure it's for a line integral, in the line,
and the Dirichlet function is, half the real numbers, even/odd, or alternating.

Then, it's very simple, real simple, that n/d for n, d natural as n -> d, and d -> oo,
results a model of a "not-a-real-function" a "not-a-standard-real-function",
that for example has least upper bound property, very trivially,
that for example "zooming" never reaches, because it's the ordered field,
not the complete ordered field, which only follows axiomatization of least-upper-bound property.
It has sigma algebras for emasure and LUB for gaplessness so the fundamental
theorems of calculus follow, it's a continuous domain.

Other "not-a-real-function's" of interest include the Dirac delta.
(As modeled by real functions and only true in the limit.)

So, in the wider framework of continuous domains, and real numbers,
are three different sets of real numbers, or frameworks of functions,
that make the equivalency function that is not a Cartesian function,
a continuous domain, and, the usual standard complete ordered field,
which is equivalence classes of sequences that are Cauchy, and for signal reals,
functions in Dirichlet functions that subset the continuous domain while
keeping rational analytical character.

https://www.youtube.com/@rossfinlayson

Then "line-drawing" is called "line-drawing" and it's a fundamental geometrical act.

So, in line reals there is a first number after f(0) = 0, it's f(1), in "iota-values".

It's real valued.

...

Ross Finlayson

2024-01-08 16:24:42 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?
.001, or .01?
Oh don't forget .00001. Not to mention .00000000001
These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

What you do is you get a compass and you get an edge,
and you throw away the compass, and get some chalk,
then you mark down the edge, the chalk, and,
you just wrote infinity points in a row.
It's sort of similar with fixed point, it's just, you know,
1, with however deep a radix in all the storage there is.
I mean, your real numbers have 1/3, right. They have n/d,
for unsigned n, d. Then just let n go from zero to d,
as sizeof(typeof(d)), or in accords with types, goes to infinity.
It's called the equivalency function, its range has a sigma algebra.
See also Ross Finlayson's 10,000's posts to sci.math,
particularly his, "slates".

Just keep in mind that the concept of a "first" real number greater than
zero doesn't exist within the framework of real numbers...

... "Standard", "standard" real numbers, indeed, the entire idea of
the "non-standard", real numbers, is these sorts things.
Then it's what's called "continuous domains", and that on the "linear continuum",
which are all, only, and everywhere real numbers, are these standard and non-standard,
real numbers.
For example if you look into Jordan measure it's for a line integral, in the line,
and the Dirichlet function is, half the real numbers, even/odd, or alternating.
Then, it's very simple, real simple, that n/d for n, d natural as n -> d, and d -> oo,
results a model of a "not-a-real-function" a "not-a-standard-real-function",
that for example has least upper bound property, very trivially,
that for example "zooming" never reaches, because it's the ordered field,
not the complete ordered field, which only follows axiomatization of least-upper-bound property.
It has sigma algebras for emasure and LUB for gaplessness so the fundamental
theorems of calculus follow, it's a continuous domain.
Other "not-a-real-function's" of interest include the Dirac delta.
(As modeled by real functions and only true in the limit.)
So, in the wider framework of continuous domains, and real numbers,
are three different sets of real numbers, or frameworks of functions,
that make the equivalency function that is not a Cartesian function,
a continuous domain, and, the usual standard complete ordered field,
which is equivalence classes of sequences that are Cauchy, and for signal reals,
functions in Dirichlet functions that subset the continuous domain while
keeping rational analytical character.
Then "line-drawing" is called "line-drawing" and it's a fundamental geometrical act.
So, in line reals there is a first number after f(0) = 0, it's f(1), in "iota-values".
It's real valued.
...

See, with such deliberate clarifications of definition, many of the intuitions of
posters like Mitch and WM can be corrected, illustrating framings where their
ideas of the objects of mathematics exist.

It takes really not so much, and indeed it's pretty simple, to arrive at these
mutually concordant definitions of repleteness in continuity.

That cranks like Mitch (the lesser, vis-a-vis stronger Mitch, fom, on sci.logic),
and WM the Hodges-hog, can simply have their definitions fixed and separated
from the usual and arrive at saying things that are actually true, is why it's
called "correcting papers" not just "marking mistakes".

So anyways there are at least three definitions of continuity, and there is
a first and central definition after line continuity, in the objects of mathematics,
where there's always a continuum.

Then, one of these models of real numbers, line reals, has each real number has
an integer part and a non-integer part, and the non-integer part ranges [0,1),
and it is a set of, "iota-values", which are real-valued, and valid in real-valued formulae.

Under my youtube's there's Philosophical Foreground, then "reading Weyl", is all standard,
then the next for "continuum infinitesimal analysis, after the first half details these things.

I.e., "Nonstandard analysis" for any kind of infinitesimals, then is for "continuum infinitesimal
analysis", nonstandard analysis with real analytical character.

--

I have a mathematics degree, and I earned one, too. "A defense of mathematics."

Ross Finlayson

2024-01-11 17:06:53 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

Real numbers, sets of, each, are generally considered "continuous domains"
of "elements of a linear continuum". There are at least three models of these,
line reals with infinitesimals [0,1], field reals the complete ordered field, and
signal reals even/odd infinitesimals.
integer continuum -> line reals <-> field reals <-> signal reals <- long-line continuum
So, in line reals there _is- a first number above zero.

There is no first real number greater than zero. Which one is first?
.001, or .01?
Oh don't forget .00001. Not to mention .00000000001
These are all reals greater than one... Now, if I define a normal unit,
say 1 / 42, then we can have a first one because I simply defined the
normal unit. Is not infinite anymore. See?

Post by Ross Finlayson
This is called, for each "definition of continuity", a "completenes", all, a "repleteness".
That is all, ....

What you do is you get a compass and you get an edge,
and you throw away the compass, and get some chalk,
then you mark down the edge, the chalk, and,
you just wrote infinity points in a row.
It's sort of similar with fixed point, it's just, you know,
1, with however deep a radix in all the storage there is.
I mean, your real numbers have 1/3, right. They have n/d,
for unsigned n, d. Then just let n go from zero to d,
as sizeof(typeof(d)), or in accords with types, goes to infinity.
It's called the equivalency function, its range has a sigma algebra.
See also Ross Finlayson's 10,000's posts to sci.math,
particularly his, "slates".

Just keep in mind that the concept of a "first" real number greater than
zero doesn't exist within the framework of real numbers...

... "Standard", "standard" real numbers, indeed, the entire idea of
the "non-standard", real numbers, is these sorts things.
Then it's what's called "continuous domains", and that on the "linear continuum",
which are all, only, and everywhere real numbers, are these standard and non-standard,
real numbers.
For example if you look into Jordan measure it's for a line integral, in the line,
and the Dirichlet function is, half the real numbers, even/odd, or alternating.
Then, it's very simple, real simple, that n/d for n, d natural as n -> d, and d -> oo,
results a model of a "not-a-real-function" a "not-a-standard-real-function",
that for example has least upper bound property, very trivially,
that for example "zooming" never reaches, because it's the ordered field,
not the complete ordered field, which only follows axiomatization of least-upper-bound property.
It has sigma algebras for emasure and LUB for gaplessness so the fundamental
theorems of calculus follow, it's a continuous domain.
Other "not-a-real-function's" of interest include the Dirac delta.
(As modeled by real functions and only true in the limit.)
So, in the wider framework of continuous domains, and real numbers,
are three different sets of real numbers, or frameworks of functions,
that make the equivalency function that is not a Cartesian function,
a continuous domain, and, the usual standard complete ordered field,
which is equivalence classes of sequences that are Cauchy, and for signal reals,
functions in Dirichlet functions that subset the continuous domain while
keeping rational analytical character.
Then "line-drawing" is called "line-drawing" and it's a fundamental geometrical act.
So, in line reals there is a first number after f(0) = 0, it's f(1), in "iota-values".
It's real valued.
...

In the above that's "measure" not "emasure",
wouldn't want it to be mistaken at all.

I intentionally spell my own words,
and know them, any mistakes are my own,
and, there aren't any.

Here then let's talk about mathematics
where light" and "dark" are simply complements,
because worldly concerns, while matters of
Man and Mind and Animal and Machine,
are only peripheral to the theory of mathematics.

Of course then mathematics upholds mathematical sciences,
according to philosophy of science.

(That "dark matter" contra "luminous matter" is just a non-scientific
placeholder in modern cosmology is matters of metaphor.)

Warm regards, and for good will toward Man, and for that
"I am not my brother's keeper, I am my brother's brother."

(Here Man is the gender neutral nominative, not the neutered androgenous,
and the filial and fraternal and paternal reflects relation and origin.)

Universe is one big dot

2024-01-08 12:23:07 UTC

Post by ***@gmail.com
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

Richard Damon

2024-01-08 12:27:46 UTC

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

No, your claim is essentially that an infinite number is not also a
finite number.

It doesn't say the numbers are dark, just too numerous list all of them
at once on finite paper.

Your logic just doens't handle inifinites.

FromTheRafters

2024-01-08 16:37:13 UTC

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.

At least that many, because there are that many as rational numbers.

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.

2024-01-10 22:07:31 UTC

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.

Every defined number does not belong to the domain covered by the smallest
ℵo unit fractions or by the largest ℵo natural numbers.

Regards, WM

zelos...@gmail.com

2024-01-09 10:03:40 UTC

Post by ***@gmail.com
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

We have been through this time and time again, all of that "individually" is meaningless.

Chris M. Thomasson

2024-01-09 23:01:53 UTC

Post by ***@gmail.com
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

We have been through this time and time again, all of that "individually" is meaningless.

The parrot is dead! ;^)

Ross Finlayson

2024-01-09 23:23:36 UTC

Post by ***@gmail.com
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

We have been through this time and time again, all of that "individually" is meaningless.

The parrot is dead! ;^)

It's just resting, ....

...

Chris M. Thomasson

2024-01-10 00:31:14 UTC

Post by ***@gmail.com
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

We have been through this time and time again, all of that "individually" is meaningless.

The parrot is dead! ;^)

It's just resting, ....
...

LOL!!! Its stunned! ROFL!!!!

mitchr...@gmail.com

2024-01-10 02:21:32 UTC

Post by ***@gmail.com
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

Above zero is first non zero fundamental infinitesimal.

Ross Finlayson

2024-01-10 02:53:38 UTC

Post by ***@gmail.com
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

Above zero is first non zero fundamental infinitesimal.

It's "nonstandard real", but, "standard infinitesimal".

Chris M. Thomasson

2024-01-10 07:47:44 UTC

Post by ***@gmail.com
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

Above zero is first non zero fundamental infinitesimal.

It's "nonstandard real", but, "standard infinitesimal".

infinitesimal is small, very small. So small that it get smaller before
you can observe it... ;^) lol. Just some jest.

2024-01-10 21:59:39 UTC

Post by ***@gmail.com
Above zero is first non zero fundamental infinitesimal.

With no doubt, there is a first unit fraction because a liner chain with
gaps has a first element:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

This 1/n is not infinitesimal like its n is not infinite. But it is dark.
The smallest ℵo unit fractions and the largest ℵo natural numbers
cannot be used as individuals. Every defined positive real number does not
belong to the smallest ℵo unit fractions or to the largest ℵo natural
numbers.

Regards, WM

Chris M. Thomasson

2024-01-13 05:14:36 UTC

Post by ***@gmail.com
Above zero is first non zero fundamental infinitesimal.

With no doubt, there is a first unit fraction because a liner chain with
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
This 1/n is not infinitesimal like its n is not infinite. But it is
dark. The smallest ℵo unit fractions and the largest ℵo natural numbers
cannot be used as individuals. Every defined positive real number does
not belong to the smallest ℵo unit fractions or to the largest ℵo
natural numbers.

1/1 is first, try to find a last smaller one...

Chris M. Thomasson

2024-01-13 05:15:45 UTC

Post by ***@gmail.com
Above zero is first non zero fundamental infinitesimal.

With no doubt, there is a first unit fraction because a liner chain
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
This 1/n is not infinitesimal like its n is not infinite. But it is
dark. The smallest ℵo unit fractions and the largest ℵo natural
numbers cannot be used as individuals. Every defined positive real
number does not belong to the smallest ℵo unit fractions or to the
largest ℵo natural numbers.

1/1 is first, try to find a last smaller one...

The smallest one is trying to feel you:

lol! Keep in mind that there is no smallest one.

mitchr...@gmail.com

2024-01-10 23:30:58 UTC

Post by ***@gmail.com
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

Above zero is first non zero fundamental infinitesimal.

It's "nonstandard real", but, "standard infinitesimal".

infinitesimal is small, very small. So small that it get smaller before
you can observe it... ;^) lol. Just some jest.

There is no smaller than inverse infinity.
Zero has no size at all. It is not a smaller.
As small it doesn't exist.

Chris M. Thomasson

2024-01-11 02:49:12 UTC

Post by ***@gmail.com
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

Above zero is first non zero fundamental infinitesimal.

It's "nonstandard real", but, "standard infinitesimal".

infinitesimal is small, very small. So small that it get smaller before
you can observe it... ;^) lol. Just some jest.

There is no smaller than inverse infinity.
Zero has no size at all. It is not a smaller.
As small it doesn't exist.

Small as is something that is greater than zero, but still small... Tack
on a negative sign, and is still small, or close to zero so to speak...
;^D lol.

By the time you take a look at .01, well, it went to .0000001,
.00000000000000000000001, on and on:

2024-01-08 10:45:42 UTC

Post by ***@gmail.com
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.

But there are ℵ real numbers above zero and smaller than every definable
positive real number. Proof: Try to define a positive number x with less
smaller numbers between 0 and x. Impossible.

Regards, WM

mitchr...@gmail.com

2024-01-08 17:55:41 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

It is the infinitesimal above zero...

Ross Finlayson

2024-01-08 23:16:10 UTC

Post by ***@gmail.com
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.

There is no _first_ real number above zero.

It is the infinitesimal above zero...

There are lots of kinds of notions of infinitesimals, just like infinities.

So, you are pointing to a particular kind, like "iota-values".

Richard Damon

2024-01-07 23:26:49 UTC

Post by ***@gmail.com
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.

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.

2024-01-08 10:56:23 UTC

Post by ***@gmail.com
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
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

Richard Damon

2024-01-08 12:21:09 UTC

Post by ***@gmail.com
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.

why do you think I can not name any of them?

Post by Richard Damon
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.

But you have explicitly said you are working with the NATURAL numbers,
and ω and such are not Natural Numbers, so apparently your dark numbers
are just Natural Numbers that are not Natural Numbers.

Post by Richard Damon
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.

WHy

Post by Richard Damon
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.

And what do you mean by "actual infinity"?

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

Your problem is you assume the existance of things that are not.

There is no "smallest" x > 0, at least not in the finite numbers.

That is just like the Barber that shaves everyone who does not shave
himself.

Your logic has just fallen into Russel's paradox.

2024-01-08 12:29:29 UTC

Post by WM
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

Fritz Feldhase

2024-01-08 23:06:05 UTC

ℵ

Kannst Du Trottel "ℵ" auch definieren?

Ross Finlayson

2024-01-08 23:14:48 UTC

Post by Fritz Feldhase

ℵ

Kannst Du Trottel "ℵ" auch definieren?

Ordnungs sind nich Machtigkeits, under aber so.

Unendlichkeit, Umbegrentzheit, ..., mehre Worten machts
fur die Scalar Umbegrentzheit.

Mann macht nich die Wortschatz schlect.

The laws (of grammar) aren't all powerset, and vice versa.

Unending, unbounded, ..., more words make
for the scalar infinity.

One doesn't pollute the vocabulary.

Richard Damon

2024-01-09 02:31:58 UTC

Post by WM
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

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, except those are individually namable, and are not members of
the finite numbers.

2024-01-09 17:23:49 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
"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.

Post by Richard Damon
It doesn't mean that any particular ones left can't be named.

It does. The last ℵ numbers will remain.

Post by Richard Damon
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

mitchr...@gmail.com

2024-01-09 19:15:59 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
"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.

Post by Richard Damon
It doesn't mean that any particular ones left can't be named.

It does. The last ℵ numbers will remain.

Post by Richard Damon
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

Infinitesimals of themselves are dark quantity by being below the finite.
Zero is no quantity to see...

zelos...@gmail.com

2024-01-17 06:20:19 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
"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.

Post by Richard Damon
It doesn't mean that any particular ones left can't be named.

It does. The last ℵ numbers will remain.

Post by Richard Damon
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

Infinitesimals of themselves are dark quantity by being below the finite.
Zero is no quantity to see...

There are no infinitesimals in real numbers, retard.

Chris M. Thomasson

2024-01-17 06:28:32 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
"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.

Post by Richard Damon
It doesn't mean that any particular ones left can't be named.

It does. The last ℵ numbers will remain.

Post by Richard Damon
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

Infinitesimals of themselves are dark quantity by being below the finite.
Zero is no quantity to see...

There are no infinitesimals in real numbers, retard.

Gotta love 10^(-n), gets pretty small and they are all real numbers, n
is an integer.

[0] = 10^(0) = 1
[1] = 10^(-1) = .1
[2] = 10^(-2) = .01
[3] = 10^(-3) = .001
...

Limit zero. On and on, never equals zero, but gets closer and closer to
it, without any bounds... Take n to infinity, 0, -1, -2, -3, ... :^)

Sum them and get, 1.(1) ;^)

Chris M. Thomasson

2024-01-17 06:30:15 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
"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.

Post by Richard Damon
It doesn't mean that any particular ones left can't be named.

It does. The last ℵ numbers will remain.

Post by Richard Damon
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

Infinitesimals of themselves are dark quantity by being below the finite.
Zero is no quantity to see...

There are no infinitesimals in real numbers, retard.

Gotta love 10^(-n), gets pretty small and they are all real numbers, n
is an integer.
[0] = 10^(0) = 1
[1] = 10^(-1) = .1
[2] = 10^(-2) = .01
[3] = 10^(-3) = .001
...
Limit zero. On and on, never equals zero, but gets closer and closer to
it, without any bounds... Take n to infinity, 0, -1, -2, -3, ... :^)
Sum them and get, 1.(1) ;^)

10^(0)+10^(-1)+10^(-3)+10^(-5)+10^(-7)+10^(-9)

;^)

Chris M. Thomasson

2024-01-17 06:31:22 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
"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.

Post by Richard Damon
It doesn't mean that any particular ones left can't be named.

It does. The last ℵ numbers will remain.

Post by Richard Damon
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

Infinitesimals of themselves are dark quantity by being below the finite.
Zero is no quantity to see...

There are no infinitesimals in real numbers, retard.

Gotta love 10^(-n), gets pretty small and they are all real numbers, n
is an integer.
[0] = 10^(0) = 1
[1] = 10^(-1) = .1
[2] = 10^(-2) = .01
[3] = 10^(-3) = .001
...
Limit zero. On and on, never equals zero, but gets closer and closer
to it, without any bounds... Take n to infinity, 0, -1, -2, -3, ... :^)
Sum them and get, 1.(1) ;^)

10^(0)+10^(-1)+10^(-3)+10^(-5)+10^(-7)+10^(-9)
;^)

Another interesting summation:

10^(-2)+10^(-3)+10^(-5)+10^(-7)+10^(-11)+10^(-13)

;^D primes anyone?

Richard Damon

2024-01-10 02:58:56 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Nope, they all have a name, we just can't express them all at once.

Post by Richard Damon
"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.

Post by Richard Damon
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.

Post by Richard Damon
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

Nope. Take ANY of your "dark numbers", if is IS finite, then make you
finite initail segment starting with the number + 1.

If it isn't finite, then it wasn't one of the Natural Numbers after the
last one you previously named.

So, you logic is just broken, because it can't handle unbounded sets.

EVERY element of the Natural Numbers is a FINITE number, and thus has a
FINITE name, and is namable.

They are unbounded, so there is no "last" number.

They set of them is INFINITE, so it isn't surprizing that we can't name
them all a once (or even more than just a few comparatively). That is
just the nature of infinite sets, and any logic that doesn't take that
into account is just incapable of working with such sets.

2024-01-10 21:51:19 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by WM
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.

Post by Richard Damon
"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.

Post by Richard Damon
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

Richard Damon

2024-01-11 00:49:28 UTC

Post by WM
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.

Post by Richard Damon
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.

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.

Post by Richard Damon
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.

That doesn't say there is a smallest number, after all, for every one of
these n, the gap below it is smaller than its spacing from 0, and in
fact there is obviously room for at least n more unit fractions below 1/n.

Post by Richard Damon
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.

Really? says what principle?

Post by WM
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.

But none of them exist, so I don't need to name them.

You of course can't name that which doesn't exist.

I guess your "dark" numbers are just figments of your imagination.

Post by Richard Damon
"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.

Where did I stop naming them?

Post by Richard Damon
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.

But I did.

Post by WM
Regards, WM

2024-01-11 21:33:24 UTC

Post by Richard Damon
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

Richard Damon

2024-01-12 03:11:03 UTC

Post by Richard Damon
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

Why?

I guess this shows that dark numbers aren't actually numbers.

It seems your problem is just that your logic system blew up on you from
misuse, and you can't see what happened.

2024-01-12 14:17:10 UTC

Post by Richard Damon
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?

You must try it. The last ℵo natural numbers and the first ℵo unit
fractions cannot be subdivided.

Regards, WM

Chris M. Thomasson

2024-01-12 20:18:42 UTC

Post by Richard Damon
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?

You must try it. The last ℵo natural numbers and the first ℵo unit
fractions cannot be subdivided.

There are no last natural numbers because they go on forever.

Richard Damon

2024-01-12 23:46:31 UTC

Post by Richard Damon
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?

You must try it. The last ℵo natural numbers and the first ℵo unit
fractions cannot be subdivided.
Regards, WM

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.

Chris M. Thomasson

2024-01-12 23:49:06 UTC

Post by Richard Damon
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?

You must try it. The last ℵo natural numbers and the first ℵo unit
fractions cannot be subdivided.
Regards, WM

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.

Faulty, cryptic, negligence, he is a teacher, right? Yikes!!!!

2024-01-13 10:23:40 UTC

Post by Richard Damon
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?

You must try it. The last ℵo natural numbers and the first ℵo unit
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.

Post by Richard Damon
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.

Post by Richard Damon
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

Richard Damon

2024-01-13 13:22:48 UTC

Post by Richard Damon
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?

You must try it. The last ℵo natural numbers and the first ℵo unit
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.

But there is no difference between the "dark" numbers and the identified
numbers.

"The Last" in non-existant, so if that is what you mean by it, you are
lying as there are ZERO dark numbers, not Aleph_0 of them.

Post by Richard Damon
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.

But since the axioms create ALL the Natural Numbers, you are just
admitting that your "Dark Numbers" are not part of them.

So, you are just admitting you are just a liar. Your Dark numbers are
defined to be part of the Natural Numbers, but are not created by the
method that creates the Natural Numbers, so are not.

Post by Richard Damon
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.

And why do I need to?

That seems to be your problem, you are assuming you can do what isn't
allowed, and therefore make up something that doesn't exist to attempt it.

Post by WM
Regards, WM

Fritz Feldhase

2024-01-13 16:56:35 UTC

Post by WM
The axioms only create the definable natnumbers.

Nonsense. Axioms don't "create" any numbers.

Post by Richard Damon
But since the axioms create ALL the Natural Numbers

Nonsense. Axioms don't "create" any numbers.

Though the axiom of infinity (in ZFC) ensures for the existence of an infinite set (which is the basis for the set IN of natural numbers).

Ross Finlayson

2024-01-13 18:10:56 UTC

Post by Fritz Feldhase

Post by WM
The axioms only create the definable natnumbers.

Nonsense. Axioms don't "create" any numbers.

Post by Richard Damon
But since the axioms create ALL the Natural Numbers

Nonsense. Axioms don't "create" any numbers.
Though the axiom of infinity (in ZFC) ensures for the existence of an infinite set (which is the basis for the set IN of natural numbers).

You mean like axiomatizing a well-founded infinite inductive set?

It sort of has its reasoning, ....

It has its reasoning that it would be an inconsistent multiplicity
to simply build one logically with quantification.

Number theorists, for example, have a point at infinity. Geometry,
for example, has a point at infinity. It's not in all theories, but,
it is in the theory.

I hope you can recognize what would be your own hypocrisy otherwise,
that "a well-founded infinite inductive set", is a non-logical constant,
and as of a restriction of comprehension, sort of like "a unique empty set".

You know, for not being wrong, ....

You know what else are objects in Peano's theories? Infinitesimals.

It seems Russell sort of laughs at Frege's honesty, and having duped most,
it's sort of insufferable. Sort of like a whimsical fairy.

The "objects of mathematics" include ordinary and extra-ordinary inductive sets,
with/without "no infinite descending epsilon chains".

When you read Cohen's Continuum Hypothesis Independence proof,
you're going to be surprised, what he finds, from what was ZFC.

2024-01-14 12:35:26 UTC

Post by WM
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.

Post by WM
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.

Post by Richard Damon
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

2024-01-17 11:36:35 UTC

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

Richard Damon

2024-01-17 12:56:00 UTC

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?

Becuase 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.
Regards, WM

because such a number doesn't exist. You don't need to name the
non-existant.

2024-01-17 14:08:06 UTC

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

2024-01-17 18:21:39 UTC

Post by WM
The set is bounded by its smallest upper bound 0.

Should read: Greatest lower bound.

Regards, WM

Richard Damon

2024-01-18 01:36:09 UTC

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.

But 0 isn't member of the set, so the set itself 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.

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

Which only shows it can't increase to a finite number at any finite
number as that number will have an unbounded number of unit fractions
below it.

Thus NUF(x) with x finite, jumps, perhaps in domain of the
infinitesimals between 0 and the bottom of (0,1]

2024-01-18 08:54:21 UTC

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.

Post by Richard Damon
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

Richard Damon

2024-01-18 12:46:47 UTC

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.

There is not need for a first element, since the chain is unbounded in
length.

Your logic is just flawed in that assumption.

Yes, ZFC says there is a "first", but that is in numbering sequence, not
value sequence, and that first is 1/1, and we number infinitely towards 0.

Post by Richard Damon
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.

But they are not elements of the finite numbers, and thus not Real,
Rational, or Unit Fractions.

Yes, if you admit your "Dark Numbers" are not part of those sets, we
might find them existing, but then we have the transfinite theories that
describe them, and they no longer stay dark.

You are just asserting contradictory claims.

If you want to call that area you do not know about "dark", go ahead,
just realize that others have explored it and mapped it, so it really
isn't unknown, except to you.

Post by WM
Regards, WM

2024-01-18 17:34:07 UTC

Post by Richard Damon
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.

Post by WM
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

Richard Damon

2024-01-19 01:26:58 UTC

Post by Richard Damon
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, so isn't the "first" of the set, or
even implies that there IS a first element.

Post by WM
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.

yes, and all unit fractions, being one divided by a Natural Number, are
individually definable by that Natural Number, and thus not "dark"

You are unable to show a set that actually show the need for "dark" numbers.

Your attempts just fail.

Post by WM
Regards, WM

2024-01-19 08:55:01 UTC

Post by Richard Damon
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.

Post by WM
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

Richard Damon

2024-01-19 15:54:20 UTC

Post by Richard Damon
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.

No, it isn't.

You words define NUF(x) to be

Post by WM
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

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,
so you can't define what your Dark Numbers are.

You are implicitly assuming that there is a "highest" defined number,
but there isn't, and thus you have a wrong definition for "dark".

2024-01-19 17:07:00 UTC

Post by WM
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.

Post by WM
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.

Post by Richard Damon
Your problem is you don't have a definiton of what "N_def" actually is,

Every number that is defined individually is visible.

Post by Richard Damon
You are implicitly assuming that there is a "highest" defined number,

Not a constant number but only temporarily.

Regards, WM

Richard Damon

2024-01-19 18:09:00 UTC

Post by WM
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.

Post by WM
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?

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by WM
Regards, WM

2024-01-19 19:18:17 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by WM
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.

Post by Richard Damon
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.

Try it.

Post by Richard Damon
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.

Post by Richard Damon
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

Post by Richard Damon
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

Richard Damon

2024-01-19 21:48:26 UTC

Post by WM
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.

Post by Richard Damon
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.

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, so that couldn't have
been the smallest unit fraction.

Post by WM
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.

Post by Richard Damon
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.

Try it.

I did. You just don't understand it.

Post by Richard Damon
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".

Post by Richard Damon
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.

Post by Richard Damon
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.

Doesn't matter if they have been actually obsevered, just that they be
observable.

You are just confusing Knowledge with Truth.

Post by WM
Regards, WM

2024-01-19 22:41:12 UTC

Post by Richard Damon
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

Post by Richard Damon
BY YOUR DEFINITON, since there is an infinte number of unit fractions
below any finite x,

Wrong premise.

Post by Richard Damon
NUF(x) has an infinite value for all finite x
greater than 0.

Wrong result.

Post by Richard Damon
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.

Post by Richard Damon
But why is it only a finite part that is visible?

Try to make more visible. Fail. Then you know it.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
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.

Post by Richard Damon
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

Richard Damon

2024-01-19 23:18:49 UTC

Post by Richard Damon
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

Which says nothing about

Post by Richard Damon
BY YOUR DEFINITON, since there is an infinte number of unit fractions
below any finite x,

Wrong premise.

So, you lied? It was your premise. There ARE an infinite (aleph) number
of unit fractions below ANY unit fraction.

Post by Richard Damon
NUF(x) has an infinite value for all finite x greater than 0.

Wrong result.

Where is it not?

Post by Richard Damon
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 what Natural Number or Unit Fraction isn't definable?

For that matter, what Rational Number isn't.

Post by Richard Damon
But why is it only a finite part that is visible?

Try to make more visible. Fail. Then you know it.

Post by Richard Damon
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.

Your lose, since they weren't lies.

You are just rejecting the truth.

Post by Richard Damon
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.

Where do Natural Numbers stop becoming visible.

You can't actually define the set of Visible Numbers, which is needed
for your theory.

Post by Richard Damon
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.

Which is all of them. You just don't understand the infinite.

Post by Richard Damon
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.

Post by WM
Regards, WM

2024-01-20 11:05:14 UTC

Post by Richard Damon
But what Natural Number or Unit Fraction isn't definable?

The smallest ones. Those which, according to you, sit at infinitesimals
between 0 and (0, 1].

Post by WM
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.

Post by Richard Damon
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.

Liar.

Regards, WM

2024-01-19 23:00:16 UTC

Post by Fritz Feldhase
After all, we already have a theory of (finite) and infinite cardinals in set
theory.

But this theory requires that if not point of (0, 1] has less than ℵ
smaller unit fractions to its left-hand side, nevertheless there are not
ℵ unit fractions to the left-hand side of the interval (0, 1], i.e. a
theory which violates geometry.

Regards, WM

FromTheRafters

2024-01-19 17:14:42 UTC

Post by Richard Damon
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.

No, it isn't.
You words define NUF(x) to be

Post by WM
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

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, so
you can't define what your Dark Numbers are.

It is a particularly sticky wicket for him because a set is a
collection of well defined objects with well defined meaning that you
can always tell whether or not a particular object is a member of said
set or not.

A set of undefinable or undefined objects is an absurdity, so he will
say it is not a set but a potentially infinite collection. This
nullifies his idea that such a collection is a subset of naturals or
any other actual set.

Post by Richard Damon
You are implicitly assuming that there is a "highest" defined number, but
there isn't, and thus you have a wrong definition for "dark".

What is the most interesting number? reframed.

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

Jim Burns

2024-01-17 20:04:48 UTC

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 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)| = ...

2024-01-18 08:48:23 UTC

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.

All sets can change by single elements.
For instance we can remove single elements from every infinite set.

Post by Jim Burns
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 .

Regards, WM

Jim Burns

2024-01-18 18:20:04 UTC

Post by Jim Burns
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 .

Thank you for the correcting "not".

I will give
cardinal which can change by 1
a shorter name: final ordinal.

"Final ordinal" is analogous to
"initial ordinal"
https://en.wikipedia.org/wiki/Von_Neumann_cardinal_assignment#Initial_ordinal_of_a_cardinal

An initial von Neumann ordinal is
the first ordinal _with that cardinality_

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

Guests from {0,1,…;ω} = ω+1 fit in
rooms from {[0],[1],[2],…} = ω in this way:
ω:[0] 0:[1] 1:[2] 2:[3] ...

ω+1 = {0,1,…;ω} is also with cardinality ℵ₀
|ω| = |ω+1|

0 and 5 are final ordinals.
ω isn't a final ordinal, it is non.final.
ω is defined as the initial non.final ordinal.

The final ordinals are also known as
the natural numbers.

With that terminology change, my proof is
much less work to write,
much less work to read.

An ordinal after each final ordinal
is not any of the final ordinals.
It is non.final.

A non.final ordinal, by definition,
is followed by
an ordinal with the same cardinality.

Post by Jim Burns
|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 .

Between unit.fraction ⅟j and
unit.fraction ⅟k: 0 < ⅟k < ⅟j
⟨⅟k,⅟k⁻¹,…,⅟j⁺¹,⅟j⟩ has a final ordinal.
|⅟k,⅟k⁻¹,…,⅟j⁺¹,⅟j| ≠ |⅟k⁺¹,⅟k,⅟k⁻¹,…,⅟j⁺¹,⅟j|

Between ⅟j and 0
for each final ordinal k
all the between.unit.fractions
do not fit in ⟨⅟j⁺ᵏ,…,⅟j⁺¹⟩

Between ⅟j and 0
all the between.unit.fractions
do not have any final ordinal.

It is a non.final ordinal.
|⅟ℕ∩(0,⅟k)| = |⅟ℕ∩(0,⅟k⁺¹)| = |⅟ℕ∩(0,⅟k⁺²)| = ...

2024-01-19 10:19:39 UTC

Post by Jim Burns
I will give
cardinal which can change by 1
a shorter name: final ordinal.

A good point.

Post by Jim Burns
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.

But ω and most of its predecessors have no FISONs. They are dark numbers.

Post by Jim Burns
0 is last with cardinality 0
5 is last with cardinality 5
ω _is not_ last with cardinality ℵ₀

ℵ₀ of set theory is a self-contradictory non-mathematical and
non-logical notion. Proof: Bob.

Post by Jim Burns
ω+1 = {0,1,…;ω} is also with cardinality ℵ₀
|ω| = |ω+1|

That is true if we simply understand "infinitely many" by ℵ₀. In order
to excorzise the bijective meaning I will henceforth use only ℵ, meaning
"infinitely many". |ℕ| = ℵ, |ℚ| = ℵ, |ℝ| = ℵ,
although |ℚ| = 2|ℕ|^2 + 1 and |ℝ| = 2B^|2ℕ| where B is the base, B
= 2 in binary notation and B = 10 in decimal notation.

Regards, WM

Jim Burns

2024-01-19 19:53:58 UTC

[...]

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
|
| (i) f(D) ∈ D
| f(D) ∈ D ⇒ f(D) ∉ D
| Contradiction.
|
| (ii) f(D) ∉ D
| f(D) ∉ D ⇒ f(D) ∈ D
| Contradiction.
|
| Thus, contradiction.

Therefore,
¬(𝒫(ℕ) ⇉ ℕ)
and
|𝒫(ℕ)| > |ℕ|

|ℕ| = ℵ, |ℚ| = ℵ, |ℝ| = ℵ,

|ℝ| > |ℚ| = |ℕ| ∉ ℕ

Fritz Feldhase

2024-01-19 20:40:29 UTC

[...]

In order to excorzise the bijective meaning

Actually, he means

"In order to be able to ignore the infinite cardinals [since I'm too dumb for them] ..."

I will henceforth use only ℵ, meaning "infinitely many".

For this he would have to

1. define ℵ (say, ℵ := ℕ)

2. redefine |.| (such that it is not a variant of /card/ any more).

Def:

|M| = card(M) if M is finite and
|M| = ℵ if M is infinite.

|ℕ| = ℵ, |ℚ| = ℵ, |ℝ| = ℵ,

On the other hand, the following

Post by Jim Burns
|ℝ| > |ℚ| = |ℕ| ∉ ℕ

would have to be (could still be) expressed this way:

card(ℝ) > card(ℚ) = card(ℕ) ∉ ℕ.

Of course, this redefinition of |.| and introduction of ℵ is a rather idiotic idea.

After all, we already have a theory of (finite) and infinite cardinals in set theory.

2024-01-19 22:46:26 UTC

[...]

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

Post by Jim Burns
|ℝ| > |ℚ|

Regards, WM

Richard Damon

2024-01-19 23:18:51 UTC

[...]

In order to excorzise the bijective meaning
I will henceforth use only ℵ,
meaning "infinitely many".

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

So you think, but can not prove.

You are just wrong, because you think with logic insufficent to handle
unbounded sets.

Post by Jim Burns
|ℝ| > |ℚ|

Regards, WM

2024-01-20 11:08:43 UTC

Post by WM
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

So you think, but can not prove.

Simple. The diagonal number has no last digit. Therefore it is undefined.

Regards, WM

Chris M. Thomasson

2024-01-19 03:55:03 UTC

[...]

Post by WM
It is caused by your inability to reduce the amounts to less than ℵ.

WM can be useful? If you accidentally drink poison, read some of WM's
thoughts and automatically puke your brains out? Might save your life?

Chris M. Thomasson

2024-01-11 02:50:39 UTC

Post by WM
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.

Post by Richard Damon
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.

Post by Richard Damon
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

Are you sure about that? There is no smallest unit fraction...

Post by WM
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
is valid for all unit fractions.
But you cannot name it.

Post by Richard Damon
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.

Post by WM
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.

Post by Richard Damon
"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.

Post by Richard Damon
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

zelos...@gmail.com

2024-01-08 06:37:51 UTC

Post by ***@gmail.com
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

Archimedes Plutonium

2024-01-10 23:02:42 UTC

Please resign Joe Biden and let Kamala finish out your term.

.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.

Good on you Kamala Harris. And here second action in presidency, is tell Netanyahu to remove all his IDF force out of Gaza tomorrow or Kamala sends in the US Marines to start to set up the Palestinian state. The next day Kamala tells Netanyahu to get all **illegal Israeli settlers out of West Bank** or she sends in the US Marines there also.

Israel is on course to spark a Nuclear World War 3 on the trajectory it is on as that of stealing land that was once Ancient King David's land. That is all Gaza is to Israel, a chance to steal that land. If Americans were living in Gaza, instead of Palestinians, there would be no difference in the behavior of Israel towards those Americans, because to Israel, nothing, nothing is more important than King David's land.

AP

P.S. Sorry Joe, you are a nice guy, but often in life, nice guys are fools that never realize their time as president is long gone. You are too old, and in 2020, you yourself even said "I be a one term president". So make full on your pledge and resign, for the good of the USA, resign. And let fresh new blood take over, let Kamala for you ran out of gas a long time ago-- those 10 Republicans rule Washington for you lost control last July. And do the honorable thing, just as the New Zealand prime minister showed the world-- she stepped down-- she ran out of gas. Biden, follow her example for you surely have lost all the gas and your tank is completely empty.

It is part of the reason the presidents of the USA have white hair after 4 years-- they never get enough sleep. They are constantly interrupted, day and night. And I myself at age 74 this year need at least 10 hours of sleep sometimes 12 hours. I can only imagine that Joe Biden age 82 gets at best 5 hours of sleep.

Resign Joe and let Kamala finish out this last year. She probably will do an excellent job and be reelected in November.

What is certain-- Joe, you failed last July as 10 Republicans run the USA now. And so be, dignified, Joe, resign for the good of USA.

AP, King of Science

Medical doctors declare Biden too old to be president--Kamala sworn in// Supreme Court for once in 30 years follows the Constitution and determines Trump is banned from public office// President Harris needs to send Ukraine 50 F-16s with volunteer ..

Read my recent posts in peace and quiet. If you, the reader, is wondering why AP posts this to a thread which is off topic in sci.math or sci.physics, is because some stalkers track AP, such as kibo, dan, jan who have been paid to stalk for 3 decades and when they see AP trying to post to his own thread that is on-topic they throw a impossible reCAPTcha suppression and repression at me that only wastes my time. From what AP can make out-- Google is not the only one using reCAPTcha, apparently the US govt rents out reCAPTcha. So if you see a AP post in a thread off topic, is because kooks of reCAPTcha are making it impossible for AP to post to the on-topic thread.

Read all of AP's post in peace and quiet in his newsgroup-- what sci.physics and sci.math should look like when govt spammers are not allowed in a newsgroup to wreck the newsgroup. Govt spammers have their agenda of drag net spam, and then their agenda of spy message codes, such as the "i sick, i cry" baloney, which only ends up ruining the newsgroups and why Google decided to close shop having fought govt bureaucrat mind sets for 30 years, and time to close shop.

AP kindly asks Google to let AP run all three, sci.math, sci.physics, PAU as he runs PAU, now--- all pure science, no spam and no govt b.s.

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
Archimedes Plutonium

Chris M. Thomasson

2024-01-11 03:57:46 UTC

Post by Archimedes Plutonium
Please resign Joe Biden and let Kamala finish out your term.
.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.
Good on you Kamala Harris. And here second action in presidency, is tell Netanyahu to remove all his IDF force out of Gaza tomorrow or Kamala sends in the US Marines to start to set up the Palestinian state. The next day Kamala tells Netanyahu to get all **illegal Israeli settlers out of West Bank** or she sends in the US Marines there also.

[..]

You want to send in US marines to help out Hamas? Are you kidding! lol.

FromTheRafters

2024-01-11 09:15:20 UTC

Post by Archimedes Plutonium
Please resign Joe Biden and let Kamala finish out your term.
.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.
Good on you Kamala Harris. And here second action in presidency, is tell
Netanyahu to remove all his IDF force out of Gaza tomorrow or Kamala sends
in the US Marines to start to set up the Palestinian state. The next day
Kamala tells Netanyahu to get all **illegal Israeli settlers out of West
Bank** or she sends in the US Marines there also.

[..]
You want to send in US marines to help out Hamas? Are you kidding! lol.

I didn't see Hamas mentioned at all.

Chris M. Thomasson

2024-01-11 19:56:51 UTC

Post by Archimedes Plutonium
Please resign Joe Biden and let Kamala finish out your term.
.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.
Good on you Kamala Harris. And here second action in presidency, is
tell Netanyahu to remove all his IDF force out of Gaza tomorrow or
Kamala sends in the US Marines to start to set up the Palestinian
state. The next day Kamala tells Netanyahu to get all **illegal
Israeli settlers out of West Bank** or she sends in the US Marines
there also.

[..]
You want to send in US marines to help out Hamas? Are you kidding! lol.

I didn't see Hamas mentioned at all.

Who do you think is the militant arm of the Gaza strip?

FromTheRafters

2024-01-11 20:33:38 UTC

Post by Archimedes Plutonium
Please resign Joe Biden and let Kamala finish out your term.
.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.
Good on you Kamala Harris. And here second action in presidency, is tell
Netanyahu to remove all his IDF force out of Gaza tomorrow or Kamala
sends in the US Marines to start to set up the Palestinian state. The
next day Kamala tells Netanyahu to get all **illegal Israeli settlers out
of West Bank** or she sends in the US Marines there also.

[..]
You want to send in US marines to help out Hamas? Are you kidding! lol.

I didn't see Hamas mentioned at all.

Who do you think is the militant arm of the Gaza strip?

Are you conflating Hamas with Palestinians?

Chris M. Thomasson

2024-01-11 20:34:56 UTC

Post by Archimedes Plutonium
Please resign Joe Biden and let Kamala finish out your term.
.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.
Good on you Kamala Harris. And here second action in presidency, is
tell Netanyahu to remove all his IDF force out of Gaza tomorrow or
Kamala sends in the US Marines to start to set up the Palestinian
state. The next day Kamala tells Netanyahu to get all **illegal
Israeli settlers out of West Bank** or she sends in the US Marines
there also.

[..]
You want to send in US marines to help out Hamas? Are you kidding! lol.

I didn't see Hamas mentioned at all.

Who do you think is the militant arm of the Gaza strip?

Are you conflating Hamas with Palestinians?

No. Afaict, Hamas is the "power" there, and the Palestinians are caught
in the middle. Is that wrong to think that way?

Chris M. Thomasson

2024-01-12 23:52:34 UTC

Post by Archimedes Plutonium
Please resign Joe Biden and let Kamala finish out your term.
.. with volunteer skilled pilots so that Ukraine war ends by Spring of 2024.
Good on you Kamala Harris. And here second action in presidency,
is tell Netanyahu to remove all his IDF force out of Gaza tomorrow
or Kamala sends in the US Marines to start to set up the
Palestinian state. The next day Kamala tells Netanyahu to get all
**illegal Israeli settlers out of West Bank** or she sends in the
US Marines there also.

[..]
You want to send in US marines to help out Hamas? Are you kidding! lol.

I didn't see Hamas mentioned at all.

Who do you think is the militant arm of the Gaza strip?

Are you conflating Hamas with Palestinians?

No. Afaict, Hamas is the "power" there, and the Palestinians are caught
in the middle. Is that wrong to think that way?

Afaict, Hamas is the evil power there. Akin to a demon...

Yikes! ;^o

Archimedes Plutonium

2024-01-11 03:56:29 UTC

Mueckenheim weighs Dr.Gerd Binnig,Dr.Horst Ludwig Stormer,Dr.Peter Grunberg for no-one at Goethe or Gottingen can do a proper Water Electrolysis and actually weigh the hydrogen and oxygen. No, these fools only look at volume.

Gottingen Uni,Dr. Sarah Köster,Dr. Reiner Kree,Dr. Matthias Krüger

Volney Physics failures..Gottingen Uni,Dr. Sarah Köster,Dr. Reiner Kree,Dr. Matthias Krüger, NSF Dr.Panchanathan,Alejandro Adem, Purdue Univ_France Cordova,

Jan Burse and Volney, why this eternal september spam cluttering up the newsgroup.

Why Volney?? Because they are so sloppy and slipshod in Physics experiment of Water Electrolysis, stopping and ceasing the experiment before weighing the mass of the hydrogen compared to mass of oxygen. Is it that they are stupid silly thinking volume and mass are the same. For AP needs to prove decisively, if Water is really H4O or H2O. And of course, this experiment would destroy the Standard Model-- that post-diction theory of physics that never gave a single prediction in all of its tenure.

And they even know that a weighing balance of Quartz Crystal MicroBalance has been around since the 1960s, what are they waiting for???

Or is it because they cannot admit the truth of math geometry that slant cut of cone is oval, not ellipse for you need the symmetry of slant cut of cylinder to yield a ellipse.

,
Volney
3
WM using AP's TEACHING TRUE MATHEMATICS to teach 13-14 year olds CALCULUS, those heading for Gottingen & Uni Berlin for AP reduced Calculus to its most simple form-- add or subtract 1 from exponent.
9:01 PM


Universitat Augsburg, Germany, rector Sabine Doering-Manteuffel
Math dept Ronald H.W.Hoppe, B. Schmidt, Sarah Friedrich, Stefan Grosskinsky, Friedrich Pukelsheim, Mirjam Dur, Ralf Werner.

Hochschule Augsburg, Wolfgang Mueckenheim

Eternal-September.org
Wolfgang M. Weyand
Berliner Strasse
Bad Homburg

Goethe Universitat Physics dept

Brigitta Wolff president

Jurgen Habermass
Horst Stocker
Gerd Binnig
Horst Ludwig Stormer
Peter Grunberg

math
Alex Kuronya
Martin Moller
Jakob Stix
Annette Werner
Andreas Bernig
Esther Cabezas-Rivas
Hans Crauel
Thomas Gerstner
Bastian von Harrach
Thomas Mettler
Tobias Weth
Amin Coja-Oghlan
Raman Sanyal
Thorsten Theobald
Yury Person

Apparently Kibo realized he was a science failure when he could not even do a proper percentage. But then one has to wonder how much he paid to bribe Rensselaer to graduate from the school in engineering unable to do a percentage properly???? For I certainly would not hire a engineer who cannot even do proper percentage.

Silly boy, that's off by more than 12.6 MeV, or 12% of the mass of a muon.
Hardly "exactly" 9 muons.
Or, 938.2720813/105.6583745 = 8.88024338572. A proton is about the mass
of 8.88 muons, not 9. About 12% short.

Why Volney?? Because they stop short of completing the Water Electrolysis Experiment by only looking at volume, when they are meant to weigh the mass of hydrogen versus oxygen?? Such shoddy minds in experimental physics and chemistry.

Rensselaer Polytechnic Institute Physics dept Dr.Martin Schmidt (ee), Dr.Ivar Giaever
Vincent Meunier, Ethan Brown, Glenn Ciolek, Julian S. Georg, Joel T. Giedt, Yong Sung Kim, Gyorgy Korniss, Toh-Ming Lu, Charles Martin, Joseph Darryl Michael, Heidi Jo Newberg, Moussa N'Gom, Peter Persans, John Schroeder, Michael Shur, Shawn-Yu Lin, Humberto Terrones, Gwo Ching Wang, Morris A Washington, Esther A. Wertz, Christian M. Wetzel, Ingrid Wilke, Shengbai Zhang

Rensselaer math department
Donald Schwendeman, Jeffrey Banks, Kristin Bennett, Mohamed Boudjelkha, Joseph Ecker, William Henshaw, Isom Herron, Mark H Holmes, David Isaacson, Elizabeth Kam, Ashwani Kapila, Maya Kiehl, Gregor Kovacic, Peter Kramer, Gina Kucinski, Rongjie Lai, Fengyan Li, Chjan Lim, Yuri V Lvov, Harry McLaughlin, John E. Mitchell, Bruce Piper, David A Schmidt, Daniel Stevenson, Yangyang Xu, Bulent Yener, Donald Drew, William Siegmann

Physics minnow

What warning is that Kibo Parry failure of science-- warning that insane persons like Kibo Parry Moroney Volney spends their entire life in a hate-mill, never doing anything in science itself. And paid to stalk hate spew

Kibo Parry Moroney-Volney blowing his cover with the CIA in 1997

Re: Archimedes Vanadium, America's most beloved poster

---quoting Wikipedia ---

Controversy
Many government and university installations blocked, threatened to block, or attempted to shut-down The World's Internet connection until Software Tool & Die was eventually granted permission by the National Science Foundation to provide public Internet access on "an experimental basis."

--- end quote ---

NATIONAL SCIENCE FOUNDATION
Dr. Panchanathan , present day
NSF Dr. Panchanathan, F. Fleming Crim, Dorothy E Aronson, Brian Stone, James S Ulvestad (math), Rebecca Lynn Keiser, Vernon D. Ross, Lloyd Whitman, John J. Veysey (physics), Scott Stanley
France Anne Cordova
Subra Suresh (bioengineer)
Arden Lee Bement Jr. (nuclear engineering)
Rita R. Colwell (microbiology)
Neal Francis Lane
John Howard Gibbons 1993
Barry Shein, kibo parry std world
Jim Frost, Joe "Spike" Ilacqua
Canada-- NSERC , Alejandro Adem (math) , Navdeep Bains, Francois-Philippe Champagne

News starting to come in that AP's Water Electrolysis Experiment proves the true formula of Water is H4O, not H2O is starting to come in.

Aug 30, 2023, 10:19:20 PM
to Plutonium Atom Universe
News starting to come in that AP's Water Electrolysis Experiment proves the true formula of Water is H4O, not H2O is starting to come in.
I received a letter today of Experiment results on Water Electrolysis of weighing the hydrogen test tube versus oxygen test tube and the result is 1/4 atomic mass units of Hydrogen compared to Oxygen.
The researcher weighing 1600 micrograms of hydrogen, using a Eisco-Brownlee-Water-Electrolysis Apparatus.
Using sulfuric acid as electrolyte on ultra pure water. Using low voltage DC of 1.5 volts, 1 amp.
I am not surprised that news of the true formula of Water is H4O comes so quickly. For not much in science is more important than knowing the truth of Water. And this means the start of the complete downfall and throwing out the sick Standard Model of Physics, for it is such an insane theory that it cannot get passed the idea of its subatomic particles as stick and ball, with no job, no function, no task. The Standard Model of Physics, is crazy insane physics for it is all postdiction, never prediction. The idea that the hydrogen atom is H2 not H, is because of the prediction of Atom Totality Theory where a atom is a proton torus with muon inside doing the Faraday law and all atoms require at least 1 capacitor. That means the one proton in H2 serves as a neutron to the other proton, storaging the electricity produced by the other proton.
The true Hydrogen Atom is H2 for all atoms need at least one capacitor, and one of the protons in H2 serves as a neutron.
Sad that chemistry and physics throughout the 20th century were too stupid to actually weigh the mass of hydrogen and oxygen in electrolysis, no, the ignorant fools stopped at looking when they saw the volume of hydrogen was twice that of oxygen. A real scientist is not that shoddy and slipshod ignorant, the real scientist then proceeds with -- let us weigh the hydrogen test tube mass versus the oxygen test tube mass.
Thanks for the news!!!!!
AP
News starting to come in that AP's Water Electrolysis Experiment proves the true formula of Water is H4O, not H2O is starting to come in.
There is another experiment that achieves the same result that Water is truly H4O and not H2O, but I suspect this second method is hugely fraught with difficulty.
The prediction of H4O comes from the Physics idea that a Atom is composed, all atoms mind you, is composed of a proton torus with muon/s inside going round and round thrusting through the torus in the Faraday law and producing electricity. So that when you have Hydrogen without a neutron, there is no way to collect the electricity produced by the Faraday law. Think of it as a automobile engine, you cannot have a engine if there is no crank shaft to collect the energy from the thrusting piston inside the crankcase.
Same thing with an Atom, it needs 3 parts-- muon as bar magnet, proton as torus of coils, and a capacitor to storage the produced electricity. If one of those parts is missing, the entity is a Subatomic particle and not a atom.
So, when we have Hydrogen as a proton with muon inside, it is not a Atom, until it has a neutron, or, has another proton of hydrogen H2, then it is a Atom.
So that H2 is not a molecule but a Atom. H alone is a subatomic particle.
Much harder than Water Electrolysis.
We need to get two identical containers.
We need to be able to make pure heavy-water with deuterium. Deuterium is proton + neutron as hydrogen. Proton + proton is H2 as hydrogen.
So we have two identical containers and we fill one with pure heavy water, deuterium water.
We have the second container and we fill it with pure (light) water.
We now weigh both of them.
If AP is correct, that water is really H4O and not H2O, then both containers should weigh almost the same. Only a tiny fraction difference because the neutron is known to be 940MeV versus proton in Old Physics as 938MeV a tiny difference of 2MeV, but we realize we have a huge number of water molecules in the two identical containers.
If water is truly H4O, the weights should be almost the same. If water is H2O, then there is a **large difference** in weights.
But the Water Electrolysis experiment is much easier to conduct and get results.
And, there is the biological processes that apparently cannot distinguish between heavy water and that of regular normal water.
Deuterium Water is the same in biology as is normal regular water. This means that water must be H4O, due to biology as proof.
Deuterium Water in atomic mass units is 16 for the oxygen and 4 for the deuterium.
Regular normal Water in atomic mass units is 16 for the oxygen and 4 for the 4 protons in H4O.
Old Physics and Old Chemistry had regular water as H2O in atomic mass units of 16 oxygen and 2 hydrogen for 2 protons.
If biology functions whether heavy water or normal water all the same, then water itself must be H4O.
Now, there maybe some animal or plant that can separate out heavy water from H4O water???
Searching the literature today for where biology needs as essential deuterium water. And not too surprised that it is a essential requirement in metabolism. In fact one web site listed the need for deuterium more than the need of many minerals and vitamins.
Now tonight I came up with two new exciting experiments to verify that Water is truly H4O and not H2O.
H4O is 4 protons with muons inside the 840MeV proton toruses.
Deuterium water is DOD. And the difference between D2O and H4O is merely the difference of 4MeV for as last reported, neutron = 940MeV and proton (with muon inside) is 938MeV, a difference of 2MeV but for water is 2+2 = 4MeV.
So these two new experiments take advantage of the fact that what we think is normal regular water is actually very close to heavy water of D2O, with only a 4MeV difference.
EXPERIMENT #3 Water layers in still pond of D2O mixed with H4O (what we thought was H2O.
So in this experiment we get a clear glass container and mix H4O with D2O. If Old Physics is correct, the heavy water should sink rapidly in the container while the light water floats to the top rapidly. And we have some sort of beam of photons that can distinguish D2O from H4O (thought of as H2O. We obtain pure D2O and pure H4O each filling 1/2 of the container. We stir and mix them. And then we observe with the EM beam for separation. If the light water is truly H4O, it takes a long time for the D2O to be on bottom and H4O on top. We measure the time of a settled container and determine this time from the theoretical 4MeV difference should take a long time, whereas if Old Physics is correct, the separation would be almost instantly and quick time.
EXPERIMENT #4 also plays on this minor difference of 4MeV. We devise a sort of squirt gun for D2O and a identical squirt gun for H4O (what we call H2O). We put pure D2O in one squirt gun and the H40 or light water in the other squirt gun. Both guns forcing the water a certain distance.
If AP is correct that light water is really H4O and not H2O as we squirt both guns, where the water lands should be almost the same distance considering H4O is only 4MeV apart from D2O.
If Old Physics and Old Chemistry is correct, then D2O water is 940 + 940 = 1880MeV apart from light water of H2O, and H4O is only 4MeV apart.
So where the squirt gun lands the D2O is a very much shorter distance than a H2O land, but a H4 land distance is nearly the same as the D2O land.
These two experiments are very exciting and would be a very nice confirming evidence to Water Electrolysis actual weighing the mass in atomic mass units.

Searching the literature today for where biology needs as essential deuterium water. And not too surprised that it is a essential requirement in metabolism. In fact one web site listed the need for deuterium more than the need of many minerals and vitamins.
Now tonight I came up with two new exciting experiments to verify that Water is truly H4O and not H2O.
H4O is 4 protons with muons inside the 840MeV proton toruses.
Deuterium water is DOD. And the difference between D2O and H4O is merely the difference of 4MeV for as last reported, neutron = 940MeV and proton (with muon inside) is 938MeV, a difference of 2MeV but for water is 2+2 = 4MeV.
So these two new experiments take advantage of the fact that what we think is normal regular water is actually very close to heavy water of D2O, with only a 4MeV difference.
EXPERIMENT #3 Water layers in still pond of D2O mixed with H4O (what we thought was H2O.
So in this experiment we get a clear glass container and mix H4O with D2O. If Old Physics is correct, the heavy water should sink rapidly in the container while the light water floats to the top rapidly. And we have some sort of beam of photons that can distinguish D2O from H4O (thought of as H2O. We obtain pure D2O and pure H4O each filling 1/2 of the container. We stir and mix them. And then we observe with the EM beam for separation. If the light water is truly H4O, it takes a long time for the D2O to be on bottom and H4O on top. We measure the time of a settled container and determine this time from the theoretical 4MeV difference should take a long time, whereas if Old Physics is correct, the separation would be almost instantly and quick time.

Apparently this Experiment is already done and called for-- There is Uniform Distribution of heavy water Deuterium Water in the Oceans, Lakes, Ponds, Streams and Rivers. Heavy Water is not layered in the oceans or lakes or ponds or streams or rivers. Uniformity means that the difference between D2O and H4O is so slight of a difference (only 4MeV, compared to 1880MeV for H2O, that Brownian motion keeps the D2O and H4O in a Uniform Distribution in all bodies of water. I was going through the research literature today and find that scientists discover Uniformity of Distribution of deuterium water. This thus closes the case on Water, for uniformity of distribution of D2O implies that Water is itself H4O and not H2O.

My 250th published book.

TEACHING TRUE CHEMISTRY; H2 is the hydrogen Atom and water is H4O, not H2O// Chemistry
by Archimedes Plutonium (Author) (Amazon's Kindle)
Prologue: This textbook is 1/2 research history and 1/2 factual textbook combined as one textbook. For many of the experiments described here-in have not yet been performed, such as water is really H4O not H2O. Written in a style of history research with date-time markers, and fact telling. And there are no problem sets. This book is intended for 1st year college. Until I include problem sets and exercises, I leave it to the professor and instructor to provide such. And also, chemistry is hugely a laboratory science, even more so than physics, so a first year college student in the lab to test whether Water is really H4O and not H2O is mighty educational.
Preface: This is my 250th book of science, and the first of my textbooks on Teaching True Chemistry. I have completed the Teaching True Physics and the Teaching True Mathematics textbook series. But had not yet started on a Teaching True Chemistry textbook series. What got me started on this project is the fact that no chemistry textbook had the correct formula for water which is actually H4O and not H2O. Leaving the true formula for hydroxyl groups as H2O and not OH. But none of this is possible in Old Chemistry, Old Physics where they had do-nothing subatomic particles that sit around and do nothing or go whizzing around the outside of balls in a nucleus, in a mindless circling. Once every subatomic particle has a job, task, function, then water cannot be H2O but rather H4O. And a hydrogen atom cannot be H alone but is actually H2. H2 is not a molecule of hydrogen but a full fledged Atom, a single atom of hydrogen.
Cover Picture: Sorry for the crude sketch work but chemistry and physics students are going to have to learn to make such sketches in a minute or less. Just as they make Lewis diagrams or ball & stick diagrams. My 4-5 minute sketch-work of the Water molecule H4O plus the subatomic particle H, and the hydrogen atom H2. Showing how one H is a proton torus with muon inside (blue color) doing the Faraday law. Protons are toruses with many windings. Protons are the coils in Faraday law while muons are the bar magnets. Neutrons are the capacitors as parallel plates, the outer skin cover of atoms.
Product details
• ASIN ‏ : ‎ B0CCLPTBDG
• Publication date ‏ : ‎ July 21, 2023
• Language ‏ : ‎ English
• File size ‏ : ‎ 788 KB
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Enabled
• Sticky notes ‏ : ‎ On Kindle Scribe
• Print length ‏ : ‎ 168 pages

Earle

2024-01-19 01:40:32 UTC

Post by ***@gmail.com
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.

*
Mitchell:

Is zero a point on the number line?

Thanks,

earle
*

Chris M. Thomasson

2024-01-20 01:37:01 UTC

Post by Earle

Post by ***@gmail.com
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.

*
Is zero a point on the number line?

[...]

(0)-->(1/4)-->(1/2)-->(3/4)-->(1)

?

What about:

(-2)-->(-1)-->(0)-->(1)-->(2)

?

Earle

2024-01-20 01:25:45 UTC