On 12/20/2023 2:07 AM, WM wrote:
> Le 19/12/2023 à 15:14, Richard Damon a écrit :
>
>> NUF actually does those finite steps at infinitesimals,
>
> Whatever you call them, these steps cannot be investigated as
> individuals. They are dark.
>> your NUF fails to be defined for x > 0.
>
> The function is defined but its values are dark for small x.

Huh? Wtf does that even mean? Are you trying to create a 100% artificial
"border" wrt infinity that the moron AP seems to like to do?

Oh wait a minute... Is small x referring to the number of active neurons
in you brain, WM?

>

On 12/21/2023 5:45 AM, WM wrote:
> Le 20/12/2023 à 23:16, Richard Damon a écrit :

>> But all ARE "accessible", just unbounded.
>
> Between NUF(0) = 0 and
> for every x > 0: NUF(x) = ℵ₀
> there are ℵ₀ unit fractions,
> in a non-vanishing part of the real axis,
> [...]

Yes.
Consider x > 0

⅟ℕₓ is the set of unit fractions in (0,x]

mₓ-1 < ⅟x ≤ mₓ ∈ ℕ
⅟mₓ is the largest unit fraction in ⅟ℕₓ

uₓ(n) = ⅟(mₓ+n)
uₓ: ℕ ↣ ⅟ℕₓ
uₓ is 1.to.1

For each number n ∈ ℕ
n has its own unit fraction uₓ(n) ∈ (0,x]

There are at least as many
unit fractions ∈ ⅟ℕₓ as
numbers ∈ ℕ

|ℕ| ≤ |⅟ℕₓ|
|ℕ| = ℵ₀

> Between NUF(0) = 0 and
> for every x > 0: NUF(x) = ℵ₀
> there are ℵ₀ unit fractions,
> in a non-vanishing part of the real axis,
> which cannot be chosen by any x.

No.
For each x > 0
for each unit.fraction u of ℵ₀.many in ⅟ℕₓ
there is an xᵤ > 0
by which u "can be chosen"
meaning
such that u ∈ (xᵤ,1]

Proof.
Consider xᵤ = u/2

Thus,
for each x > 0
there are 0.many
unit.fractions in ⅟ℕₓ which "cannot be chosen".

----
A quantifier shift is unreliable.

Consider a table.setting for four:
four glasses 1234, four plates NSEW

If
for each glass,
there is a plate such that
the glass is with the plate,

then
the glasses could be placed like
N1234 S E W
there is a plate such that,
for each glass,
the glass is with the plate

and
the glasses could be placed like
N1 S2 E3 W4
there ISN'T a plate such that,
for each glass,
the glass is with the plate

Le 21/12/2023 à 21:08, Richard Damon a écrit :
> On 12/21/23 2:35 PM, WM wrote:

>>>>> But all ARE "accessible", just unbounded.
>>>>
>>>> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
>>>> fractions, in a non-vanishing part of the real axis, which cannot be
>>>> chosen by any x. How would you choose one of them? Don't claim that
>>>> it could be done. Show how it can be done!
>>>>
>>>
>>> What unit fraction can't be chosen?
>>
>> Those which cannot be chosen, almost all.
>
> If you can't show them, you can't say they exist.

The proof of existence is taken from the assumption of actual infinity and
of the mathematical formula here:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
>
> Your "proof" is you can't see them, so they must be "dark", but they
> don't actually need to exist.
>
> All your arguement shows is that they are not bounded, but that was
> already known.

The fractions have a greatest lower bound, namely zero. It is called their
infimum.

> Your logic system just can't handle unbounded sets,

The unit fractions, like all positive fractions have an infimum, that is a
bound.

>>
>> Of course, but ℵo unit fractions, almost all unit fractions, remain
>> unchosen.
>
> Please describe which one can't be?
>
> Remember, ALL Natural Numbers ARE describable,

You are wrong. It can be seen by the fact that
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
The set of undefined natnumbers will never be emptied.

> But if there was, then there is also a smalller one (since x/2 will also
> be a unit fraction, and smaller than it)

That is provable for visible unit fractions only.

Regards, WM

Le 22/12/2023 à 05:19, Jim Burns a écrit :
> On 12/21/2023 5:45 AM, WM wrote:

>
>> Between NUF(0) = 0 and
>> for every x > 0: NUF(x) = ℵ₀
>> there are ℵ₀ unit fractions,
>> in a non-vanishing part of the real axis,
>> which cannot be chosen by any x.
>
> No.
> For each x > 0
> for each unit.fraction u of ℵ₀.many in ⅟ℕₓ
> there is an xᵤ > 0
> by which u "can be chosen"
> meaning
> such that u ∈ (xᵤ,1]

ℵ₀ unit fractions populate at least ℵ₀ points which cannot be
left-hand of all points, because they are just these points. "There does
not exist a point of (0, 1] with less than ℵ₀ smaller unit fractions"
is so simple to prove wrong (like disappearing Bob), that it should no
longer be claimed. Therefore all your arguing in favour of this nonsense
will no longer be read by me.

Regards, WM

On 12/22/23 3:23 AM, WM wrote:
> Le 21/12/2023 à 21:08, Richard Damon a écrit :
>> On 12/21/23 2:35 PM, WM wrote:
>
>>>>>> But all ARE "accessible", just unbounded.
>>>>>
>>>>> Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀
>>>>> unit fractions, in a non-vanishing part of the real axis, which
>>>>> cannot be chosen by any x. How would you choose one of them? Don't
>>>>> claim that it could be done. Show how it can be done!
>>>>>
>>>>
>>>> What unit fraction can't be chosen?
>>>
>>> Those which cannot be chosen, almost all.
>>
>> If you can't show them, you can't say they exist.
>
> The proof of existence is taken from the assumption of actual infinity
> and of the mathematical formula here:
> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .

Please define your assumption of "actual infinity".

If you mean that some n ∈ ℕ is actually infinite, that is a false statement.

If you mean that ℕ is infinite in size, how does that imply what you mean.

And How do you claim "proof" from that statement?

>>
>> Your "proof" is you can't see them, so they must be "dark", but they
>> don't actually need to exist.
>>
>> All your arguement shows is that they are not bounded, but that was
>> already known.
>
> The fractions have a greatest lower bound, namely zero. It is called
> their infimum.

Nope, since Zero isn't a member of the set, it isn't a "Bound" but a limit.

>
>> Your logic system just can't handle unbounded sets,
>
> The unit fractions, like all positive fractions have an infimum, that is
> a bound.

Nope.

Quoting a definition:

In mathematics, the infimum (abbreviated inf; plural infima) of a subset
S of a partially ordered set P is the greatest element in P that is less
than or equal to each element of S *if such an element exists.*

Note also:

Infima and suprema do not necessarily exist. Existence of an infimum of
a subset S of P can fail if S has no lower bound at all, or if the set
of lower bounds does not contain a greatest element.

Since Unit Fractions are UNBOUNDED, that lower bound does not exist as a
member of the set, so on infimum exists.

>
>>>
>>> Of course, but ℵo unit fractions, almost all unit fractions, remain
>>> unchosen.
>>
>> Please describe which one can't be?
>>
>> Remember, ALL Natural Numbers ARE describable,
>
> You are wrong. It can be seen by the fact that
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> The set of undefined natnumbers will never be emptied.

So? That is just a property of using Bounded operations on Unbounded sets.

>
>> But if there was, then there is also a smalller one (since x/2 will
>> also be a unit fraction, and smaller than it)
>
> That is provable for visible unit fractions only.

But you can't actually prove that your "dark" numbers exist, and your
proof requires showing the existance of some lower bound on the visible
numbers to create the dark numbers.

Your selection criteria is just invalid.

It is like saying that while many cats are "selectable", there exists a
set of "dark" cats, that are also dogs, that are only usable as a
collection.

The selection criteria, (cats that are dogs) while showing that no
member is selectable, doesn't show that there are actually members of
that set.

This is the problem of Russel's Teapot, which your proof is dependent on
existing.

>
> Regards, WM
>
>

On 12/22/2023 3:30 AM, WM wrote:
> Le 22/12/2023 à 05:19, Jim Burns a écrit :
>> On 12/21/2023 5:45 AM, WM wrote:

>>> Between NUF(0) = 0 and
>>> for every x > 0: NUF(x) = ℵ₀
>>> there are ℵ₀ unit fractions,
>>> in a non-vanishing part of the real axis,
>>> which cannot be chosen by any x.
>>
>> No.
>> For each x > 0
>> for each unit.fraction u of ℵ₀.many in ⅟ℕₓ
>> there is an xᵤ > 0
>> by which u "can be chosen"
>> meaning
>> such that u ∈ (xᵤ,1]
>>
>> Proof.
<> Consider xᵤ = u/2
>
> ℵ₀ unit fractions populate
> at least ℵ₀ points

at least and at most ℵ₀ points

> which cannot be left-hand of all points,
> because they are just these points.

These points are always these points.
How does that connect to
being left-hand of all points?

What I suspect is that
you (WM) cope with ℵ₀.many points by
fading them into or out of darknessᵂᴹ,
Cheshire.cat.like,
which, through logical explosion,
provides you with means to "justify"
whatever your intuition tells you.

What I suspect is that
points which are just.these.pointsᵂᴹ
are points you've finished fading.
Completedᵂᴹ points.

None of this works like that. Definitely none.
But you might think it does work like that, and
that anything opposed to it violates logic.

> "There does not exist
> a point of (0, 1] with [fewer] than
> ℵ₀ smaller unit fractions"

¬∃x ∈ (0,1]:
¬∀n ∈ ℕ₁:
⅟(n+⌊⅟x⌋) ∈ (0,x)∩⅟ℕ₁ ∧
¬∃n′≠n: ⅟(n′+⌊⅟x⌋) = ⅟(n+⌊⅟x⌋)

Because arithmetic and 0 < ⅟n⁺¹ < ⅟n

> is so simple to prove wrong
> (like disappearing Bob),

Your simple and wrong proof
uses an unjustified quantifier shift.

You doing the equivalent of:
from being told that,
| for each glass, there is a plate such that
| the glass is by the plate
| N1 S2 E3 W4
| N1234 S E W
| N12 S E34 W
| N S1 E W234
| ...
"inferring" that
| there is a plate such that, for each glass,
| the glass is by the plate
| N1234 S E W
| N S1234 E W
| N S E1234 W
| N S E W1234

That "inference" is unjustified.

> that it should no longer be claimed.
> Therefore all your arguing in favour of
> this nonsense will no longer be read by me.

In a finite sequence of claims,
if we see each claim is
not.first.false about visibleᵂᴹ and darkᵂᴹ
then we know each claim is
true about visibleᵂᴹ and darkᵂᴹ

Fading visibleᵂᴹ.darkᵂᴹ and darkᵂᴹ.visibleᵂᴹ points,
by logical explosion, gives us anything.we.ask.for.

Anything.we.ask.for is the sugar.high of
confirming our pre.analyzed beliefs.
Sugar.high is followed by sugar.crash of
finding out that real is going to real.

Le 22/12/2023 à 14:56, Richard Damon a écrit :
> On 12/22/23 3:23 AM, WM wrote:

>>>>> What unit fraction can't be chosen?
>>>>
>>>> Those which cannot be chosen, almost all.
>>>
>>> If you can't show them, you can't say they exist.
>>
>> The proof of existence is taken from the assumption of actual infinity
>> and of the mathematical formula here:
>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
>
> Please define your assumption of "actual infinity".

The elements of a set exist such that none can be added.

> If you mean that some n ∈ ℕ is actually infinite

No n is infinite and no 1/n is zero. Therefore NUF(x) increases only after
0 for the first time.

> Since Unit Fractions are UNBOUNDED, that lower bound does not exist as a
> member of the set,

That would be a minimum. The infimum is explained here:
https://en.wikipedia.org/wiki/Infimum_and_supremum
The set of unit fractions has the infimum 0.

>> The set of undefined natnumbers will never be emptied.
>
> So? That is just a property of using Bounded operations on Unbounded sets.

No, the set of unit fractions is bounded.

Regards, WM

Le 22/12/2023 à 16:44, Jim Burns a écrit :
> Your simple and wrong proof
> uses an unjustified quantifier shift.

No! If every point of (0, 1] has ℵo smaller unit fractions,
then no point is existing in (0, 1] with less than ℵo smaller unit
fractions,
then the interval (0, 1] has ℵo smaller unit fractions.

This is applied mathematics. Your quantifier delusions may be accepted by
weak mathematicians in the infinite but it is inapplicable in concrete
mathematics.
>
> You doing the equivalent of:

No, the unit fractions are in linear order. If no point of the set has
less than ℵo smaller unit fractions, then there is no such point. Then
we can sort all points into the set (0, 1]. But in fact only visible
points are concerned, such that only the interval (eps, 1) has ℵo
smaller unit fractions.

Regards, WM

On 12/23/2023 6:37 AM, WM wrote:

> [...]
> with less than ℵo smaller unit fractions,
> [...]

with fewer than ℵo smaller unit fractions,
or
with less than ℵo smaller unit fraction,

https://en.wikipedia.org/wiki/Mass_noun
similar to the German
https://de.wikipedia.org/wiki/Stoffname

On 12/23/23 6:18 AM, WM wrote:
> Le 22/12/2023 à 14:56, Richard Damon a écrit :
>> On 12/22/23 3:23 AM, WM wrote:
>
>>>>>> What unit fraction can't be chosen?
>>>>>
>>>>> Those which cannot be chosen, almost all.
>>>>
>>>> If you can't show them, you can't say they exist.
>>>
>>> The proof of existence is taken from the assumption of actual
>>> infinity and of the mathematical formula here:
>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
>>
>> Please define your assumption of "actual infinity".
>
> The elements of a set exist such that none can be added.

So, you ADMIT that you are just admitting the assumption that "Dark
Numbers" exist.

Please show HOW a Natural Number can exist, that can not be "added".

All Natural Numbers can be chosen and added to the set.

>
>> If you mean that some n ∈ ℕ is actually infinite
>
> No n is infinite and no 1/n is zero. Therefore NUF(x) increases only
> after 0 for the first time.

No, you are assuming that NUF is definable by this set of words. Since
at ANY 1/n, there are an infinite number of unit fractions with values
below it, it never has a finite value.

Your definition assumes an upper bound that just doesn't exist.

>
>> Since Unit Fractions are UNBOUNDED, that lower bound does not exist as
>> a member of the set,
>
> That would be a minimum. The infimum is explained here:
> https://en.wikipedia.org/wiki/Infimum_and_supremum
> The set of unit fractions has the infimum 0.

It CAN'T be because the infimum must be a member of the set, and you
admtted that 0 is not a unit fraction.

So, you LIE.

>
>>> The set of undefined natnumbers will never be emptied.
>>
>> So? That is just a property of using Bounded operations on Unbounded
>> sets.
>
> No, the set of unit fractions is bounded.

Yes, At 1/1, but what is the other bound?

You "Theory" is based on a LIE, and thus is totally unsound.
>
> Regards, WM

Le 23/12/2023 à 14:29, Richard Damon a écrit :
>
> you are assuming that NUF is definable by this set of words.

Number of Unit Fractions between 0 and x.

> Since
> at ANY 1/n, there are an infinite number of unit fractions with values
> below it, it never has a finite value.

NUF(0) = 0, with increasing x it grows. The question is whether it can
grow by more than one unit fractions at one and the same x. The answer is
no, given by mathematics
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>
>>> Since Unit Fractions are UNBOUNDED, that lower bound does not exist as
>>> a member of the set,
>>
>> That would be a minimum. The infimum is explained here:
>> https://en.wikipedia.org/wiki/Infimum_and_supremum
>> The set of unit fractions has the infimum 0.
>
> It CAN'T be because the infimum must be a member of the set,

Try to correct Wikipedia. You are obliged to correct the lie.

Regards, WM

On 12/23/23 9:53 AM, WM wrote:
> Le 23/12/2023 à 14:29, Richard Damon a écrit :
>>
>> you are assuming that NUF is definable by this set of words.
>
> Number of Unit Fractions between 0 and x.
>
>> Since at ANY 1/n, there are an infinite number of unit fractions with
>> values below it, it never has a finite value.
>
> NUF(0) = 0, with increasing x it grows. The question is whether it can
> grow by more than one unit fractions at one and the same x. The answer
> is no, given by mathematics
> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0

It can only grow by one at any finite number, but since there is a
discontuity in the fationals between 0 and the rationals x > 0, it can
jump at that point.

The problem is that you definition of NUF just isn't actually realizable
with the logic you are trying to use, because of a mismatch of your
logic and your system of numbers.

>>
>>>> Since Unit Fractions are UNBOUNDED, that lower bound does not exist
>>>> as a member of the set,
>>>
>>> That would be a minimum. The infimum is explained here:
>>> https://en.wikipedia.org/wiki/Infimum_and_supremum
>>> The set of unit fractions has the infimum 0.
>>
>> It CAN'T be because the infimum must be a member of the set,
>
> Try to correct Wikipedia. You are obliged to correct the lie.

What "Lie"?

That bounds might not exsit?
or that bounds must be a member of the set in question?

THat isn't a lie, it is your misconception.

The set of Unit Fractions can only have a bound of 0 if 0 is a member of
the set of Unit Fractions.

Since 0 isn't a Unit Fraction, it can't be a bound of the Unit Fractions.

Or, what Natural Number generates 0 as a unit fraction?

You are caught in a LIE based on a contradiction of your own making.

If you disagree with the DEFINITION of the term, you can't use that
term, or need to redefine it and admit you are working in a DIFFERENT
field of study, as fields of study include the definitions they are
based on.

>
> Regards, WM
>

On 12/23/2023 5:29 AM, Richard Damon wrote:
> On 12/23/23 6:18 AM, WM wrote:
>> Le 22/12/2023 à 14:56, Richard Damon a écrit :
>>> On 12/22/23 3:23 AM, WM wrote:
>>
>>>>>>> What unit fraction can't be chosen?
>>>>>>
>>>>>> Those which cannot be chosen, almost all.
>>>>>
>>>>> If you can't show them, you can't say they exist.
>>>>
>>>> The proof of existence is taken from the assumption of actual
>>>> infinity and of the mathematical formula here:
>>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
>>>
>>> Please define your assumption of "actual infinity".
>>
>> The elements of a set exist such that none can be added.
>
> So, you ADMIT that you are just admitting the assumption that "Dark
> Numbers" exist.
>
> Please show HOW a Natural Number can exist, that can not be "added".
>
> All Natural Numbers can be chosen and added to the set.
>
>>
>>> If you mean that some n ∈ ℕ is actually infinite
>>
>> No n is infinite and no 1/n is zero. Therefore NUF(x) increases only
>> after 0 for the first time.
>
> No, you are assuming that NUF is definable by this set of words. Since
> at ANY 1/n, there are an infinite number of unit fractions with values
> below it, it never has a finite value.
>
> Your definition assumes an upper bound that just doesn't exist.
>
>>
>>> Since Unit Fractions are UNBOUNDED, that lower bound does not exist
>>> as a member of the set,
>>
>> That would be a minimum. The infimum is explained here:
>> https://en.wikipedia.org/wiki/Infimum_and_supremum
>> The set of unit fractions has the infimum 0.
>
> It CAN'T be because the infimum must be a member of the set, and you
> admtted that 0 is not a unit fraction.
>
> So, you LIE.
>
>>
>>>> The set of undefined natnumbers will never be emptied.
>>>
>>> So? That is just a property of using Bounded operations on Unbounded
>>> sets.
>>
>> No, the set of unit fractions is bounded.
>
>
> Yes, At 1/1, but what is the other bound?
>
> You "Theory" is based on a LIE, and thus is totally unsound.

He is a damn teacher, right? Humm... Poor students! YIKES!

[...]

Le 23/12/2023 à 18:07, Richard Damon a écrit :
> On 12/23/23 9:53 AM, WM wrote:

>> NUF(0) = 0, with increasing x it grows. The question is whether it can
>> grow by more than one unit fractions at one and the same x. The answer
>> is no, given by mathematics
>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
>
> It can only grow by one at any finite number, but since there is a
> discontuity in the fationals between 0 and the rationals x > 0, it can
> jump at that point.

It cannot jump by more than 1.
>
> The problem is that you definition of NUF just isn't actually realizable

It is true for all unit fractions.

>>>>> Since Unit Fractions are UNBOUNDED, that lower bound does not exist
>>>>> as a member of the set,
>>>>
>>>> That would be a minimum. The infimum is explained here:
>>>> https://en.wikipedia.org/wiki/Infimum_and_supremum
>>>> The set of unit fractions has the infimum 0.
>>>
>>> It CAN'T be because the infimum must be a member of the set,
>>
>> Try to correct Wikipedia. You are obliged to correct the lie.
>
> What "Lie"?
>
> That bounds might not exsit?

The infimum of the sequence of unit fractions is 0.

> or that bounds must be a member of the set in question?

That is wrong as I teach already in the first semester.
Try to correct Wikipedia in order to receive information.

Regards, WM

Your game of billiards does not provide an algorithm that defines A.

On 25.12.2023 19:04, William wrote:
>
> Your game of billiards does not provide an algorithm that defines A.

There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.
It stems from Cantor and yields
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... .

Mine is the same 1/k = (m + n - 1)(m + n - 2)/2 + m.

But whether or not A is defined, the matrix
XOOO...
XOOO...
XOOO...
XOOO...
....
will never be covered by X. This shows that never all matrix positions
will be indeXed.

Regards, WM

On 12/26/23 4:52 AM, WM wrote:
> On 25.12.2023 19:04, William wrote:
>>
>> Your game of billiards does not provide an algorithm that defines A.
>
> There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.
> It stems from Cantor and yields
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
> 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... .
>
> Mine is the same 1/k = (m + n - 1)(m + n - 2)/2 + m.
>
> But whether or not A is defined, the matrix
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
> will never be covered by X. This shows that never all matrix positions
> will be indeXed.
>
> Regards, WM
>
>

Because it is not following the actual Algorithm.

The algorithm is DEFINED as drawing k, from the LINE of the Natural
Numbers, and showing that this covers the MATRIX indexed by m,n.

You are holding the gun backwards, no wonder you keep shooting yourself.

On 12/26/2023 4:35 AM, Richard Damon wrote:
> On 12/26/23 4:52 AM, WM wrote:
>> On 25.12.2023 19:04, William wrote:
>>>
>>> Your game of billiards does not provide an algorithm that defines A.
>>
>> There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.
>> It stems from Cantor and yields
>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
>> 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... .
>>
>> Mine is the same 1/k = (m + n - 1)(m + n - 2)/2 + m.
>>
>> But whether or not A is defined, the matrix
>> XOOO...
>> XOOO...
>> XOOO...
>> XOOO...
>> ...
>> will never be covered by X. This shows that never all matrix positions
>> will be indeXed.
>>
>> Regards, WM
>>
>>
>
> Because it is not following the actual Algorithm.
>
> The algorithm is DEFINED as drawing k, from the LINE of the Natural
> Numbers, and showing that this covers the MATRIX indexed by m,n.
>
> You are holding the gun backwards, no wonder you keep shooting yourself.

No shit! God damn.

On Tuesday, December 26, 2023 at 5:52:43 AM UTC-4, WM wrote:
> On 25.12.2023 19:04, William wrote:
> >
> > Your game of billiards does not provide an algorithm that defines A.
> There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.

Correct, This defines B. Your game of billiards does not define A.

Le 26/12/2023 à 13:35, Richard Damon a écrit :
> The algorithm is DEFINED as drawing k, from the LINE of the Natural
> Numbers, and showing that this covers the MATRIX indexed by m,n.

Precisely this happens after the integer fractions have been replaced by
the natural numbers:

1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
...

Should this simple replacement destroy the process, then the whole process
is impossible.
Regards, WM

On 27.12.2023 00:29, William wrote:
> On Tuesday, December 26, 2023 at 5:52:43 AM UTC-4, WM wrote:
>> On 25.12.2023 19:04, William wrote:
>>>
>>> Your game of billiards does not provide an algorithm that defines A.
>> There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.
>
> Correct, This defines B. Your game of billiards does not define A.
>
Why do you mention this boring idea? My game shows that the matrix A is
larger than B and containing not indexed fractions.

But whether or not A is defined, the matrix
XOOO...
XOOO...
XOOO...
XOOO...
....
will never be covered by X. This shows that never all matrix positions
will be indeXed.

Regards, WM

On 12/27/23 3:20 AM, WM wrote:
> Le 26/12/2023 à 13:35, Richard Damon a écrit :
>> The algorithm is DEFINED as drawing k, from the LINE of the Natural
>> Numbers, and showing that this covers the MATRIX indexed by m,n.
>
> Precisely this happens after the integer fractions have been replaced by
> the natural numbers:
>
> 1, 1/2, 1/3, 1/4, ...
> 2, 2/2, 2/3, 2/4, ...
> 3, 3/2, 3/3, 3/4, ...
> 4, 4/2, 4/3, 4/4, ...
> 5, 5/2, 5/3, 5/4, ...
> ..
>
> Should this simple replacement destroy the process, then the whole
> process is impossible.
>
> Regards, WM
>

Nope, you just don't understand how the process actual works.

the entire matrix gets covered by the m,n pairs generated by the process.

The k terms, of your first column, are the INPUTS, not the OUTPUTS of
the process.

So, as I said, you are holding the gun backwards and shooting yourself
with your ignorance.

Le 27/12/2023 à 14:12, Richard Damon a écrit :
> On 12/27/23 3:20 AM, WM wrote:
>> Le 26/12/2023 à 13:35, Richard Damon a écrit :
>>> The algorithm is DEFINED as drawing k, from the LINE of the Natural
>>> Numbers, and showing that this covers the MATRIX indexed by m,n.
>>
>> Precisely this happens after the integer fractions have been replaced by
>> the natural numbers:
>>
>> 1, 1/2, 1/3, 1/4, ...
>> 2, 2/2, 2/3, 2/4, ...
>> 3, 3/2, 3/3, 3/4, ...
>> 4, 4/2, 4/3, 4/4, ...
>> 5, 5/2, 5/3, 5/4, ...
>> ..
>>
>> Should this simple replacement destroy the process, then the whole
>> process is impossible.
>
> The k terms, of your first column, are the INPUTS,

Of course. And everybody sees that they will never cover the whole matrix.

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 27. Dezember 2023 um 16:22:14 UTC+1:

Nonsense about quantifiers.

My proof has nothing to do with quantifier shift.

The proof is this: If before and after every unit fraction 1/n there is
a distance (containing uncountably many points) up to the next unit
fraction, then over this distance the Number of Unit Fractions between 0
and x, NUF(x), is constant. Unless any unit fraction evades this law,
NUF(x) must increase in steps of height 1, starting from NUF(0) = 0.

Regards, WM

On 12/27/23 11:54 AM, WM wrote:
> Le 27/12/2023 à 14:12, Richard Damon a écrit :
>> On 12/27/23 3:20 AM, WM wrote:
>>> Le 26/12/2023 à 13:35, Richard Damon a écrit :
>>>> The algorithm is DEFINED as drawing k, from the LINE of the Natural
>>>> Numbers, and showing that this covers the MATRIX indexed by m,n.
>>>
>>> Precisely this happens after the integer fractions have been replaced
>>> by the natural numbers:
>>>
>>> 1, 1/2, 1/3, 1/4, ...
>>> 2, 2/2, 2/3, 2/4, ...
>>> 3, 3/2, 3/3, 3/4, ...
>>> 4, 4/2, 4/3, 4/4, ...
>>> 5, 5/2, 5/3, 5/4, ...
>>> ..
>>>
>>> Should this simple replacement destroy the process, then the whole
>>> process is impossible.
>>
>> The k terms, of your first column, are the INPUTS,
>
> Of course. And everybody sees that they will never cover the whole matrix.
>
> Regards, WM
>

WHo needs the INPUT to cover, the test was if the OUTPUT cover.

You just shot yourself again, killing your arguement.