Le 16/12/2023 à 02:15, Richard Damon a écrit :
> On 12/15/23 5:32 PM, WM wrote:

>>>> That is not the problem!
>>>> Simply try to distinguish
>>>> any unit fraction which has
>>>> not infinitely many smaller ones.
>>>> Such unit fractions must exist,
>>>> because
>>>> they occupy separate points in a linear order.
>>>
>>> No.
>>
>> Yes. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n.
>> ∀x ∈ (0, 1]: NUF(x) = ℵo is wrong.

> Then name a value that contradicts it!
>
There are ℵo values which cannot be named as individuals.

> Your concept of the function says it only has "finite values>0" for
> input values that can not be used as individual numbers, and thus,
> nowhere is your NUF finite and not 0.

You despise mathematics and logic. An increase from 0 to more must pass 1.

Regards, WM

Le 16/12/2023 à 02:21, Richard Damon a écrit :
> On 12/15/23 5:43 PM, WM wrote:

>> Show us how you do that for the ℵ₀ unit fractions which are smaller than
>> every chosen x.
>>
>> Given n such that 1/n+1 < x <= 1/n, (or n == 0 if 1 < x) then the ℵ₀
> unit fractions smaller than that x are the set of 1/(n+k) for k = 1, 2,
> 3, ...
>
> That IS an unbounded enumeration that visits EVERY unit fraction smaller
> than x. It enumerates them from highest to lowest, since the unit
> fractions form a decreasing series.

In every case ℵ₀ unit fractions are left. Show us a unit fraction
having less than ℵ₀ smaller unit fractions. Fail.
Note that by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 the increase from 0
happens in steps of height 1. If you deny that then you are outside of
mathematics and logic in the realm of glowing faith but irrelevant for
mathematical work.

Regards, WM

Le 16/12/2023 à 02:29, Richard Damon a écrit :
> On 12/15/23 5:48 PM, WM wrote:

> I said:
> The problem is that there are no positions that can not be chosen.
>
> And that means that there is no element of ℕ that can't become a member
> of ℕ_def,

That is wrong.

> and thus, there are no member of ℕ that are always left
> behind, and thus no elements left to be "dark".

You despise mathematics which proves the contrary:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>
> You say ℕ_def is "potentially" infinite, which means "Dark Numbers" are
> "potentially non-existent",

Dark numbers are the matter that is required to make numbers visible. But
almost all dark numbers cannot be made visible because otherwise nothing
would be left between n and ω.

Regards, WM

WM always keeps driveling his abhoring, unworthy, filthy bullshit:

> Le 16/12/2023 à 02:29, Richard Damon a écrit :
>> On 12/15/23 5:48 PM, WM wrote:
>
>> I said:
>> The problem is that there are no positions that can not be chosen.
>>
>> And that means that there is no element of ℕ that can't become a member
>> of ℕ_def,
>
>
> That is wrong.
>
>
>> and thus, there are no member of ℕ that are always left
>> behind, and thus no elements left to be "dark".
>
> You despise mathematics which proves the contrary:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>
>> You say ℕ_def is "potentially" infinite, which means "Dark Numbers" are
>> "potentially non-existent",
>
> Dark numbers are the matter that is required to make numbers visible.

For the identification of any number it's NOT required to "make these visible",
same as it's not required to make each constituent of your abhoring idiocy visible.

On 12/16/23 3:37 AM, WM wrote:
> Le 16/12/2023 à 02:29, Richard Damon a écrit :
>> On 12/15/23 5:48 PM, WM wrote:
>
>> I said:
>> The problem is that there are no positions that can not be chosen.
>>
>> And that means that there is no element of ℕ that can't become a
>> member of ℕ_def,
>
>
> That is wrong.

Why?

Show the element that can't be.

>
>> and thus, there are no member of ℕ that are always left behind, and
>> thus no elements left to be "dark".
>
> You despise mathematics which proves the contrary:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

But you can't define the actual set ℕ_def, so that doesn't mean anything.

>>
>> You say ℕ_def is "potentially" infinite, which means "Dark Numbers"
>> are "potentially non-existent",
>
> Dark numbers are the matter that is required to make numbers visible.
> But almost all dark numbers cannot be made visible because otherwise
> nothing would be left between n and ω.
>

Which is meaningless gobbledygook.

Note, your "n" is an poorly defined term, you confuse it between A
natural number, and the "last" final natural number. Since no "last"
Natural Nubmer exists, your definition of n doesn't exist.

Your logic just doesn't handle Unboundedness

> Regards, WM
>
>

On 12/16/23 3:32 AM, WM wrote:
> Le 16/12/2023 à 02:21, Richard Damon a écrit :
>> On 12/15/23 5:43 PM, WM wrote:
>
>>> Show us how you do that for the ℵ₀ unit fractions which are smaller
>>> than every chosen x.
>>>
>>> Given n such that 1/n+1 < x <= 1/n,  (or n == 0 if 1 < x) then the ℵ₀
>> unit fractions smaller than that x are the set of 1/(n+k) for k = 1,
>> 2, 3, ...
>>
>> That IS an unbounded enumeration that visits EVERY unit fraction
>> smaller than x. It enumerates them from highest to lowest, since the
>> unit fractions form a decreasing series.
>
> In every case ℵ₀ unit fractions are left. Show us a unit fraction having
> less than ℵ₀ smaller unit fractions. Fail.
> Note that by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 the increase from 0 happens
> in steps of height 1. If you deny that then you are outside of
> mathematics and logic in the realm of glowing faith but irrelevant for
> mathematical work.
>
> Regards, WM
>
>

I don't need to, that is bounded logic, so invalid to be applied to an
unbounded set.

You are the one that is going "outside" of mathematics, not looking at
the actual observable behavior of the system.

Please show where your requirements are actually requirements in a
system that can generate the Natural Numbers.

Since you can't generate the Natural Numbers with your logic, you can't
use it to talk about them.

On 12/15/23 5:48 PM, WM wrote:
> Le 15/12/2023 à 13:55, Richard Damon a écrit :
>> On 12/15/23 5:17 AM, WM wrote:
>
>>> The index k of the position (m, n) is defined by the choice of m and n.
>>> For dark positions (that cannot b chosen) no index can be defined.
>>> Therefore the formula does ot define the whole matrix B.
>>>
>> The problem is that there are no positions that can not be chosen.
>
> You don't see them. But they are there.
> Try to identify those which I cannot identify because they are larger
> than every identified natural number:
>
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
>> That is your false assumption.
>
> No that is mathematics, contrary to the belief of matheologial cranks.
>
> Regards, WM
>
>

WM was thinking very hard :
> Le 15/12/2023 à 21:52, Jim Reburns a écrit :
>> Le 15/12/2023 à 18:27, FromTheRafters a écrit :
>>> WM has brought this to us :
>>>> On 15.12.2023 05:49, Jim Burns wrote:
>>>>> On 12/14/2023 2:31 PM, WM wrote:
>>>>
>>>>> Because we cannot perform a supertask,
>>>>> there are only finitely.many claims we've made,
>>>>
>>>> How do you enumerate the algebraic numbers?
>>>
>>> All at once.
>>
>> Not true, impossible. All at once is an oximoronic statement.
>
> Enumerating is a stepwise process.

No it isn't. This is just a bijective function being applied to the
whole set. Nothing about steps at all.

https://en.wikipedia.org/wiki/Enumeration#Countable_vs._uncountable

> For algebraic numbers the polynomials of
> height n have to be searched one after the other.

Have they found them all yet?

Le 16/12/2023 à 13:13, FromTheRafters a écrit :
> WM was thinking very hard :
>> Le 15/12/2023 à 21:52, Jim Reburns a écrit :

>>> Not true, impossible. All at once is an oximoronic statement.
>>
>> Enumerating is a stepwise process.
>
> No it isn't. This is just a bijective function being applied to the
> whole set. Nothing about steps at all.

This answer shows that unrealistic fools have seized power and try to shut
down mathematics. Counting is done step by step.
>
>> For algebraic numbers the polynomials of
>> height n have to be searched one after the other.
>
> Have they found them all yet?

No infinite set will ever be counted.

Regards, WM

Le 16/12/2023 à 12:56, Richard Damon a écrit :
> On 12/16/23 3:37 AM, WM wrote:

>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>
> But you can't define the actual set ℕ_def, so that doesn't mean anything.
>
There is no set but only a potentially infinite collection. Just
like the collection of known prime numbers.

Regards, WM

Le 16/12/2023 à 12:56, Richard Damon a écrit :
> On 12/16/23 3:32 AM, WM wrote:
>> Le 16/12/2023 à 02:21, Richard Damon a écrit :
>>> On 12/15/23 5:43 PM, WM wrote:
>>
>>>> Show us how you do that for the ℵ₀ unit fractions which are smaller
>>>> than every chosen x.
>>>>
>>>> Given n such that 1/n+1 < x <= 1/n,  (or n == 0 if 1 < x) then the ℵ₀
>>> unit fractions smaller than that x are the set of 1/(n+k) for k = 1,
>>> 2, 3, ...
>>>
>>> That IS an unbounded enumeration that visits EVERY unit fraction
>>> smaller than x. It enumerates them from highest to lowest, since the
>>> unit fractions form a decreasing series.
>>
>> In every case ℵ₀ unit fractions are left. Show us a unit fraction having
>> less than ℵ₀ smaller unit fractions. Fail.
>> Note that by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 the increase from 0 happens
>> in steps of height 1. If you deny that then you are outside of
>> mathematics and logic in the realm of glowing faith but irrelevant for
>> mathematical work.
> I don't need to, that is bounded logic,

It is mathematics.

> You are the one that is going "outside" of mathematics,

No. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.

Regards, WM

On 12/16/23 12:47 PM, WM wrote:
> Le 16/12/2023 à 13:13, FromTheRafters a écrit :
>> WM was thinking very hard :
>>> Le 15/12/2023 à 21:52, Jim Reburns a écrit :
>
>>>> Not true, impossible. All at once is an oximoronic statement.
>>>
>>> Enumerating is a stepwise process.
>>
>> No it isn't. This is just a bijective function being applied to the
>> whole set. Nothing about steps at all.
>
> This answer shows that unrealistic fools have seized power and try to
> shut down mathematics. Counting is done step by step.

Then you can't have the full "Counting Numbers" as you never get to the end.

>>
>>> For algebraic numbers the polynomials of height n have to be searched
>>> one after the other.
>>
>> Have they found them all yet?
>
> No infinite set will ever be counted.

Which is why you can't insist on "counting" step by step sets that are
infinite.

>
> Regards, WM
>
>

On 12/16/23 12:53 PM, WM wrote:
> Le 16/12/2023 à 12:56, Richard Damon a écrit :
>> On 12/16/23 3:32 AM, WM wrote:
>>> Le 16/12/2023 à 02:21, Richard Damon a écrit :
>>>> On 12/15/23 5:43 PM, WM wrote:
>>>
>>>>> Show us how you do that for the ℵ₀ unit fractions which are smaller
>>>>> than every chosen x.
>>>>>
>>>>> Given n such that 1/n+1 < x <= 1/n,  (or n == 0 if 1 < x) then the ℵ₀
>>>> unit fractions smaller than that x are the set of 1/(n+k) for k = 1,
>>>> 2, 3, ...
>>>>
>>>> That IS an unbounded enumeration that visits EVERY unit fraction
>>>> smaller than x. It enumerates them from highest to lowest, since the
>>>> unit fractions form a decreasing series.
>>>
>>> In every case ℵ₀ unit fractions are left. Show us a unit fraction
>>> having less than ℵ₀ smaller unit fractions. Fail.
>>> Note that by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 the increase from 0
>>> happens in steps of height 1. If you deny that then you are outside
>>> of mathematics and logic in the realm of glowing faith but irrelevant
>>> for mathematical work.
>
>> I don't need to, that is bounded logic,
>
> It is mathematics.

No, Mathematics handles the conditions. It is your logic that is broken,

>
>> You are the one that is going "outside" of mathematics,
>
> No. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.
>
> Regards, WM
>

Which doesn't say what you want it to say,

The LOGIC you apply to that statement can't handle unbounded values

On 12/16/23 12:50 PM, WM wrote:
> Le 16/12/2023 à 12:56, Richard Damon a écrit :
>> On 12/16/23 3:37 AM, WM wrote:
>
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>
>> But you can't define the actual set ℕ_def, so that doesn't mean anything.
>>
> There is no set but only a potentially infinite collection. Just like
> the collection of known prime numbers.
> Regards, WM

Which makes it NOT a "Set of Numbers" and thus can't be removed from a
"Set of Numbers"

This makes your "Dark Numbers" not numbers that CAN'T be used
individually, just those that haven't yet been, but still could be in
the future.

Your logic is just broken.

The Clown WM always keeps driveling his abhoring, filthy bullshit for about 20 years:

> Le 16/12/2023 à 12:56, Richard Damon a écrit :
>> On 12/16/23 3:37 AM, WM wrote:
>
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>
>> But you can't define the actual set ℕ_def, so that doesn't mean anything.
>>
> There is no set but only a potentially infinite collection. Just
> like the collection of known prime numbers.

For us and with our math everything is fine while for you insulated Eunuch the
insane shit that you have read somewhere else takes place right for you alone.

Piss off already - 20+ years full of you clownish but militant bullshit is enough...

Le 16/12/2023 à 19:02, Richard Damon a écrit :
> On 12/16/23 12:47 PM, WM wrote:
>> Le 16/12/2023 à 13:13, FromTheRafters a écrit :
>>> WM was thinking very hard :
>>>> Le 15/12/2023 à 21:52, Jim Reburns a écrit :
>>
>>>>> Not true, impossible. All at once is an oximoronic statement.
>>>>
>>>> Enumerating is a stepwise process.
>>>
>>> No it isn't. This is just a bijective function being applied to the
>>> whole set. Nothing about steps at all.
>>
>> This answer shows that unrealistic fools have seized power and try to
>> shut down mathematics. Counting is done step by step.
>
> Then you can't have the full "Counting Numbers" as you never get to the end.

Bingo!
>>>
>>>> For algebraic numbers the polynomials of height n have to be searched
>>>> one after the other.
>>>
>>> Have they found them all yet?
>>
>> No infinite set will ever be counted.
>
> Which is why you can't insist on "counting" step by step sets that are
> infinite.

The desire to count infinite sets does not relieve the rules of counting.

Regards, WM

Le 16/12/2023 à 19:03, Richard Damon a écrit :
> On 12/16/23 12:53 PM, WM wrote:
>
>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.

> Which doesn't say what you want it to say,

It says that NUF(x) cannot grow to 2 without having paused at 1.

Regards, WM

Le 16/12/2023 à 19:05, Richard Damon a écrit :
> On 12/16/23 12:50 PM, WM wrote:
>> Le 16/12/2023 à 12:56, Richard Damon a écrit :
>>> On 12/16/23 3:37 AM, WM wrote:
>>
>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>
>>> But you can't define the actual set ℕ_def, so that doesn't mean anything.
>>>
>> There is no set but only a potentially infinite collection. Just like
>> the collection of known prime numbers.
>> Regards, WM
>
> Which makes it NOT a "Set of Numbers" and thus can't be removed from a
> "Set of Numbers"

Yes, it is hard to delimit ℕ_def.
>
> This makes your "Dark Numbers" not numbers that CAN'T be used
> individually, just those that haven't yet been, but still could be in
> the future.

Many can be used, but almost all cannot. Because you cannot use a unit
fraction that has only few smaller unit fractions.

Regards, WM

On 12/15/2023 5:35 AM, WM wrote:
> On 15.12.2023 05:49, Jim Burns wrote:

>> Irony:
>> It is because we are finite that we know that
>> our claims are correct about infinity.
>
> No,
> because your are stupid, you believe that.

We know some facts about finite sequences.

In a finite sequence of claims,
if any claim is false
then some claim is first.false.

In a finite sequence of claims,
if each claim is not.first.false
then each claim is not.false.

Some claims in some sequences
are visibly not-first-false.
For example, we can see that
Q(x) is not.first.false about x in
⟨ P(x) Q(x)∨¬P(x) Q(x) ⟩
   t       t       t
   f       t       t
   t       f       f
   f       t       f

For some sequences, we can see that
they have only not-first-false claims.

Even if
I don't know which x refers to
out of infinitely.many,
if I can see Q(x) is in a finite claim.sequence
with only not-first-false claims about x
I can see that Q(x) is true about x

Not because of what I know about x
(which isn't much, here).
Because of what I know about finite sequences.

On 12/16/23 2:09 PM, WM wrote:
> Le 16/12/2023 à 19:02, Richard Damon a écrit :
>> On 12/16/23 12:47 PM, WM wrote:
>>> Le 16/12/2023 à 13:13, FromTheRafters a écrit :
>>>> WM was thinking very hard :
>>>>> Le 15/12/2023 à 21:52, Jim Reburns a écrit :
>>>
>>>>>> Not true, impossible. All at once is an oximoronic statement.
>>>>>
>>>>> Enumerating is a stepwise process.
>>>>
>>>> No it isn't. This is just a bijective function being applied to the
>>>> whole set. Nothing about steps at all.
>>>
>>> This answer shows that unrealistic fools have seized power and try to
>>> shut down mathematics. Counting is done step by step.
>>
>> Then you can't have the full "Counting Numbers" as you never get to
>> the end.
>
> Bingo!

So, you are admitting your logic system can't handle it, so any clams
you make are just invalid

>>>>
>>>>> For algebraic numbers the polynomials of height n have to be
>>>>> searched one after the other.
>>>>
>>>> Have they found them all yet?
>>>
>>> No infinite set will ever be counted.
>>
>> Which is why you can't insist on "counting" step by step sets that are
>> infinite.
>
> The desire to count infinite sets does not relieve the rules of counting.

It meas that any logicsystem that insists on "rules of counting" is
insufficient to handle the Natural Numbers (or any bigger set).

Thus, your "Dark Numbers" don't exist, because they are predicated on a
number system that can't be handled by your logic system.\

Making your logic just the logic of a fool.

>
> Regards, WM
>

On 12/16/23 2:11 PM, WM wrote:
> Le 16/12/2023 à 19:03, Richard Damon a écrit :
>> On 12/16/23 12:53 PM, WM wrote:
>>
>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.
>
>> Which doesn't say what you want it to say,
>
> It says that NUF(x) cannot grow to 2 without having paused at 1.
>
> Regards, WM
>

Which it does before moving any finite distance, since it happens in the
ubounded extreame.

That is one of the strange properties of unboundely small values, while
any distance is finite, you can have an infinite number of them before
any finite distance.

Your logic just can;t handle the unbounded,

On 12/16/23 2:15 PM, WM wrote:
> Le 16/12/2023 à 19:05, Richard Damon a écrit :
>> On 12/16/23 12:50 PM, WM wrote:
>>> Le 16/12/2023 à 12:56, Richard Damon a écrit :
>>>> On 12/16/23 3:37 AM, WM wrote:
>>>
>>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>>
>>>> But you can't define the actual set ℕ_def, so that doesn't mean
>>>> anything.
>>>>
>>> There is no set but only a potentially infinite collection. Just like
>>> the collection of known prime numbers.
>>> Regards, WM
>>
>> Which makes it NOT a "Set of Numbers" and thus can't be removed from a
>> "Set of Numbers"
>
> Yes, it is hard to delimit ℕ_def.

No, ℕ_def is just ℕ, since all of ℕ can be chosen to make a finite
initial sequence.

>>
>> This makes your "Dark Numbers" not numbers that CAN'T be used
>> individually, just those that haven't yet been, but still could be in
>> the future.
>
> Many can be used, but almost all cannot. Because you cannot use a unit
> fraction that has only few smaller unit fractions.
>
No, None can be used because none exist.

NO unit fraction exists that only has a finite number of smaller unit
fractions,

> Regards, WM
>
>

On 12/16/2023 12:37 AM, WM wrote:
> Le 16/12/2023 à 02:29, Richard Damon a écrit :
>> On 12/15/23 5:48 PM, WM wrote:
>
>> I said:
>> The problem is that there are no positions that can not be chosen.
>>
>> And that means that there is no element of ℕ that can't become a
>> member of ℕ_def,
>
>
> That is wrong.
>
>> and thus, there are no member of ℕ that are always left behind, and
>> thus no elements left to be "dark".
>
> You despise mathematics which proves the contrary:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>
>> You say ℕ_def is "potentially" infinite, which means "Dark Numbers"
>> are "potentially non-existent",
>
> Dark numbers are the matter that is required to make numbers visible.

Oh boy.... Forgot your meds, or did your doctor forget to give you a refill?

> But almost all dark numbers cannot be made visible because otherwise
> nothing would be left between n and ω.
>
> Regards, WM
>
>

On 12/16/2023 3:56 AM, Richard Damon wrote:
> On 12/16/23 3:37 AM, WM wrote:
>> Le 16/12/2023 à 02:29, Richard Damon a écrit :
>>> On 12/15/23 5:48 PM, WM wrote:
>>
>>> I said:
>>> The problem is that there are no positions that can not be chosen.
>>>
>>> And that means that there is no element of ℕ that can't become a
>>> member of ℕ_def,
>>
>>
>> That is wrong.
>
> Why?
>
> Show the element that can't be.

WM cannot ever do this. This is the folly of his moronic idea.

>
>>
>>> and thus, there are no member of ℕ that are always left behind, and
>>> thus no elements left to be "dark".
>>
>> You despise mathematics which proves the contrary:
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>
> But you can't define the actual set ℕ_def, so that doesn't mean anything.
>
>
>>>
>>> You say ℕ_def is "potentially" infinite, which means "Dark Numbers"
>>> are "potentially non-existent",
>>
>> Dark numbers are the matter that is required to make numbers visible.
>> But almost all dark numbers cannot be made visible because otherwise
>> nothing would be left between n and ω.
>>
>
> Which is meaningless gobbledygook.
>
> Note, your "n" is an poorly defined term, you confuse it between A
> natural number, and the "last" final natural number. Since no "last"
> Natural Nubmer exists, your definition of n doesn't exist.
>
> Your logic just doesn't handle Unboundedness
>
>> Regards, WM
>>
>>
>

On 12/16/2023 11:15 AM, WM wrote:
> Le 16/12/2023 à 19:05, Richard Damon a écrit :
>> On 12/16/23 12:50 PM, WM wrote:
>>> Le 16/12/2023 à 12:56, Richard Damon a écrit :
>>>> On 12/16/23 3:37 AM, WM wrote:
>>>
>>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>>
>>>> But you can't define the actual set ℕ_def, so that doesn't mean
>>>> anything.
>>>>
>>> There is no set but only a potentially infinite collection. Just like
>>> the collection of known prime numbers.
>>> Regards, WM
>>
>> Which makes it NOT a "Set of Numbers" and thus can't be removed from a
>> "Set of Numbers"
>
> Yes, it is hard to delimit ℕ_def.
>>
>> This makes your "Dark Numbers" not numbers that CAN'T be used
>> individually, just those that haven't yet been, but still could be in
>> the future.
>
> Many can be used, but almost all cannot. Because you cannot use a unit
> fraction that has only few smaller unit fractions.

Not sure what that even means!

1/4->1/2

Want the middle? Try:

1/4->1/8->1/2

This goes on forever.

1/4->1/8

Want the middle? Try:

1/4->1/16->1/8

Keep in mind that rationals are not reals, however its fun to
approximate reals with rationals with a given precision using convergent
s of continued fractions.