On 11/18/2022 4:43 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 18. November 2022 um 06:26:16 UTC+1:
>> On 11/17/2022 11:42 AM, WM wrote:
>>> Sergi o schrieb am Donnerstag,
>>> 17. November 2022 um 14:44:43 UTC+1:

>>>> you continue to use a "step by step" process
>>>> in an infinite set,
>>>> than try to blame Cantor for your failure.
>>>
>>> I use a step-by-step process only for
>>> a finite number of steps.
>>
>> You are insisting that finitely-many
>> be treated as infinitely-many.
>
> Up to every finite index
> the number of steps is finite, isn't it?

A step for which
there is no split before that step
which doesn't have a one-by-one move
across that split
is a _finite step_

A move across a split is _one-by-one_ if
there is a last step before
and a first step after
and nothing between them.

Yes,
each finite step
is a finite step.

Also,
for each finite step,
there is a step next after it.

For each step next after a finite step,
there is no split before that next step
without a one-by-one move across that split

Thus,
each step next after a finite step
is also a finite step.

However,
any step after all finite steps
is NOT a finite step.

To be a finite step,
there would need to be,
for each split before it,
a one-by-one move across that split.

A one-by-one move requires
a last before
and a first after
and nothing between them.

However,
each step next after a finite step
is a finite step.

There is no
last finite before all finites.
So
there is no
one-by-one move from before to after
all finites.
So,
a step after all finite steps
is not a finite step.

Note that this arises from
what we mean by "finite".
not from your dark numbers.

>> You (WM) have this broken version of "all".
>
> That is Cantor's version.

If Cantor said
"all gizmos made in this factory"
includes gizmos made in different factories,
then Cantor would be wrong.

Cantor didn't say anything even vaguely like that.
Set aside the issue of who said what.
_Whoever says that_ is wrong.

Cantor is not our Pope.
Things are not correct *because*
Cantor or Frege or Leibnitz or Euler
said them.

In mathematics,
sola scriptura speaks ex cathedra.

>>> But I know that, if Cantor is right,
>>
>> If Cantor claimed what you claim for Cantor,
>> then Cantor would be wrong.
>
> He claimed this:
> "every number p/q comes at an absolutely fixed
> place of a simple infinite sequence",
> "The infinite sequence thus defined has the
> peculiar property to contain all positive
> rational numbers and each of them only once
> at a determined place."

Cantor's fraction-sequence 𝐶 begins at 1/1
For each split of 𝐶
some fraction p/q is last before that split and
some fraction p'/q' such that
| if q ≠ 1
| then p'/q' = p⁺⁺/q⁻⁻
| else p'/q' = 1/p⁺⁺
is first after that split.

For each fraction p/q
there is a unique index k
s = p+q
k = (s-1)(s-2)/2+p

For each index k
there is a unique fraction p/q
s = ⌈(8k+1)¹ᐟ²+1)/2)⌉
p = k-(s-1)(s-2)/2
q = s-p

> That means all (cancelled) fractions.

We describe one of some things (fractions, indexes).
The descriptive claims are true of
each of those things,
in the same why that
each gizmo made in this factory
is made in this factory, not elsewhere.

We reason from those descriptive claims
to further claims,
but only by claim-to-claim steps which we know
cannot lose the true-of-each-one property.

Jim Burns schrieb am Mittwoch, 23. November 2022 um 01:49:14 UTC+1:
> On 11/18/2022 4:43 AM, WM wrote:
> > Jim Burns schrieb am Freitag,

> > Up to every finite index
> > the number of steps is finite, isn't it?
>
> A step for which
> there is no split before that step
> which doesn't have a one-by-one move
> across that split
> is a _finite step_

A definable finite step is the last step of a FISON.
An undefinable finite step belongs to the ℵo-infinit set of dark numbers which cannot be discerned and hence cannot be put in order. No first no last, but only darkness.
>
> A move across a split is _one-by-one_ if
> there is a last step before
> and a first step after
> and nothing between them.

Definable steps.
>
> Yes,
> each finite step
> is a finite step.
>
Yes. In the darkness there are no steps.

> each step next after a finite step
> is also a finite step.

Yes.
>
> However,
> any step after all finite steps
> is NOT a finite step.

There are not "all" finite steps. Potential infinity.
>
> A one-by-one move requires
> a last before
> and a first after
> and nothing between them.

Yes.
>
> However,
> each step next after a finite step
> is a finite step.

Yes.

> So
> there is no
> one-by-one move from before to after
> all finites.
> So,
> a step after all finite steps
> is not a finite step.
>
> Note that this arises from
> what we mean by "finite".
> not from your dark numbers.

All that is true. But to believe that an inclusion monotonic sequence of infinite sets like endsegments has an empty intersection is wrong. To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

> Cantor is not our Pope.
> Things are not correct *because*
> Cantor or Frege or Leibnitz or Euler
> said them.

So it is. Therefore I am entitled to show where he went astray.
>
> We reason from those descriptive claims
> to further claims,
> but only by claim-to-claim steps which we know
> cannot lose the true-of-each-one property.

To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

Regards, WM

After serious thinking Chris M. Thomasson wrote :
> On 11/22/2022 2:26 PM, Sergi o wrote:
>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
>>>> Chris M. Thomasson used his keyboard to write :
>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
>>>>>>> WM formulated on Thursday :
>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um 20:46:44
>>>>>>>> UTC+1:
>>>>>>>>> After serious thinking WM wrote :
>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which
>>>>>>
>>>>>> [...]
>>>>>>
>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and 1/0?
>>>>>> Humm...
>>>>>>
>>>>>
>>>>> Oops! I responded to FromTheRafters instead of WM.
>>>>>
>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings. There is
>>>>> 0/3, 2/0, ect...
>>>>>
>>>>> Why does WM seem to think that a Cantor pair must be a proper fraction?
>>>>> Well, it sure seems that way.
>>>>
>>>> In this specific case, we are dealing with rationals whose numerator and
>>>> denominator are positive, not non-negative.
>>>
>>> Signed zero for fun:
>>>
>>> 1/+0 vs 1/-0
>>>
>>> ;^)
>>
>>
>>
>> like a plot of  1/x^2, or 1/x^(even)
>
> Fwiw, I forgot one of the states of zero.
>
> -0, 0, +0
>
> We can define it as ternary. :^)

Ones' Complement binary representations have a zero and a negative
zero.

What makes your plus zero 'different' from your zero such that it would
be needed? Two ways to say nothing is somehow not enough? :)

WM expressed precisely :
> Jim Burns schrieb am Mittwoch, 23. November 2022 um 01:49:14 UTC+1:
>> On 11/18/2022 4:43 AM, WM wrote:
>>> Jim Burns schrieb am Freitag,
>
>>> Up to every finite index
>>> the number of steps is finite, isn't it?
>>
>> A step for which
>> there is no split before that step
>> which doesn't have a one-by-one move
>> across that split
>> is a _finite step_
>
> A definable finite step is the last step of a FISON.
> An undefinable finite step belongs to the ℵo-infinit set of dark numbers
> which cannot be discerned and hence cannot be put in order. No first no last,
> but only darkness.

Please don't abuse the steps, they are for demonstration purposes only.
A bijective mapping takes 'very little' time and should work for sets
that aren't ordinal numbers too. It is convenient to *show* a bijection
of naturals by matching two known ordinal numbers at a time 'checking'
each 'step' of the 'way' that they conform to a specific generation
algorithm but these stepwise checks are not definitive.

onsdag 23 november 2022 kl. 13:06:13 UTC+1 skrev WM:
> Jim Burns schrieb am Mittwoch, 23. November 2022 um 01:49:14 UTC+1:
> > On 11/18/2022 4:43 AM, WM wrote:
> > > Jim Burns schrieb am Freitag,
> > > Up to every finite index
> > > the number of steps is finite, isn't it?
> >
> > A step for which
> > there is no split before that step
> > which doesn't have a one-by-one move
> > across that split
> > is a _finite step_
> A definable finite step is the last step of a FISON.
> An undefinable finite step belongs to the ℵo-infinit set of dark numbers which cannot be discerned and hence cannot be put in order. No first no last, but only darkness.
> >
> > A move across a split is _one-by-one_ if
> > there is a last step before
> > and a first step after
> > and nothing between them.
> Definable steps.
> >
> > Yes,
> > each finite step
> > is a finite step.
> >
> Yes. In the darkness there are no steps.
> > each step next after a finite step
> > is also a finite step.
> Yes.
> >
> > However,
> > any step after all finite steps
> > is NOT a finite step.
> There are not "all" finite steps. Potential infinity.
> >
> > A one-by-one move requires
> > a last before
> > and a first after
> > and nothing between them.
> Yes.
> >
> > However,
> > each step next after a finite step
> > is a finite step.
> Yes.
> > So
> > there is no
> > one-by-one move from before to after
> > all finites.
> > So,
> > a step after all finite steps
> > is not a finite step.
> >
> > Note that this arises from
> > what we mean by "finite".
> > not from your dark numbers.
> All that is true. But to believe that an inclusion monotonic sequence of infinite sets like endsegments has an empty intersection is wrong. To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.
> > Cantor is not our Pope.
> > Things are not correct *because*
> > Cantor or Frege or Leibnitz or Euler
> > said them.
> So it is. Therefore I am entitled to show where he went astray.
> >
> > We reason from those descriptive claims
> > to further claims,
> > but only by claim-to-claim steps which we know
> > cannot lose the true-of-each-one property.
> To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.
>
> Regards, WM
All of this boils down to bullshit!

For fuck sake we don't even need to use the fucking function given!

we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
so |N^2| = |N|
We also have |c-n|=c
we then also have |Q+|=|Nx(N/{0})|=|NxN|=|N^2|=|N|

and by definition of how cardinal arithmetic works, we have a bijection from N to Q+

Now shut the fuck up with your bullshit.

FromTheRafters schrieb am Mittwoch, 23. November 2022 um 13:53:33 UTC+1:
> WM expressed precisely :
>
> Please don't abuse the steps, they are for demonstration purposes only.

They are defined by the formula
k = (m + n - 1)(m + n - 2)/2 + m
and be the sequence
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, ...

> A bijective mapping takes 'very little' time and should work for sets
> that aren't ordinal numbers too. It is convenient to *show* a bijection
> of naturals by matching two known ordinal numbers at a time 'checking'
> each 'step' of the 'way' that they conform to a specific generation
> algorithm but these stepwise checks are not definitive.

This however is definitive and does not require checking single steps but holds for every k in above formula: To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 23. November 2022 um 13:59:43 UTC+1:
> onsdag 23 november 2022 kl. 13:06:13 UTC+1 skrev WM:

> For fuck sake we don't even need to use the fucking function given!
>
> we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
> so |N^2| = |N|

and I have proved that that is nonsense.

To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.

Regards, WM

On 11/23/2022 7:02 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 23. November 2022 um 13:53:33 UTC+1:
>> WM expressed precisely :
>>
>> Please don't abuse the steps, they are for demonstration purposes only.
>
> They are defined by the formula
> k = (m + n - 1)(m + n - 2)/2 + m
> and be the sequence
> 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, ...

However, your three dots ... , to you, indicate the start of your dark numbers.

>> A bijective mapping takes 'very little' time and should work for sets
>> that aren't ordinal numbers too. It is convenient to *show* a bijection
>> of naturals by matching two known ordinal numbers at a time 'checking'
>> each 'step' of the 'way' that they conform to a specific generation
>> algorithm but these stepwise checks are not definitive.
>
> This however is definitive and does not require checking single steps but holds for every k in above formula:

so, you now believe in "variables" ? how can that be ?

>
> Regards, WM

On 11/23/2022 7:04 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 23. November 2022 um 13:59:43 UTC+1:
>> onsdag 23 november 2022 kl. 13:06:13 UTC+1 skrev WM:
>
>> For fuck sake we don't even need to use the fucking function given!
>>
>> we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
>> so |N^2| = |N|
>
> and I have proved that that is nonsense.

you have stated previously you do not believe in proofs, even simple ones, and you have never provided a proof.

why should we believe you now?

>

>
> Regards, WM

onsdag 23 november 2022 kl. 14:04:36 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 23. November 2022 um 13:59:43 UTC+1:
> > onsdag 23 november 2022 kl. 13:06:13 UTC+1 skrev WM:
>
> > For fuck sake we don't even need to use the fucking function given!
> >
> > we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
> > so |N^2| = |N|
> and I have proved that that is nonsense.

no you haven't, all you have proven is that your fucking matrix is bullshit. The theorem I state is more powerful than your shit

> To believe that exchanging the X's and O's in my example will delete all O's in a definable way but only after all definable terms of the sequence have kept infinitely many O's, is wrong.
>
> Regards, WM

On Wednesday, November 23, 2022 at 3:33:19 PM UTC+1, Sergi o wrote:

> why should we believe you now?

Because he's a psychotic crank full of shit?

On Wednesday, November 23, 2022 at 2:04:36 PM UTC+1, WM wrote:

> and I have proved that that is nonsense.

All you have proved is that you are a full blown crank full of shit.

On 11/23/2022 7:06 AM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 23. November 2022 um 01:49:14 UTC+1:
>> On 11/18/2022 4:43 AM, WM wrote:

>>> Up to every finite index
>>> the number of steps is finite, isn't it?
>>
>> A step for which
>> there is no split before that step
>> which doesn't have a one-by-one move
>> across that split
>> is a _finite step_
>
> A definable finite step is
> the last step of a FISON.

Yes but
that description doesn't make us
any better off than we were before.
𝐅𝐈𝐒𝐎𝐍 ==
𝐅inite 𝐈nitial 𝐒egment 𝐎f 𝐍aturals.

"Finite is finite." So?

Our over-arching strategy is to _start_ with
claims true of each one of whatever we're
reasoning about, and then _keep_ that
being-true-of-each as we advance to further
claims.

The more true-of-each detail we start with,
the more we have to work with, when we try to
find claims we care about to which we can
advance in true-of-each-keeping ways.

> An undefinable finite step belongs to the
> ℵo-infinit set of dark numbers which cannot be
> discerned and hence cannot be put in order.
> No first no last, but only darkness.

Your undefinable finites are not in FISONs.

We can start with being in a FISON as
a description of one of what we're reasoning
about, and then none of the claims we start with
or advance to are about undefinable finites.

If I say
| _All_ the gizmos made in this factory
| have new-style thingummies.
then
pointing to a gizmo made elsewhere
_is not a counter-example_

If I say
| _All_ the FISON-ends >= 1 have
| unique prime factorizations.
then
pointing to a dark number
_is not a counter-example_

>> However,
>> any step after all finite steps
>> is NOT a finite step.
>
> There are not "all" finite steps.
> Potential infinity.

Because you decided to call "finite steps"
things which I am NOT talking about.

However,
I am STILL NOT talking about your other things.

An ordered collection
which begins at 0 and ends somewhere and
which, for each split, there is a
predecessor-successor pair across it
is
an ordered collection
which begins at 0 and ends somewhere and
which, for each split, there is a
predecessor-successor pair across it,
no matter what you or I or Georg Cantor
call it.

We reason _from the description_
You call something else a blablablah
finite step, but you haven't changed
_the description_ from which I'm
reasoning.

On 11/23/2022 5:02 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 23. November 2022 um 13:53:33 UTC+1:
>> WM expressed precisely :
>>
>> Please don't abuse the steps, they are for demonstration purposes only.
>
> They are defined by the formula
> k = (m + n - 1)(m + n - 2)/2 + m
> and be the sequence
> 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, ...

[...]

You are missing a lot of pairs here...

On 11/23/2022 9:09 AM, Fritz Feldhase wrote:
> On Wednesday, November 23, 2022 at 3:33:19 PM UTC+1, Sergi o wrote:
>
>> why should we believe you now?
>
> Because he's a psychotic crank full of shit?

Ditto.

On 11/23/2022 4:13 AM, FromTheRafters wrote:
> After serious thinking Chris M. Thomasson wrote :
>> On 11/22/2022 2:26 PM, Sergi o wrote:
>>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
>>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
>>>>> Chris M. Thomasson used his keyboard to write :
>>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
>>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
>>>>>>>> WM formulated on Thursday :
>>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um
>>>>>>>>> 20:46:44 UTC+1:
>>>>>>>>>> After serious thinking WM wrote :
>>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ...
>>>>>>>>>>> which
>>>>>>>
>>>>>>> [...]
>>>>>>>
>>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and
>>>>>>> 1/0? Humm...
>>>>>>>
>>>>>>
>>>>>> Oops! I responded to FromTheRafters instead of WM.
>>>>>>
>>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings.
>>>>>> There is 0/3, 2/0, ect...
>>>>>>
>>>>>> Why does WM seem to think that a Cantor pair must be a proper
>>>>>> fraction? Well, it sure seems that way.
>>>>>
>>>>> In this specific case, we are dealing with rationals whose
>>>>> numerator and denominator are positive, not non-negative.
>>>>
>>>> Signed zero for fun:
>>>>
>>>> 1/+0 vs 1/-0
>>>>
>>>> ;^)
>>>
>>>
>>>
>>> like a plot of  1/x^2, or 1/x^(even)
>>
>> Fwiw, I forgot one of the states of zero.
>>
>> -0, 0, +0
>>
>> We can define it as ternary. :^)
>
> Ones' Complement binary representations have a zero and a negative zero.
>
> What makes your plus zero 'different' from your zero such that it would
> be needed? Two ways to say nothing is somehow not enough? :)

:^) Well, humm... perhaps... We can say a step-by-step process whose
limit is zero, say:

..1, .01, .001, .0001, ...

Gets to "positive zero", while:

-1, -.01, -.001, -.0001, ...

Gets to "negative zero".

And zero, is just "zero with no sign"?

..1, .01, .001, .0001, ... = +0
-1, -.01, -.001, -.0001, ... = -0
0 = 0

Make any sense to you at all? Or, is that just stepping deep into the
heart of kookville? Humm...

Jim Burns schrieb am Mittwoch, 23. November 2022 um 19:04:17 UTC+1:

> Our over-arching strategy is to _start_ with
> claims true of each one of whatever we're
> reasoning about, and then _keep_ that
> being-true-of-each as we advance to further
> claims.

Exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O.
>
> Your undefinable finites are not in FISONs.

So it is.
>
> We reason _from the description_

Then you should understand this description: Exchanging the X's and O's will never delete any O.

Regards, WM

On 11/23/2022 4:09 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 23. November 2022 um 19:04:17 UTC+1:
>
>> Our over-arching strategy is to _start_ with
>> claims true of each one of whatever we're
>> reasoning about, and then _keep_ that
>> being-true-of-each as we advance to further
>> claims.
>
> Exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O.

as WM for each fraction, will remove AND delete the O stickie on the fraction, and replace it with a X pastie.

>>
>> Your undefinable finites are not in FISONs.
>
> So it is.

It is so.

>>
>> We reason _from the description_
>
> Then you should understand this description: Exchanging the X's and O's will never delete any O.

then you failed to delete the O pastie as you said you would above. Fail

>
> Regards, WM

On 11/23/2022 1:20 PM, Chris M. Thomasson wrote:
> On 11/23/2022 4:13 AM, FromTheRafters wrote:
>> After serious thinking Chris M. Thomasson wrote :
>>> On 11/22/2022 2:26 PM, Sergi o wrote:
>>>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
>>>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
>>>>>> Chris M. Thomasson used his keyboard to write :
>>>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
>>>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
>>>>>>>>> WM formulated on Thursday :
>>>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um 20:46:44 UTC+1:
>>>>>>>>>>> After serious thinking WM wrote :
>>>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... which
>>>>>>>>
>>>>>>>> [...]
>>>>>>>>
>>>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and 1/0? Humm...
>>>>>>>>
>>>>>>>
>>>>>>> Oops! I responded to FromTheRafters instead of WM.
>>>>>>>
>>>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings. There is 0/3, 2/0, ect...
>>>>>>>
>>>>>>> Why does WM seem to think that a Cantor pair must be a proper fraction? Well, it sure seems that way.
>>>>>>
>>>>>> In this specific case, we are dealing with rationals whose numerator and denominator are positive, not non-negative.
>>>>>
>>>>> Signed zero for fun:
>>>>>
>>>>> 1/+0 vs 1/-0
>>>>>
>>>>> ;^)
>>>>
>>>>
>>>>
>>>> like a plot of  1/x^2, or 1/x^(even)
>>>
>>> Fwiw, I forgot one of the states of zero.
>>>
>>> -0, 0, +0
>>>
>>> We can define it as ternary. :^)
>>
>> Ones' Complement binary representations have a zero and a negative zero.
>>
>> What makes your plus zero 'different' from your zero such that it would be needed? Two ways to say nothing is somehow not enough? :)
>
> :^) Well, humm... perhaps... We can say a step-by-step process whose limit is zero, say:
>
> .1, .01, .001, .0001, ...
>
> Gets to "positive zero", while:
>
> -1, -.01, -.001, -.0001, ...
>
> Gets to "negative zero".
>
> And zero, is just "zero with no sign"?
>
>
> .1, .01, .001, .0001, ... = +0
> -1, -.01, -.001, -.0001, ... = -0
> 0 = 0
>
> Make any sense to you at all? Or, is that just stepping deep into the heart of kookville? Humm...

makes a difference on the approach, as with 1/x^2.

onsdag 23 november 2022 kl. 23:09:05 UTC+1 skrev WM:
> Jim Burns schrieb am Mittwoch, 23. November 2022 um 19:04:17 UTC+1:
>
> > Our over-arching strategy is to _start_ with
> > claims true of each one of whatever we're
> > reasoning about, and then _keep_ that
> > being-true-of-each as we advance to further
> > claims.
> Exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O.
> >
> > Your undefinable finites are not in FISONs.
> So it is.
> >
> > We reason _from the description_
> Then you should understand this description: Exchanging the X's and O's will never delete any O.
>
> Regards, WM

we know that for any infinite cardinality c that for a finite number n, we have |c^n|=|c|
so |N^2| = |N|
We also have |c-n|=c
we then also have |Q+|=|Nx(N/{0})|=|NxN|=|N^2|=|N|

and by definition of how cardinal arithmetic works, we have a bijection from N to Q+

Now shut the fuck up with your bullshit.

On 11/23/2022 5:09 PM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 23. November 2022 um 19:04:17 UTC+1:

>> Your undefinable finites are not in FISONs.
>
> So it is.
>
>> We reason _from the description_
>
> Then you should understand this description:
> Exchanging the X's and O's will never delete any O.

Exchanging X's and O's (a bijection)
*within any FISON* will never delete any O

Each exchange is within some FISON.
It does not delete O's

Each FISON does not have all exchanges within it.
O's might be deleted by all the exchanges.
For example,
by all of this sequence of exchanges:
⟨1/1 1/1⟩
⟨1/2 2/1⟩
⟨2/1 3/1⟩
⟨1/3 4/1⟩
⟨2/2 5/1⟩
⟨3/1 6/1⟩
....

It is a _sequence_ of exchanges.
It begins at ⟨1/1 1/1⟩
For each split of the sequence,
there is an exchange ⟨p/q k/1⟩ last before
and an exchange ⟨p'/q' k'/1⟩ first after
such that
k' = k⁺⁺
if q ≠ 1
then p'/q' = p⁺⁺/q⁻⁻
else p'/q' = 1/p⁺⁺

All the O's are deleted.

There is no dark p q or k in any exchange
in the sequence.

| Assume otherwise.
| Assume some ⟨p/q k/1⟩ in the sequence
| contains a dark p q or k
| | There is a _first_ ⟨p₁/q₁ k₁/1⟩
| in which p₁ q₁ or k₁ is dark
| and, in its predecessor ⟨p₀/q₀ k₀/1⟩
| p₀ q₀ and k₀ are NOT dark
| | However,
| p₀ q₀ and k₀ are NOT dark
| and
| k₁ = k₀⁺⁺
| if q₀ ≠ 1
| then p₁/q₁ = p₀⁺⁺/q₀⁻⁻
| else p₁/q₁ = 1/p₀⁺⁺
| So
| p₁ q₁ and k₁ are NOT dark
| Contradiction.

Therefore,
there is no dark p q or k in any exchange
in the sequence.
Nonetheless, all O's are deleted by
all the exchanges.

Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:
> On 11/23/2022 5:09 PM, WM wrote:

> > Then you should understand this description:
> > Exchanging the X's and O's will never delete any O.
> Exchanging X's and O's (a bijection)
> *within any FISON* will never delete any O
>
> Each exchange is within some FISON.
> It does not delete O's

Fine.
>
> Each FISON does not have all exchanges within it.

But all definable exchanges occur within FISONs.

> O's might be deleted by all the exchanges.

Maybe, but not in a definable way.

> For example,
> by all of this sequence of exchanges:
> ⟨1/1 1/1⟩
> ⟨1/2 2/1⟩
> ⟨2/1 3/1⟩
> ⟨1/3 4/1⟩
> ⟨2/2 5/1⟩
> ⟨3/1 6/1⟩
> ...
>
Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.

> It is a _sequence_ of exchanges.
> It begins at ⟨1/1 1/1⟩
> For each split of the sequence,
> there is an exchange ⟨p/q k/1⟩ last before
> and an exchange ⟨p'/q' k'/1⟩ first after
> such that
> k' = k⁺⁺
> if q ≠ 1
> then p'/q' = p⁺⁺/q⁻⁻
> else p'/q' = 1/p⁺⁺
> All the O's are deleted.

Not in a definable way. That is proved by the fact, that never anybody will be able to find a furst O deleted.

> Therefore,
> there is no dark p q or k in any exchange
> in the sequence.
> Nonetheless, all O's are deleted by
> all the exchanges.

That is wrong. Find the first O that is deleted in the visible domain.

Regards, WM

fredag 25 november 2022 kl. 10:55:25 UTC+1 skrev WM:
> Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:
> > On 11/23/2022 5:09 PM, WM wrote:
>
> > > Then you should understand this description:
> > > Exchanging the X's and O's will never delete any O.
> > Exchanging X's and O's (a bijection)
> > *within any FISON* will never delete any O
> >
> > Each exchange is within some FISON.
> > It does not delete O's
> Fine.
> >
> > Each FISON does not have all exchanges within it.
> But all definable exchanges occur within FISONs.
> > O's might be deleted by all the exchanges.
> Maybe, but not in a definable way.
> > For example,
> > by all of this sequence of exchanges:
> > ⟨1/1 1/1⟩
> > ⟨1/2 2/1⟩
> > ⟨2/1 3/1⟩
> > ⟨1/3 4/1⟩
> > ⟨2/2 5/1⟩
> > ⟨3/1 6/1⟩
> > ...
> >
> Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.
> > It is a _sequence_ of exchanges.
> > It begins at ⟨1/1 1/1⟩
> > For each split of the sequence,
> > there is an exchange ⟨p/q k/1⟩ last before
> > and an exchange ⟨p'/q' k'/1⟩ first after
> > such that
> > k' = k⁺⁺
> > if q ≠ 1
> > then p'/q' = p⁺⁺/q⁻⁻
> > else p'/q' = 1/p⁺⁺
> > All the O's are deleted.
> Not in a definable way. That is proved by the fact, that never anybody will be able to find a furst O deleted.
> > Therefore,
> > there is no dark p q or k in any exchange
> > in the sequence.
> > Nonetheless, all O's are deleted by
> > all the exchanges.
> That is wrong. Find the first O that is deleted in the visible domain.
>
> Regards, WM

We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!

zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 12:38:39 UTC+1:
> fredag 25 november 2022 kl. 10:55:25 UTC+1 skrev WM:
> > Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:

> > > For example,
> > > by all of this sequence of exchanges:
> > > ⟨1/1 1/1⟩
> > > ⟨1/2 2/1⟩
> > > ⟨2/1 3/1⟩
> > > ⟨1/3 4/1⟩
> > > ⟨2/2 5/1⟩
> > > ⟨3/1 6/1⟩
> > > ...
> > >
> > Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.

> We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!

Find the first O that is deleted. Every remianing O proves the existence of a fraction that is not indexed such that the index can be known.

Regards, WM

fredag 25 november 2022 kl. 13:09:58 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 12:38:39 UTC+1:
> > fredag 25 november 2022 kl. 10:55:25 UTC+1 skrev WM:
> > > Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:
>
> > > > For example,
> > > > by all of this sequence of exchanges:
> > > > ⟨1/1 1/1⟩
> > > > ⟨1/2 2/1⟩
> > > > ⟨2/1 3/1⟩
> > > > ⟨1/3 4/1⟩
> > > > ⟨2/2 5/1⟩
> > > > ⟨3/1 6/1⟩
> > > > ...
> > > >
> > > Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.
> > We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!
> Find the first O that is deleted. Every remianing O proves the existence of a fraction that is not indexed such that the index can be known.
>
> Regards, WM
Your O and X does not change the fact that I can find those two injections and thus proves you wrong.