A(0)=0
A(n)=(A(n-1)+1)/2 ... n∈ℕ

Density property::= Given two different number A(n), and 1, there always
exists another different number A(n+1) such that A(n)<A(n+1)<1

When and how A(n)=1 and the density property still holds?

(note: when n is indefinitely large, A(n)=0.999... is the only way to express
in 'conventional' fractional notation.)

wij pretended :
> A(0)=0
> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>
> Density property::= Given two different number A(n), and 1, there always
> exists another different number A(n+1) such that A(n)<A(n+1)<1

No.

On Saturday, 18 December 2021 at 20:06:33 UTC+8, FromTheRafters wrote:
> wij pretended :
> > A(0)=0
> > A(n)=(A(n-1)+1)/2 ... n∈ℕ
> >
> > Density property::= Given two different number A(n), and 1, there always
> > exists another different number A(n+1) such that A(n)<A(n+1)<1
> No.

Do 0.999...=1 and the density property still holds?
No one need vote or opinion here, idiot.
Q: Is the density property broken if 0.999...=1

on 12/18/2021, wij supposed :
> On Saturday, 18 December 2021 at 20:06:33 UTC+8, FromTheRafters wrote:
>> wij pretended :
>>> A(0)=0
>>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>>>
>>> Density property::= Given two different number A(n), and 1, there always
>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>> No.
>
> Do 0.999...=1 and the density property still holds?
> No one need vote or opinion here, idiot.
>
> Q: Is the density property broken if 0.999...=1

I already answered. The density property as laid out requires two
*DIFFERENT* numbers. You have suggested two representations of the
*SAME* number, so the answer is no.

On 18/12/2021 11:27, wij wrote:
> A(0)=0
> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>
> Density property::= Given two different number A(n), and 1, there always
> exists another different number A(n+1) such that A(n)<A(n+1)<1

Correct. The density property is not broken.

>
> When and how A(n)=1 and the density property still holds?

There is no n such that A(n) = 1.

>
> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> in 'conventional' fractional notation.)

n is always a number, like 7 or 159 or 372743777. A(n) is always less
than 1. There is no n for which A(n)=0.999... = 1. The problem is your
understanding of the phrase "when n is indefinitely large...". No
number is "indefinitely large"!

The density property is not broken. You appeared to have agreed to this
a few days ago in comp.theory! :

----- from comp.theory thread -----
On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
05:19:39 UTC+8, Mike Terry wrote:
>> On 13/12/2021 19:35, wij wrote:
>>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
wrote:<..snip..>

>>>> No the density property holds. It is a basic property of the real
>>>> numbers. Remember, the property says:
>>>>
>>>> if a < b, then there exists x such that a < x < b
>>>>
>>>> So, what do you think are the a, b in that statement that break the
>>>> density property?
>>>
>>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
>> Yes.
>>
>>> If ∃m, A(m)=1 (x=A(m)=1)
>>
>> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
>> asked what is the point. Do you see now?
>>> then, x=b => "x<b" is false (Density property is broken)
>> Given there is no such m, this part of your statement is irrelevant,
>> because the IF condition is false.
>>
>> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
>> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
logic.)
>>
>> So still no density property is broken.
>>
>>
>> Mike.
>
> Agree, no such m exists that x=A(m)=1. *Density property remain valid*
>

On 12/18/2021 5:27 AM, wij wrote:
> A(0)=0
> A(n)=(A(n-1)+1)/2 ... n∈ℕ

A(0) = 0
A(1) = (0+1)/2 = 1/2
A(2) = (1/2 + 1)/2 = 3/4
..
..
..
A(huge) => 1

>
> Density property::= Given two different number A(n), and 1, there always
> exists another different number A(n+1) such that A(n)<A(n+1)<1

mistake
that really isn't the density property, it is result of your first 2 equations.

>
> When and how A(n)=1 and the density property still holds?

mistake;
A(n)=1 never holds, but A(huge) => 1 does, density property is not involved.

>
> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> in 'conventional' fractional notation.)

no need for note

On 12/18/2021 9:12 AM, Mike Terry wrote:
> On 18/12/2021 11:27, wij wrote:
>> A(0)=0
>> A(n)=(A(n-1)+1)/2  ... n∈ℕ
>>
>> Density property::= Given two different number A(n), and 1, there always
>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>
> Correct.  The density property is not broken.

the density property is independent of equations.

On 18/12/2021 16:10, Serg io wrote:
> On 12/18/2021 9:12 AM, Mike Terry wrote:
>> On 18/12/2021 11:27, wij wrote:
>>> A(0)=0
>>> A(n)=(A(n-1)+1)/2  ... n∈ℕ
>>>
>>> Density property::= Given two different number A(n), and 1, there always
>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>
>> Correct.  The density property is not broken.
>
> the density property is independent of equations.
>

Of course. To claim "the density property is broken" you would need to
exhibit two numbers a,b with a < b, such that there is no number
strictly between a and b. I asked wij what his proposed a and b were,
over in comp.theory, and he seemed to have changed his mind about the
density property being broken, but not for long it seems! :)

[Yes, of course (a+b)/2 is always an example of such a number... The
puzzle is working out what wij has in mind.]

Mike.

On 12/18/2021 10:20 AM, Mike Terry wrote:
> On 18/12/2021 16:10, Serg io wrote:
>> On 12/18/2021 9:12 AM, Mike Terry wrote:
>>> On 18/12/2021 11:27, wij wrote:
>>>> A(0)=0
>>>> A(n)=(A(n-1)+1)/2  ... n∈ℕ
>>>>
>>>> Density property::= Given two different number A(n), and 1, there always
>>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>>
>>> Correct.  The density property is not broken.
>>
>> the density property is independent of equations.
>>
>
> Of course.  To claim "the density property is broken" you would need to exhibit two numbers a,b  with a < b, such that there is no number strictly
> between a and b.  I asked wij what his proposed a and b were, over in comp.theory, and he seemed to have changed his mind about the density property
> being broken, but not for long it seems!  :)
>
> [Yes, of course (a+b)/2 is always an example of such a number... The puzzle is working out what wij has in mind.]
>
> Mike.

Extra Credit: is it only the average that works ?

given
a < b

a < (a+b)/n < b

what are the bounds on n ?

n * a < a + b < n * b

if a = b,

n * a < 2 * a < a * n

so n = 2

....................

if a = 0

n * 0 < 0 + b < n * b

0 < b < n * b (n postive and greater than 1)

>>>>>>>>>>>>>

so looks like the average is the only thing that works, n = 2

On Saturday, 18 December 2021 at 23:12:54 UTC+8, Mike Terry wrote:
> On 18/12/2021 11:27, wij wrote:
> > A(0)=0
> > A(n)=(A(n-1)+1)/2 ... n∈ℕ
> >
> > Density property::= Given two different number A(n), and 1, there always
> > exists another different number A(n+1) such that A(n)<A(n+1)<1
> Correct. The density property is not broken.
> >
> > When and how A(n)=1 and the density property still holds?
> There is no n such that A(n) = 1.
> >
> > (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> > in 'conventional' fractional notation.)
> n is always a number, like 7 or 159 or 372743777. A(n) is always less
> than 1. There is no n for which A(n)=0.999... = 1. The problem is your
> understanding of the phrase "when n is indefinitely large...". No
> number is "indefinitely large"!
>
> The density property is not broken. You appeared to have agreed to this
> a few days ago in comp.theory! :
>
> ----- from comp.theory thread -----
> On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
> 05:19:39 UTC+8, Mike Terry wrote:
> >> On 13/12/2021 19:35, wij wrote:
> >>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
> wrote:<..snip..>
>
> >>>> No the density property holds. It is a basic property of the real
> >>>> numbers. Remember, the property says:
> >>>>
> >>>> if a < b, then there exists x such that a < x < b
> >>>>
> >>>> So, what do you think are the a, b in that statement that break the
> >>>> density property?
> >>>
> >>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
> >> Yes.
> >>
> >>> If ∃m, A(m)=1 (x=A(m)=1)
> >>
> >> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
> >> asked what is the point. Do you see now?
> >>> then, x=b => "x<b" is false (Density property is broken)
> >> Given there is no such m, this part of your statement is irrelevant,
> >> because the IF condition is false.
> >>
> >> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
> >> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
> logic.)
> >>
> >> So still no density property is broken.
> >>
> >>
> >> Mike.
> >
> > Agree, no such m exists that x=A(m)=1. *Density property remain valid*
> >

I think there is some cultural problems here.
E.g. "Don't A like B?" The answer in English is opposite from many others.
I am from the culture that the yes/no answer to such question means the
opposite in English. Many people now in the world speaks English may
have different meaning to such question (by default).
I am still not sure what you mean exactly.

On 12/18/2021 11:42 AM, wij wrote:
> On Saturday, 18 December 2021 at 23:12:54 UTC+8, Mike Terry wrote:
>> On 18/12/2021 11:27, wij wrote:
>>> A(0)=0
>>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>>>
>>> Density property::= Given two different number A(n), and 1, there always
>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>> Correct. The density property is not broken.
>>>
>>> When and how A(n)=1 and the density property still holds?
>> There is no n such that A(n) = 1.
>>>
>>> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
>>> in 'conventional' fractional notation.)
>> n is always a number, like 7 or 159 or 372743777. A(n) is always less
>> than 1. There is no n for which A(n)=0.999... = 1. The problem is your
>> understanding of the phrase "when n is indefinitely large...". No
>> number is "indefinitely large"!
>>
>> The density property is not broken. You appeared to have agreed to this
>> a few days ago in comp.theory! :
>>
>> ----- from comp.theory thread -----
>> On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
>> 05:19:39 UTC+8, Mike Terry wrote:
>>>> On 13/12/2021 19:35, wij wrote:
>>>>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
>> wrote:<..snip..>
>>
>>>>>> No the density property holds. It is a basic property of the real
>>>>>> numbers. Remember, the property says:
>>>>>>
>>>>>> if a < b, then there exists x such that a < x < b
>>>>>>
>>>>>> So, what do you think are the a, b in that statement that break the
>>>>>> density property?
>>>>>
>>>>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
>>>> Yes.
>>>>
>>>>> If ∃m, A(m)=1 (x=A(m)=1)
>>>>
>>>> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
>>>> asked what is the point. Do you see now?
>>>>> then, x=b => "x<b" is false (Density property is broken)
>>>> Given there is no such m, this part of your statement is irrelevant,
>>>> because the IF condition is false.
>>>>
>>>> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
>>>> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
>> logic.)
>>>>
>>>> So still no density property is broken.
>>>>
>>>>
>>>> Mike.
>>>
>>> Agree, no such m exists that x=A(m)=1. *Density property remain valid*
>>>
>
> I think there is some cultural problems here.
> E.g. "Don't A like B?" The answer in English is opposite from many others.
> I am from the culture that the yes/no answer to such question means the
> opposite in English. Many people now in the world speaks English may
> have different meaning to such question (by default).
> I am still not sure what you mean exactly.

the density property is independent of equations.

On Saturday, 18 December 2021 at 23:33:02 UTC+8, Serg io wrote:
> On 12/18/2021 5:27 AM, wij wrote:
> > A(0)=0
> > A(n)=(A(n-1)+1)/2 ... n∈ℕ
> A(0) = 0
> A(1) = (0+1)/2 = 1/2
> A(2) = (1/2 + 1)/2 = 3/4
> .
> .
> .
> A(huge) => 1
> >
> > Density property::= Given two different number A(n), and 1, there always
> > exists another different number A(n+1) such that A(n)<A(n+1)<1
> mistake
> that really isn't the density property, it is result of your first 2 equations.
> >
> > When and how A(n)=1 and the density property still holds?
> mistake;
> A(n)=1 never holds, but A(huge) => 1 does, density property is not involved.

You can count n infinitely, forever, in A(n). This is the 'infinity' I meant for the moment.

> >
> > (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> > in 'conventional' fractional notation.)
> no need for note

There is some important issue hidden, this note can be ignored for the moment.
It may not be the right time to explain, yet.

On Sunday, 19 December 2021 at 00:38:17 UTC+8, Serg io wrote:
> On 12/18/2021 10:20 AM, Mike Terry wrote:
> > On 18/12/2021 16:10, Serg io wrote:
> >> On 12/18/2021 9:12 AM, Mike Terry wrote:
> >>> On 18/12/2021 11:27, wij wrote:
> >>>> A(0)=0
> >>>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
> >>>>
> >>>> Density property::= Given two different number A(n), and 1, there always
> >>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
> >>>
> >>> Correct. The density property is not broken.
> >>
> >> the density property is independent of equations.
> >>
> >
> > Of course. To claim "the density property is broken" you would need to exhibit two numbers a,b with a < b, such that there is no number strictly
> > between a and b. I asked wij what his proposed a and b were, over in comp.theory, and he seemed to have changed his mind about the density property
> > being broken, but not for long it seems! :)
> >
> > [Yes, of course (a+b)/2 is always an example of such a number... The puzzle is working out what wij has in mind.]
> >
> > Mike.
> Extra Credit: is it only the average that works ?
>
> given
> a < b
>
> a < (a+b)/n < b
>
> what are the bounds on n ?
>
> n * a < a + b < n * b
>
> if a = b,
>
> n * a < 2 * a < a * n
>
> so n = 2
>

So n<2<n , which is false.

What do you mean by saying "n=2" ?

> ...................
>
> if a = 0
>
> n * 0 < 0 + b < n * b
>
> 0 < b < n * b (n postive and greater than 1)
>
> >>>>>>>>>>>>>
>
> so looks like the average is the only thing that works, n = 2

On Sunday, 19 December 2021 at 01:50:45 UTC+8, Serg io wrote:
> [cut]
> the density property is independent of equations.

I suggested A(n)= 0.999... when n is indefinitely large (assume this meaning
for the moment). Because this is the number in fractional notation we typically
use.
Basic consensus to make: The number of "0.999..." is indefinite.

On Saturday, December 18, 2021 at 3:27:33 AM UTC-8, wij wrote:
> A(0)=0
> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>
> Density property::= Given two different number A(n), and 1, there always
> exists another different number A(n+1) such that A(n)<A(n+1)<1
>
> When and how A(n)=1 and the density property still holds?
>
> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> in 'conventional' fractional notation.)

some quantities can be infinitely close...
like .9 repeating and first integer...

Mitchell Raemsch

On 18/12/2021 16:38, Serg io wrote:
> On 12/18/2021 10:20 AM, Mike Terry wrote:
>> On 18/12/2021 16:10, Serg io wrote:
>>> On 12/18/2021 9:12 AM, Mike Terry wrote:
>>>> On 18/12/2021 11:27, wij wrote:
>>>>> A(0)=0
>>>>> A(n)=(A(n-1)+1)/2  ... n∈ℕ
>>>>>
>>>>> Density property::= Given two different number A(n), and 1, there
>>>>> always
>>>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>>>
>>>> Correct.  The density property is not broken.
>>>
>>> the density property is independent of equations.
>>>
>>
>> Of course.  To claim "the density property is broken" you would need
>> to exhibit two numbers a,b  with a < b, such that there is no number
>> strictly between a and b.  I asked wij what his proposed a and b were,
>> over in comp.theory, and he seemed to have changed his mind about the
>> density property being broken, but not for long it seems!  :)
>>
>> [Yes, of course (a+b)/2 is always an example of such a number... The
>> puzzle is working out what wij has in mind.]
>>
>> Mike.
>
>
> Extra Credit:  is it only the average that works ?
>
We can use other properly "weighted averages"

ja + kb

where j,k > 0 and j+k = 1

The normal average is the case j = k = 1/2.

Mike.

On 18/12/2021 17:42, wij wrote:
> On Saturday, 18 December 2021 at 23:12:54 UTC+8, Mike Terry wrote:
>> On 18/12/2021 11:27, wij wrote:
>>> A(0)=0
>>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>>>
>>> Density property::= Given two different number A(n), and 1, there always
>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>> Correct. The density property is not broken.
>>>
>>> When and how A(n)=1 and the density property still holds?
>> There is no n such that A(n) = 1.
>>>
>>> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
>>> in 'conventional' fractional notation.)
>> n is always a number, like 7 or 159 or 372743777. A(n) is always less
>> than 1. There is no n for which A(n)=0.999... = 1. The problem is your
>> understanding of the phrase "when n is indefinitely large...". No
>> number is "indefinitely large"!
>>
>> The density property is not broken. You appeared to have agreed to this
>> a few days ago in comp.theory! :
>>
>> ----- from comp.theory thread -----
>> On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
>> 05:19:39 UTC+8, Mike Terry wrote:
>>>> On 13/12/2021 19:35, wij wrote:
>>>>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
>> wrote:<..snip..>
>>
>>>>>> No the density property holds. It is a basic property of the real
>>>>>> numbers. Remember, the property says:
>>>>>>
>>>>>> if a < b, then there exists x such that a < x < b
>>>>>>
>>>>>> So, what do you think are the a, b in that statement that break the
>>>>>> density property?
>>>>>
>>>>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
>>>> Yes.
>>>>
>>>>> If ∃m, A(m)=1 (x=A(m)=1)
>>>>
>>>> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
>>>> asked what is the point. Do you see now?
>>>>> then, x=b => "x<b" is false (Density property is broken)
>>>> Given there is no such m, this part of your statement is irrelevant,
>>>> because the IF condition is false.
>>>>
>>>> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
>>>> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
>> logic.)
>>>>
>>>> So still no density property is broken.
>>>>
>>>>
>>>> Mike.
>>>
>>> Agree, no such m exists that x=A(m)=1. *Density property remain valid*
>>>
>
> I think there is some cultural problems here.
> E.g. "Don't A like B?" The answer in English is opposite from many others.
> I am from the culture that the yes/no answer to such question means the
> opposite in English. Many people now in the world speaks English may
> have different meaning to such question (by default).
> I am still not sure what you mean exactly.
>

Out of curiosity, I wonder where you're from. I'm sure that "no such m
existst that x=A(m)=1" means what everyone thinks it means - that there
is no number m with the property A(m) = 1. Perhaps "Density property
remains valid" has a different meaning, or the meaning is qualified
somehow so that you still think it might be broken or something.

The answer to all this is for you to explain how you believe the density
property might be broken! JUST GIVE THE values of a and b that you
think satisfy:

a < b
there is no x satisfying a < x < b

I asked you for the a, b before, and you gave the example of a = A(n), b
= A(n+1), and then immediately admitted (giving a concrete example)
there is an x strictly between a and b !! That's no good :)

You need to give the a and b which you believe breaks the density
property if 0.999... = 1 [which it does, so go on: what are your a and b?]

Or perhaps you mean something quite different by "the density property"?
Either way you'll have to explain what you think the problem is.

Mike.

On Saturday, December 18, 2021 at 12:04:44 PM UTC-8, Mike Terry wrote:
> On 18/12/2021 17:42, wij wrote:
> > On Saturday, 18 December 2021 at 23:12:54 UTC+8, Mike Terry wrote:
> >> On 18/12/2021 11:27, wij wrote:
> >>> A(0)=0
> >>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
> >>>
> >>> Density property::= Given two different number A(n), and 1, there always
> >>> exists another different number A(n+1) such that A(n)<A(n+1)<1
> >> Correct. The density property is not broken.
> >>>
> >>> When and how A(n)=1 and the density property still holds?
> >> There is no n such that A(n) = 1.
> >>>
> >>> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> >>> in 'conventional' fractional notation.)
> >> n is always a number, like 7 or 159 or 372743777. A(n) is always less
> >> than 1. There is no n for which A(n)=0.999... = 1. The problem is your
> >> understanding of the phrase "when n is indefinitely large...". No
> >> number is "indefinitely large"!
> >>
> >> The density property is not broken. You appeared to have agreed to this
> >> a few days ago in comp.theory! :
> >>
> >> ----- from comp.theory thread -----
> >> On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
> >> 05:19:39 UTC+8, Mike Terry wrote:
> >>>> On 13/12/2021 19:35, wij wrote:
> >>>>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
> >> wrote:<..snip..>
> >>
> >>>>>> No the density property holds. It is a basic property of the real
> >>>>>> numbers. Remember, the property says:
> >>>>>>
> >>>>>> if a < b, then there exists x such that a < x < b
> >>>>>>
> >>>>>> So, what do you think are the a, b in that statement that break the
> >>>>>> density property?
> >>>>>
> >>>>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
> >>>> Yes.
> >>>>
> >>>>> If ∃m, A(m)=1 (x=A(m)=1)
> >>>>
> >>>> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
> >>>> asked what is the point. Do you see now?
> >>>>> then, x=b => "x<b" is false (Density property is broken)
> >>>> Given there is no such m, this part of your statement is irrelevant,
> >>>> because the IF condition is false.
> >>>>
> >>>> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
> >>>> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
> >> logic.)
> >>>>
> >>>> So still no density property is broken.
> >>>>
> >>>>
> >>>> Mike.
> >>>
> >>> Agree, no such m exists that x=A(m)=1. *Density property remain valid*
> >>>
> >
> > I think there is some cultural problems here.
> > E.g. "Don't A like B?" The answer in English is opposite from many others.
> > I am from the culture that the yes/no answer to such question means the
> > opposite in English. Many people now in the world speaks English may
> > have different meaning to such question (by default).
> > I am still not sure what you mean exactly.
> >
> Out of curiosity, I wonder where you're from. I'm sure that "no such m
> existst that x=A(m)=1" means what everyone thinks it means - that there
> is no number m with the property A(m) = 1. Perhaps "Density property
> remains valid" has a different meaning, or the meaning is qualified
> somehow so that you still think it might be broken or something.
>
>
> The answer to all this is for you to explain how you believe the density
> property might be broken! JUST GIVE THE values of a and b that you
> think satisfy:
>
> a < b
> there is no x satisfying a < x < b
>
> I asked you for the a, b before, and you gave the example of a = A(n), b
> = A(n+1), and then immediately admitted (giving a concrete example)
> there is an x strictly between a and b !! That's no good :)
>
> You need to give the a and b which you believe breaks the density
> property if 0.999... = 1 [which it does, so go on: what are your a and b?]
>
> Or perhaps you mean something quite different by "the density property"?
> Either way you'll have to explain what you think the problem is.
>
>
> Mike.

The infinitely small fundamental is infinitely close together...
Parallel lines can be infinitely close to being parallel.
Radius's are the same...

Mitchell Raemsch

On Sunday, 19 December 2021 at 04:04:44 UTC+8, Mike Terry wrote:
> On 18/12/2021 17:42, wij wrote:
> > On Saturday, 18 December 2021 at 23:12:54 UTC+8, Mike Terry wrote:
> >> On 18/12/2021 11:27, wij wrote:
> >>> A(0)=0
> >>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
> >>>
> >>> Density property::= Given two different number A(n), and 1, there always
> >>> exists another different number A(n+1) such that A(n)<A(n+1)<1
> >> Correct. The density property is not broken.
> >>>
> >>> When and how A(n)=1 and the density property still holds?
> >> There is no n such that A(n) = 1.
> >>>
> >>> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
> >>> in 'conventional' fractional notation.)
> >> n is always a number, like 7 or 159 or 372743777. A(n) is always less
> >> than 1. There is no n for which A(n)=0.999... = 1. The problem is your
> >> understanding of the phrase "when n is indefinitely large...". No
> >> number is "indefinitely large"!
> >>
> >> The density property is not broken. You appeared to have agreed to this
> >> a few days ago in comp.theory! :
> >>
> >> ----- from comp.theory thread -----
> >> On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
> >> 05:19:39 UTC+8, Mike Terry wrote:
> >>>> On 13/12/2021 19:35, wij wrote:
> >>>>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
> >> wrote:<..snip..>
> >>
> >>>>>> No the density property holds. It is a basic property of the real
> >>>>>> numbers. Remember, the property says:
> >>>>>>
> >>>>>> if a < b, then there exists x such that a < x < b
> >>>>>>
> >>>>>> So, what do you think are the a, b in that statement that break the
> >>>>>> density property?
> >>>>>
> >>>>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
> >>>> Yes.
> >>>>
> >>>>> If ∃m, A(m)=1 (x=A(m)=1)
> >>>>
> >>>> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
> >>>> asked what is the point. Do you see now?
> >>>>> then, x=b => "x<b" is false (Density property is broken)
> >>>> Given there is no such m, this part of your statement is irrelevant,
> >>>> because the IF condition is false.
> >>>>
> >>>> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
> >>>> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
> >> logic.)
> >>>>
> >>>> So still no density property is broken.
> >>>>
> >>>>
> >>>> Mike.
> >>>
> >>> Agree, no such m exists that x=A(m)=1. *Density property remain valid*
> >>>
> >
> > I think there is some cultural problems here.
> > E.g. "Don't A like B?" The answer in English is opposite from many others.
> > I am from the culture that the yes/no answer to such question means the
> > opposite in English. Many people now in the world speaks English may
> > have different meaning to such question (by default).
> > I am still not sure what you mean exactly.
> >
> Out of curiosity, I wonder where you're from.

I am from Republic Of China, R.O.C (Taiwan), Earth. I am told most languages
reply the question "Don't A like B" the opposite meaning as in English.

> I'm sure that "no such m
> existst that x=A(m)=1" means what everyone thinks it means - that there
> is no number m with the property A(m) = 1. Perhaps "Density property
> remains valid" has a different meaning, or the meaning is qualified
> somehow so that you still think it might be broken or something.
>
>
> The answer to all this is for you to explain how you believe the density
> property might be broken! JUST GIVE THE values of a and b that you
> think satisfy:
>
> a < b
> there is no x satisfying a < x < b
>
> I asked you for the a, b before, and you gave the example of a = A(n), b
> = A(n+1), and then immediately admitted (giving a concrete example)
> there is an x strictly between a and b !! That's no good :)
>
> You need to give the a and b which you believe breaks the density
> property if 0.999... = 1 [which it does, so go on: what are your a and b?]
>
> Or perhaps you mean something quite different by "the density property"?
> Either way you'll have to explain what you think the problem is.
>
>
> Mike.

I gave you "Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds."
which is not you showed above.
So, I assume you misunderstood first, then I don't have to answer the rest
questions (sure, unless you are not satisfied).

On 12/18/2021 11:55 AM, wij wrote:
> On Sunday, 19 December 2021 at 00:38:17 UTC+8, Serg io wrote:
>> On 12/18/2021 10:20 AM, Mike Terry wrote:
>>> On 18/12/2021 16:10, Serg io wrote:
>>>> On 12/18/2021 9:12 AM, Mike Terry wrote:
>>>>> On 18/12/2021 11:27, wij wrote:
>>>>>> A(0)=0
>>>>>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>>>>>>
>>>>>> Density property::= Given two different number A(n), and 1, there always
>>>>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>>>>
>>>>> Correct. The density property is not broken.
>>>>
>>>> the density property is independent of equations.
>>>>
>>>
>>> Of course. To claim "the density property is broken" you would need to exhibit two numbers a,b with a < b, such that there is no number strictly
>>> between a and b. I asked wij what his proposed a and b were, over in comp.theory, and he seemed to have changed his mind about the density property
>>> being broken, but not for long it seems! :)
>>>
>>> [Yes, of course (a+b)/2 is always an example of such a number... The puzzle is working out what wij has in mind.]
>>>
>>> Mike.
>> Extra Credit: is it only the average that works ?
>>
>> given
>> a < b
>>
>> a < (a+b)/n < b
>>
>> what are the bounds on n ?
>>
>> n * a < a + b < n * b
>>
>> if a = b,
>>
>> n * a < 2 * a < a * n
>>
>> so n = 2
>>
>
> So n<2<n , which is false.

nope, I didnt use <=, should have, that is all.

>
> What do you mean by saying "n=2" ?

n * a <= 2 * a <= a * n

divide by a

n <= 2 <= n

>
>> ...................
>>
>> if a = 0
>>
>> n * 0 < 0 + b < n * b
>>
>> 0 < b < n * b (n postive and greater than 1)
>>
>>>>>>>>>>>>>>>
>>
>> so looks like the average is the only thing that works, n = 2

On 12/18/2021 1:45 PM, Mike Terry wrote:
> On 18/12/2021 16:38, Serg io wrote:
>> On 12/18/2021 10:20 AM, Mike Terry wrote:
>>> On 18/12/2021 16:10, Serg io wrote:
>>>> On 12/18/2021 9:12 AM, Mike Terry wrote:
>>>>> On 18/12/2021 11:27, wij wrote:
>>>>>> A(0)=0
>>>>>> A(n)=(A(n-1)+1)/2  ... n∈ℕ
>>>>>>
>>>>>> Density property::= Given two different number A(n), and 1, there always
>>>>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>>>>
>>>>> Correct.  The density property is not broken.
>>>>
>>>> the density property is independent of equations.
>>>>
>>>
>>> Of course.  To claim "the density property is broken" you would need to exhibit two numbers a,b  with a < b, such that there is no number strictly
>>> between a and b.  I asked wij what his proposed a and b were, over in comp.theory, and he seemed to have changed his mind about the density property
>>> being broken, but not for long it seems!  :)
>>>
>>> [Yes, of course (a+b)/2 is always an example of such a number... The puzzle is working out what wij has in mind.]
>>>
>>> Mike.
>>
>>
>> Extra Credit:  is it only the average that works ?
>>
> We can use other properly "weighted averages"
>
>      ja + kb
>
> where j,k > 0  and j+k = 1
>
> The normal average is the case j = k = 1/2.
>
> Mike.

j = 100 k = 1

(100 * a + 1 * b) /(101) {like that?}

so is a < 100/101 * a + b/101 <b given a < b ?

hmmm...

On 12/18/2021 12:01 PM, wij wrote:
> On Sunday, 19 December 2021 at 01:50:45 UTC+8, Serg io wrote:
>> [cut]
>> the density property is independent of equations.
>
> I suggested A(n)= 0.999... when n is indefinitely large (assume this meaning
> for the moment). Because this is the number in fractional notation we typically
> use.
> Basic consensus to make: The number of "0.999..." is indefinite.

no, it is = 1 as there is no distance between 0.999... and 1

On 12/18/2021 2:54 PM, wij wrote:
> On Sunday, 19 December 2021 at 04:04:44 UTC+8, Mike Terry wrote:
>> On 18/12/2021 17:42, wij wrote:
>>> On Saturday, 18 December 2021 at 23:12:54 UTC+8, Mike Terry wrote:
>>>> On 18/12/2021 11:27, wij wrote:
>>>>> A(0)=0
>>>>> A(n)=(A(n-1)+1)/2 ... n∈ℕ
>>>>>
>>>>> Density property::= Given two different number A(n), and 1, there always
>>>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>>> Correct. The density property is not broken.
>>>>>
>>>>> When and how A(n)=1 and the density property still holds?
>>>> There is no n such that A(n) = 1.
>>>>>
>>>>> (note: when n is indefinitely large, A(n)=0.999... is the only way to express
>>>>> in 'conventional' fractional notation.)
>>>> n is always a number, like 7 or 159 or 372743777. A(n) is always less
>>>> than 1. There is no n for which A(n)=0.999... = 1. The problem is your
>>>> understanding of the phrase "when n is indefinitely large...". No
>>>> number is "indefinitely large"!
>>>>
>>>> The density property is not broken. You appeared to have agreed to this
>>>> a few days ago in comp.theory! :
>>>>
>>>> ----- from comp.theory thread -----
>>>> On 13/12/2021 22:14, wij wrote:> On Tuesday, 14 December 2021 at
>>>> 05:19:39 UTC+8, Mike Terry wrote:
>>>>>> On 13/12/2021 19:35, wij wrote:
>>>>>>> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry
>>>> wrote:<..snip..>
>>>>
>>>>>>>> No the density property holds. It is a basic property of the real
>>>>>>>> numbers. Remember, the property says:
>>>>>>>>
>>>>>>>> if a < b, then there exists x such that a < x < b
>>>>>>>>
>>>>>>>> So, what do you think are the a, b in that statement that break the
>>>>>>>> density property?
>>>>>>>
>>>>>>> Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds.
>>>>>> Yes.
>>>>>>
>>>>>>> If ∃m, A(m)=1 (x=A(m)=1)
>>>>>>
>>>>>> But THERE DOES NOT EXIST SUCH AN m! That's what I said above and you
>>>>>> asked what is the point. Do you see now?
>>>>>>> then, x=b => "x<b" is false (Density property is broken)
>>>>>> Given there is no such m, this part of your statement is irrelevant,
>>>>>> because the IF condition is false.
>>>>>>
>>>>>> It's like I say "if 10 < 0 then 11 < 1". That is a valid argument, but
>>>>>> 10 < 0 is false, so the conclusion 11 < 1 does not follow. (Basic
>>>> logic.)
>>>>>>
>>>>>> So still no density property is broken.
>>>>>>
>>>>>>
>>>>>> Mike.
>>>>>
>>>>> Agree, no such m exists that x=A(m)=1. *Density property remain valid*
>>>>>
>>>
>>> I think there is some cultural problems here.
>>> E.g. "Don't A like B?" The answer in English is opposite from many others.
>>> I am from the culture that the yes/no answer to such question means the
>>> opposite in English. Many people now in the world speaks English may
>>> have different meaning to such question (by default).
>>> I am still not sure what you mean exactly.
>>>
>> Out of curiosity, I wonder where you're from.
>
> I am from Republic Of China, R.O.C (Taiwan), Earth. I am told most languages
> reply the question "Don't A like B" the opposite meaning as in English.
>
>> I'm sure that "no such m
>> existst that x=A(m)=1" means what everyone thinks it means - that there
>> is no number m with the property A(m) = 1. Perhaps "Density property
>> remains valid" has a different meaning, or the meaning is qualified
>> somehow so that you still think it might be broken or something.
>>
>>
>> The answer to all this is for you to explain how you believe the density
>> property might be broken! JUST GIVE THE values of a and b that you
>> think satisfy:
>>
>> a < b
>> there is no x satisfying a < x < b
>>
>> I asked you for the a, b before, and you gave the example of a = A(n), b
>> = A(n+1), and then immediately admitted (giving a concrete example)
>> there is an x strictly between a and b !! That's no good :)
>>
>> You need to give the a and b which you believe breaks the density
>> property if 0.999... = 1 [which it does, so go on: what are your a and b?]
>>
>> Or perhaps you mean something quite different by "the density property"?
>> Either way you'll have to explain what you think the problem is.
>>
>>
>> Mike.
>
> I gave you "Let a=A(n), b=1, then x=A(n+1) and a<x<b. Density property holds."
> which is not you showed above.
> So, I assume you misunderstood first, then I don't have to answer the rest
> questions (sure, unless you are not satisfied).

the density property is not dependent on your equations, it is independent.

in english your above statement means this

"Let a=A(n), b=1, then x=A(n+1) and a<x<b. (implies that) Density property holds."

however the two are unrelated.

On 19/12/2021 00:18, Serg io wrote:
> On 12/18/2021 1:45 PM, Mike Terry wrote:
>> On 18/12/2021 16:38, Serg io wrote:
>>> On 12/18/2021 10:20 AM, Mike Terry wrote:
>>>> On 18/12/2021 16:10, Serg io wrote:
>>>>> On 12/18/2021 9:12 AM, Mike Terry wrote:
>>>>>> On 18/12/2021 11:27, wij wrote:
>>>>>>> A(0)=0
>>>>>>> A(n)=(A(n-1)+1)/2  ... n∈ℕ
>>>>>>>
>>>>>>> Density property::= Given two different number A(n), and 1, there
>>>>>>> always
>>>>>>> exists another different number A(n+1) such that A(n)<A(n+1)<1
>>>>>>
>>>>>> Correct.  The density property is not broken.
>>>>>
>>>>> the density property is independent of equations.
>>>>>
>>>>
>>>> Of course.  To claim "the density property is broken" you would need
>>>> to exhibit two numbers a,b  with a < b, such that there is no number
>>>> strictly between a and b.  I asked wij what his proposed a and b
>>>> were, over in comp.theory, and he seemed to have changed his mind
>>>> about the density property being broken, but not for long it seems!  :)
>>>>
>>>> [Yes, of course (a+b)/2 is always an example of such a number... The
>>>> puzzle is working out what wij has in mind.]
>>>>
>>>> Mike.
>>>
>>>
>>> Extra Credit:  is it only the average that works ?
>>>
>> We can use other properly "weighted averages"
>>
>>       ja + kb
>>
>> where j,k > 0  and j+k = 1
>>
>> The normal average is the case j = k = 1/2.
>>
>> Mike.
>
> j = 100 k = 1
>
> (100 * a + 1 * b) /(101)     {like that?}
>
>
> so is    a < 100/101 * a  + b/101 <b      given a < b ?
>
> hmmm...

Dude... :(
j=100, k=1 doesn't satisfy j+k = 1.
(duh! I'd like to pretend that was an understandable oversight for
some reason...)

Mike.

Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:

> On 19/12/2021 00:18, Serg io wrote:
>> On 12/18/2021 1:45 PM, Mike Terry wrote:
>>> On 18/12/2021 16:38, Serg io wrote:

>>>> Extra Credit:  is it only the average that works ?
>>>>
>>> We can use other properly "weighted averages"
>>>
>>>       ja + kb
>>>
>>> where j,k > 0  and j+k = 1
>>>
>>> The normal average is the case j = k = 1/2.
>>>
>>> Mike.
>> j = 100 k = 1
>> (100 * a + 1 * b) /(101)     {like that?}
>>
>> so is    a < 100/101 * a  + b/101 <b      given a < b ?
>> hmmm...
>
> Dude... :(
> j=100, k=1 doesn't satisfy j+k = 1.
> (duh! I'd like to pretend that was an understandable oversight for
> some reason...)

It seems likely to be a slip-up since the actual calculation uses the
weights 100/101 and 1/101. (The power of names might also be involved:
j and k are often integers in maths. I wonder if you'd used

ua + vb where u,v > 0 and u+v = 1

the slip-up would not have happened?)

--
Ben.