Example 1:
Let A≡ Σ(n=1,∞) 9/10^n= 0.999...= 999.../1000...= (3*3*(11...1))/(5*2)^n
If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
If A=1, what is the 3rd digit after the decimal point of A?
If A=1, density property (of ℚ and ℝ) is broken (false).

Example 2:
1/3≈0.333... + x (the conversion never divides completely, non-zero remainder x
always exists as the prerequisite of 'infinite' repeating)

Note that A is DEFINED exactly 0.999..., not 1.
Since the non-zero fractional pattern of the number (repeating decimal) is
defined to repeat infinitely, no p,q∈ℕ such that A=p/q, therefor, A (repeating
decimal) is irrational.

----
[Tip] More about "repeating decimals":
A0= 0.9 9 9 9 ...
A1= 0.99 99 99 99 ...
A2= 0.9 99 999 9999 ...
A3= 0.999 9 9999 9 99999 ...
A4= lim(n->∞) 1-1/n
A5= lim(n->∞) 1-2/n
A6= lim(n->∞) 1-3/10^n
A7= lim(n->∞) n/(n+1)
....
This is just tip of the iceberg, "0.999..." is an infinite set of numbers.
Actually, card("0.999...") is greater than ℵ1,ℵ2,ℵ3...., and more:
https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

On 12/12/21 3:48 AM, wij wrote:
> Example 1:
> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...= 999.../1000...= (3*3*(11...1))/(5*2)^n

Which is the value 1

> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?

There is not number 11..1, (if by that notation you mean in infinite
number of 1s

> If A=1, what is the 3rd digit after the decimal point of A?

0, not a problem, as 0.999999.... isn't a proper representation

> If A=1, density property (of ℚ and ℝ) is broken (false).

Why. For what numbers is 0.9999... needed to exist to have a number in
between? In fact if 0.9999... existed you would break the density
property as no number could exist between it and 1,

>
> Example 2:
> 1/3≈0.333... + x (the conversion never divides completely, non-zero remainder x
> always exists as the prerequisite of 'infinite' repeating)

But since 1/3 * 3 = 1 (by definition) but 3 * 3 = 9, your 0.999999....
is shown to be equal to 1

>
> Note that A is DEFINED exactly 0.999..., not 1.

Which isn't a distinct number. It isn't a distinct element of R
(different from 1)

> Since the non-zero fractional pattern of the number (repeating decimal) is
> defined to repeat infinitely, no p,q∈ℕ such that A=p/q, therefor, A (repeating
> decimal) is irrational.

>
> ----
> [Tip] More about "repeating decimals":
> A0= 0.9 9 9 9 ...
> A1= 0.99 99 99 99 ...
> A2= 0.9 99 999 9999 ...
> A3= 0.999 9 9999 9 99999 ...
> A4= lim(n->∞) 1-1/n
> A5= lim(n->∞) 1-2/n
> A6= lim(n->∞) 1-3/10^n
> A7= lim(n->∞) n/(n+1)
> ...
> This is just tip of the iceberg, "0.999..." is an infinite set of numbers.
> Actually, card("0.999...") is greater than ℵ1,ℵ2,ℵ3..., and more:
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>

On 2021-12-12 08:48:35 +0000, wij said:
Let A≡ Σ(n=1,∞) 9/10^n= 0.999...= 999.../1000...= (3*3*(11...1))/(5*2)^n

A = Σ(n=1,∞) 9/10^n
= 9 / 10 + Σ(n=2,∞) 9/10^n
= 9 / 10 + Σ(n=1,∞) 9/10^(n+1)
= 9 / 10 + (Σ(n=1,∞) 9/10^n) / 10
= 9 / 10 + A / 10
=> A - A / 10 = 9 / 10
=> 10 A - A = 9
=> 9 A = 9
=> A = 1

Mikko

On 12/12/2021 08:48, wij wrote:
> Example 1:
> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...

ok. So A = 1

> = 999.../1000...= (3*3*(11...1))/(5*2)^n

Does not compute. 999... and 1000... are not numbers.

> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?

No, of course not. (Does not compute)

> If A=1, what is the 3rd digit after the decimal point of A?

Real numbers may have one or two decimal representations, a bit like +0
= -0, or 3/5 = 6/10. A has two representations: 0.999... and 1.000....
The 3rd digit after the decimal point of representation 0.999 is 9,
while for the representation 1.000 it is 0.

> If A=1, density property (of ℚ and ℝ) is broken (false).

What density property is that? (And how do you think it is broken?)

>
> Example 2:
> 1/3≈0.333... + x (the conversion never divides completely, non-zero remainder x
> always exists as the prerequisite of 'infinite' repeating)

No, 1/3 = 0.333..., so x = 0. Talk about conversions never dividing
completely is just gibberish! (The kind of talk we here from PO.)

>
> Note that A is DEFINED exactly 0.999..., not 1.

No. A is DEFINED to be the limit of a particular infinite series, which
is equal to 1. (or 0.999...).,

> Since the non-zero fractional pattern of the number (repeating decimal) is
> defined to repeat infinitely, no p,q∈ℕ such that A=p/q, therefor, A (repeating
> decimal) is irrational.

No. A = 1/1. You can take p=1, q=1. So A is rational, duh.

Dude, this is all off topic, plus my advice would be to sign up for some
basic maths lessons, where they explain about numbers, and decimal
notation. You've already posted to sci.math, where it was on topic, so
no need to post here. (Sheesh! You're as bad as PO for deliberately
cross-posting to irrelevent newsgroups...)

Mike.

On 12/12/2021 8:24 AM, Mike Terry wrote:
> On 12/12/2021 08:48, wij wrote:
>> Example 1:
>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>
> ok.  So A = 1
>
>>   = 999.../1000...= (3*3*(11...1))/(5*2)^n
>
> Does not compute.  999... and 1000... are not numbers.
>
>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>
> No, of course not.  (Does not compute)
>
>> If A=1, what is the 3rd digit after the decimal point of A?
>
> Real numbers may have one or two decimal representations, a bit like +0
> = -0, or 3/5 = 6/10.  A has two representations:  0.999... and 1.000....
>  The 3rd digit after the decimal point of representation 0.999 is 9,
> while for the representation 1.000 it is 0.
>
>> If A=1, density property (of ℚ and ℝ) is broken (false).
>
> What density property is that?  (And how do you think it is broken?)
I think he means if there are two reals, r1 /= r2, then there must exist
a real, r, such that either r1>r>r2 or r2>r>r1. There is some other name
for this property but I don't remember at the moment. I think it might
have been Tarski or Dedekind that used it as one of the axioms to define
the real. I wrote a fluff piece awhile back that put this axiom to work;
it's available at http://notatt.com/calibration.pdf - not required reading.
--
Jeffrey Barnett

On 12/12/21 10:24 AM, Mike Terry wrote:
> On 12/12/2021 08:48, wij wrote:
>> Example 1:
>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>
> ok.  So A = 1
>
>>   = 999.../1000...= (3*3*(11...1))/(5*2)^n
>
> Does not compute.  999... and 1000... are not numbers.
>
>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>
> No, of course not.  (Does not compute)
>
>> If A=1, what is the 3rd digit after the decimal point of A?
>
> Real numbers may have one or two decimal representations, a bit like +0
> = -0, or 3/5 = 6/10.  A has two representations:  0.999... and 1.000....
>  The 3rd digit after the decimal point of representation 0.999 is 9,
> while for the representation 1.000 it is 0.
>
>> If A=1, density property (of ℚ and ℝ) is broken (false).
>
> What density property is that?  (And how do you think it is broken?)

I believe he means the property that between any two members of the
Real, or the Rationals, there will ALWAYS be another member of that set
between them. I.E., there is NOT a 'next' value from a given value.

One value between x and y will be (x+y)/2

The problem with thinking of 0.9999.... as something distinct from 1 is
THAT breaks the density property, as there can be no number bigger than
0.9999... and less than 1.0000

> Mike.

On 12/12/2021 18:15, Richard Damon wrote:
> On 12/12/21 10:24 AM, Mike Terry wrote:
>> On 12/12/2021 08:48, wij wrote:
>>> Example 1:
>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>
>> ok.  So A = 1
>>
>>>   = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>
>> Does not compute.  999... and 1000... are not numbers.
>>
>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>
>> No, of course not.  (Does not compute)
>>
>>> If A=1, what is the 3rd digit after the decimal point of A?
>>
>> Real numbers may have one or two decimal representations, a bit like
>> +0 = -0, or 3/5 = 6/10.  A has two representations:  0.999... and
>> 1.000....   The 3rd digit after the decimal point of representation
>> 0.999 is 9, while for the representation 1.000 it is 0.
>>
>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>
>> What density property is that?  (And how do you think it is broken?)
>
> I believe he means the property that between any two members of the
> Real, or the Rationals, there will ALWAYS be another member of that set
> between them. I.E., there is NOT a 'next' value from a given value.
>

Yes, I thought he might mean that. I wouldn't call that "density"
myself, as "density" has a different meaning. Perhaps the "denseness"
property?

But if A=1, how does this break the property? It doesn't - the property
breaks if A != 1, so that would make wij's claim plain Wrong, like
everything else he said. :) [no surprise, I guess.]

Mike.

> One value between x and y will be (x+y)/2
>
> The problem with thinking of 0.9999.... as something distinct from 1 is
> THAT breaks the density property, as there can be no number bigger than
> 0.9999... and less than 1.0000
>
>> Mike.
>

On 12/12/2021 11:38 AM, Mike Terry wrote:
> On 12/12/2021 18:15, Richard Damon wrote:
>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>> On 12/12/2021 08:48, wij wrote:
>>>> Example 1:
>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>
>>> ok.  So A = 1
>>>
>>>>   = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>
>>> Does not compute.  999... and 1000... are not numbers.
>>>
>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>
>>> No, of course not.  (Does not compute)
>>>
>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>
>>> Real numbers may have one or two decimal representations, a bit like
>>> +0 = -0, or 3/5 = 6/10.  A has two representations:  0.999... and
>>> 1.000....   The 3rd digit after the decimal point of representation
>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>
>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>
>>> What density property is that?  (And how do you think it is broken?)
>>
>> I believe he means the property that between any two members of the
>> Real, or the Rationals, there will ALWAYS be another member of that
>> set between them. I.E., there is NOT a 'next' value from a given value.
>>
>
> Yes, I thought he might mean that.  I wouldn't call that "density"
> myself, as "density" has a different meaning.  Perhaps the "denseness"
> property?
>
> But if A=1, how does this break the property?  It doesn't - the property
> breaks if A != 1, so that would make wij's claim plain Wrong, like
> everything else he said.  :)   [no surprise, I guess.]
>
> Mike.
>
>> One value between x and y will be (x+y)/2
>>
>> The problem with thinking of 0.9999.... as something distinct from 1
>> is THAT breaks the density property, as there can be no number bigger
>> than 0.9999... and less than 1.0000
It finally came back to me I think: sets with such order relations are
called "dense in themselves".
--
Jeff Barnett

On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> On 12/12/2021 18:15, Richard Damon wrote:
> > On 12/12/21 10:24 AM, Mike Terry wrote:
> >> On 12/12/2021 08:48, wij wrote:
> >>> Example 1:
> >>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>
> >> ok. So A = 1
> >>
> >>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>
> >> Does not compute. 999... and 1000... are not numbers.
> >>
> >>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>
> >> No, of course not. (Does not compute)
> >>
> >>> If A=1, what is the 3rd digit after the decimal point of A?
> >>
> >> Real numbers may have one or two decimal representations, a bit like
> >> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >> 1.000.... The 3rd digit after the decimal point of representation
> >> 0.999 is 9, while for the representation 1.000 it is 0.
> >>
> >>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>
> >> What density property is that? (And how do you think it is broken?)
> >
> > I believe he means the property that between any two members of the
> > Real, or the Rationals, there will ALWAYS be another member of that set
> > between them. I.E., there is NOT a 'next' value from a given value.
> >
> Yes, I thought he might mean that. I wouldn't call that "density"
> myself, as "density" has a different meaning. Perhaps the "denseness"
> property?
>

Amazing! What a phenomenon!

google "density property".

Density Property:: For any two different numbers, there exists another different number in
between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
i<k<j.
https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

> But if A=1, how does this break the property? It doesn't - the property
> breaks if A != 1, so that would make wij's claim plain Wrong, like
> everything else he said. :) [no surprise, I guess.]
>
> Mike.
> > One value between x and y will be (x+y)/2
> >
> > The problem with thinking of 0.9999.... as something distinct from 1 is
> > THAT breaks the density property, as there can be no number bigger than
> > 0.9999... and less than 1.0000
> >
> >> Mike.
> >

Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ

Given two different numbser A(n), and 1, there always exists another different
number A(n+1) such that A(n)<A(n+1)<1

When A(n)=1? Infinity?

On 12/13/21 7:27 AM, wij wrote:
> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>> On 12/12/2021 18:15, Richard Damon wrote:
>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>> On 12/12/2021 08:48, wij wrote:
>>>>> Example 1:
>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>
>>>> ok. So A = 1
>>>>
>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>
>>>> Does not compute. 999... and 1000... are not numbers.
>>>>
>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>
>>>> No, of course not. (Does not compute)
>>>>
>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>
>>>> Real numbers may have one or two decimal representations, a bit like
>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>
>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>
>>>> What density property is that? (And how do you think it is broken?)
>>>
>>> I believe he means the property that between any two members of the
>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>
>> Yes, I thought he might mean that. I wouldn't call that "density"
>> myself, as "density" has a different meaning. Perhaps the "denseness"
>> property?
>>
>
> Amazing! What a phenomenon!
>
> google "density property".
>
> Density Property::=
> For any two different numbers, there exists another different number in
> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> i<k<j.
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>
>> But if A=1, how does this break the property? It doesn't - the property
>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>> everything else he said. :) [no surprise, I guess.]
>>
>> Mike.
>>> One value between x and y will be (x+y)/2
>>>
>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>> THAT breaks the density property, as there can be no number bigger than
>>> 0.9999... and less than 1.0000
>>>
>>>> Mike.
>>>
>
> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>
> Given two different numbser A(n), and 1, there always exists another different
> number A(n+1) such that A(n)<A(n+1)<1
>
> When A(n)=1? Infinity?
>

There is no FINITE n where A(n) is equal to 1

Note, for the Reals, Naturals, etc., 'Infinity' isn't a value, only a
'limiting case'

Thus the limit(n->infinity) A(n) is 1, even though no individual A(n) is 1.

This is a common property of limits.

Just like 0.9999... for any finite number of 9s isn't equal to 1, but
the limiting case with the endless 9s is.

On Monday, 13 December 2021 at 20:53:29 UTC+8, richar...@gmail.com wrote:
> On 12/13/21 7:27 AM, wij wrote:
> > On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> >> On 12/12/2021 18:15, Richard Damon wrote:
> >>> On 12/12/21 10:24 AM, Mike Terry wrote:
> >>>> On 12/12/2021 08:48, wij wrote:
> >>>>> Example 1:
> >>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>>>
> >>>> ok. So A = 1
> >>>>
> >>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>>>
> >>>> Does not compute. 999... and 1000... are not numbers.
> >>>>
> >>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>>>
> >>>> No, of course not. (Does not compute)
> >>>>
> >>>>> If A=1, what is the 3rd digit after the decimal point of A?
> >>>>
> >>>> Real numbers may have one or two decimal representations, a bit like
> >>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >>>> 1.000.... The 3rd digit after the decimal point of representation
> >>>> 0.999 is 9, while for the representation 1.000 it is 0.
> >>>>
> >>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>>>
> >>>> What density property is that? (And how do you think it is broken?)
> >>>
> >>> I believe he means the property that between any two members of the
> >>> Real, or the Rationals, there will ALWAYS be another member of that set
> >>> between them. I.E., there is NOT a 'next' value from a given value.
> >>>
> >> Yes, I thought he might mean that. I wouldn't call that "density"
> >> myself, as "density" has a different meaning. Perhaps the "denseness"
> >> property?
> >>
> >
> > Amazing! What a phenomenon!
> >
> > google "density property".
> >
> > Density Property::=
> > For any two different numbers, there exists another different number in
> > between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> > i<k<j.
> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >
> >> But if A=1, how does this break the property? It doesn't - the property
> >> breaks if A != 1, so that would make wij's claim plain Wrong, like
> >> everything else he said. :) [no surprise, I guess.]
> >>
> >> Mike.
> >>> One value between x and y will be (x+y)/2
> >>>
> >>> The problem with thinking of 0.9999.... as something distinct from 1 is
> >>> THAT breaks the density property, as there can be no number bigger than
> >>> 0.9999... and less than 1.0000
> >>>
> >>>> Mike.
> >>>
> >
> > Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
> >
> > Given two different numbser A(n), and 1, there always exists another different
> > number A(n+1) such that A(n)<A(n+1)<1
> >
> > When A(n)=1? Infinity?
> >
> There is no FINITE n where A(n) is equal to 1
>

Neither a FINITE n is in limit.
What is the n in "lim(n->∞) A(n)=1"? Finite, infinite, or not a number?

> Note, for the Reals, Naturals, etc., 'Infinity' isn't a value, only a
> 'limiting case'
>
> Thus the limit(n->infinity) A(n) is 1, even though no individual A(n) is 1.
>
> This is a common property of limits.
>

"No individual A(n) is 1. But limit(n->infinity) A(n) is 1".
So limit theory turns 'approaching' to 'equal' in term of the limit smoke.
Where I can find evidence that A(∞)=1 but from the 'approaching is equal'
theory is the problem.

>
> Just like 0.9999... for any finite number of 9s isn't equal to 1, but
> the limiting case with the endless 9s is.

I am not talking about 'limiting case'. Limit theory is full of inconsistency.
(Every one learned 'limit method' should have a sense of this. I do not what to dig into this shit deep)
We should be interested in the case that 0.999... equal to 1 or not, not the "limiting case".

And, here, right now, the density property in this thread.
Does not 'density property' mean to hold infinitely?
The problem is: The density property procedure can go on infinitely. Can not?

On 13/12/2021 12:27, wij wrote:
> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>> On 12/12/2021 18:15, Richard Damon wrote:
>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>> On 12/12/2021 08:48, wij wrote:
>>>>> Example 1:
>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>
>>>> ok. So A = 1
>>>>
>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>
>>>> Does not compute. 999... and 1000... are not numbers.
>>>>
>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>
>>>> No, of course not. (Does not compute)
>>>>
>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>
>>>> Real numbers may have one or two decimal representations, a bit like
>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>
>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>
>>>> What density property is that? (And how do you think it is broken?)
>>>
>>> I believe he means the property that between any two members of the
>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>
>> Yes, I thought he might mean that. I wouldn't call that "density"
>> myself, as "density" has a different meaning. Perhaps the "denseness"
>> property?
>>
>
> Amazing! What a phenomenon!
>
> google "density property".
>
> Density Property::=
> For any two different numbers, there exists another different number in
> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> i<k<j.
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>

So we were right!

So if A=1, how does this break the density property?

>> But if A=1, how does this break the property? It doesn't - the property
>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>> everything else he said. :) [no surprise, I guess.]
>>
>> Mike.
>>> One value between x and y will be (x+y)/2
>>>
>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>> THAT breaks the density property, as there can be no number bigger than
>>> 0.9999... and less than 1.0000
>>>
>>>> Mike.
>>>
>
> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>
> Given two different numbser A(n), and 1, there always exists another different
> number A(n+1) such that A(n)<A(n+1)<1

Correct.

>
> When A(n)=1? Infinity?

A(n) < 1 for all n. (n = oo is not in the range of n)

Mike.

On 12/12/2021 21:03, Jeff Barnett wrote:
> On 12/12/2021 11:38 AM, Mike Terry wrote:
>> On 12/12/2021 18:15, Richard Damon wrote:
>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>> On 12/12/2021 08:48, wij wrote:
>>>>> Example 1:
>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>
>>>> ok.  So A = 1
>>>>
>>>>>   = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>
>>>> Does not compute.  999... and 1000... are not numbers.
>>>>
>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>
>>>> No, of course not.  (Does not compute)
>>>>
>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>
>>>> Real numbers may have one or two decimal representations, a bit like
>>>> +0 = -0, or 3/5 = 6/10.  A has two representations:  0.999... and
>>>> 1.000....   The 3rd digit after the decimal point of representation
>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>
>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>
>>>> What density property is that?  (And how do you think it is broken?)
>>>
>>> I believe he means the property that between any two members of the
>>> Real, or the Rationals, there will ALWAYS be another member of that
>>> set between them. I.E., there is NOT a 'next' value from a given value.
>>>
>>
>> Yes, I thought he might mean that.  I wouldn't call that "density"
>> myself, as "density" has a different meaning.  Perhaps the "denseness"
>> property?
>>
>> But if A=1, how does this break the property?  It doesn't - the
>> property breaks if A != 1, so that would make wij's claim plain Wrong,
>> like everything else he said.  :)   [no surprise, I guess.]
>>
>> Mike.
>>
>>> One value between x and y will be (x+y)/2
>>>
>>> The problem with thinking of 0.9999.... as something distinct from 1
>>> is THAT breaks the density property, as there can be no number bigger
>>> than 0.9999... and less than 1.0000
>
> It finally came back to me I think: sets with such order relations are
> called "dense in themselves".

As I've heard the term used that's slightly different. It's from the
field of topology, meaning that every point of the set is a limit point
of the set. So it's more to do with neighbourhoods (closeness) than
order - but now I've remembered we also have "densely ordered" sets
which is exactly what we want, sets having the "in between" property.

But just to complicate things, a total ordering induces a topology for
the space (the "order topology" based on open intervals) so for a
totally ordered set like Q, both definitions make sense. And Q is both
dense in itself and densely ordered. Same for R.

Going even further OT, that makes me wonder whether for totally ordered
sets, the two definitions amount to the same thing? But a little
thought tells me that for such sets:

densely ordered implies dense in itself

dense in itself does not imply densely ordered!

so the definitions aren't equivalent. [Umm, I guess we need to also
insist the sets have at least two elements to eliminate pathelogical
edge cases...]

Mike.

On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
> On 13/12/2021 12:27, wij wrote:
> > On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> >> On 12/12/2021 18:15, Richard Damon wrote:
> >>> On 12/12/21 10:24 AM, Mike Terry wrote:
> >>>> On 12/12/2021 08:48, wij wrote:
> >>>>> Example 1:
> >>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>>>
> >>>> ok. So A = 1
> >>>>
> >>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>>>
> >>>> Does not compute. 999... and 1000... are not numbers.
> >>>>
> >>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>>>
> >>>> No, of course not. (Does not compute)
> >>>>
> >>>>> If A=1, what is the 3rd digit after the decimal point of A?
> >>>>
> >>>> Real numbers may have one or two decimal representations, a bit like
> >>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >>>> 1.000.... The 3rd digit after the decimal point of representation
> >>>> 0.999 is 9, while for the representation 1.000 it is 0.
> >>>>
> >>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>>>
> >>>> What density property is that? (And how do you think it is broken?)
> >>>
> >>> I believe he means the property that between any two members of the
> >>> Real, or the Rationals, there will ALWAYS be another member of that set
> >>> between them. I.E., there is NOT a 'next' value from a given value.
> >>>
> >> Yes, I thought he might mean that. I wouldn't call that "density"
> >> myself, as "density" has a different meaning. Perhaps the "denseness"
> >> property?
> >>
> >
> > Amazing! What a phenomenon!
> >
> > google "density property".
> >
> > Density Property::=
> > For any two different numbers, there exists another different number in
> > between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> > i<k<j.
> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >
> So we were right!
>
> So if A=1, how does this break the density property?

When there ever exists a time a n such that A(n)=1, "A(n)<A(n+1)<1"
then, the density property does not hold.
This leads to 0.999...≠1.

> >> But if A=1, how does this break the property? It doesn't - the property
> >> breaks if A != 1, so that would make wij's claim plain Wrong, like
> >> everything else he said. :) [no surprise, I guess.]
> >>
> >> Mike.
> >>> One value between x and y will be (x+y)/2
> >>>
> >>> The problem with thinking of 0.9999.... as something distinct from 1 is
> >>> THAT breaks the density property, as there can be no number bigger than
> >>> 0.9999... and less than 1.0000
> >>>
> >>>> Mike.
> >>>
> >
> > Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
> >
> > Given two different numbser A(n), and 1, there always exists another different
> > number A(n+1) such that A(n)<A(n+1)<1
> Correct.
>
> >
> > When A(n)=1? Infinity?
>
> A(n) < 1 for all n. (n = oo is not in the range of n)
>
>
> Mike.

Correct.

On 13/12/2021 15:53, wij wrote:
> On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
>> On 13/12/2021 12:27, wij wrote:
>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>>> Example 1:
>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>>
>>>>>> ok. So A = 1
>>>>>>
>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>>
>>>>>> Does not compute. 999... and 1000... are not numbers.
>>>>>>
>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>>
>>>>>> No, of course not. (Does not compute)
>>>>>>
>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>>
>>>>>> Real numbers may have one or two decimal representations, a bit like
>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>>
>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>>
>>>>>> What density property is that? (And how do you think it is broken?)
>>>>>
>>>>> I believe he means the property that between any two members of the
>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>>>
>>>> Yes, I thought he might mean that. I wouldn't call that "density"
>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
>>>> property?
>>>>
>>>
>>> Amazing! What a phenomenon!
>>>
>>> google "density property".
>>>
>>> Density Property::=
>>> For any two different numbers, there exists another different number in
>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
>>> i<k<j.
>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>
>> So we were right!
>>
>> So if A=1, how does this break the density property?
>
> When there ever exists a time a n such that A(n)=1,

Where does A(n) come from? I'll take it that you're using

A(n) = 1 - (1/10)^n

so A(n) is the sequence (0.9, 0.99, 0.999, ...)

Well, there IS NO n SUCH THAT A(n) = 1

> "A(n)<A(n+1)<1"
> then, the density property does not hold.

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?

> This leads to 0.999...≠1.

But your argument above is nonsense, so it doesn't lead to that at all. :)

Mike.

>
>>>> But if A=1, how does this break the property? It doesn't - the property
>>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>>>> everything else he said. :) [no surprise, I guess.]
>>>>
>>>> Mike.
>>>>> One value between x and y will be (x+y)/2
>>>>>
>>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>>>> THAT breaks the density property, as there can be no number bigger than
>>>>> 0.9999... and less than 1.0000
>>>>>
>>>>>> Mike.
>>>>>
>>>
>>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>>>
>>> Given two different numbser A(n), and 1, there always exists another different
>>> number A(n+1) such that A(n)<A(n+1)<1
>> Correct.
>>
>>>
>>> When A(n)=1? Infinity?
>>
>> A(n) < 1 for all n. (n = oo is not in the range of n)
>>
>>
>> Mike.
>
> Correct.
>

On 12/13/2021 8:35 AM, Mike Terry wrote:
> On 12/12/2021 21:03, Jeff Barnett wrote:
>> On 12/12/2021 11:38 AM, Mike Terry wrote:
>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>> Example 1:
>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>
>>>>> ok.  So A = 1
>>>>>
>>>>>>   = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>
>>>>> Does not compute.  999... and 1000... are not numbers.
>>>>>
>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>
>>>>> No, of course not.  (Does not compute)
>>>>>
>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>
>>>>> Real numbers may have one or two decimal representations, a bit
>>>>> like +0 = -0, or 3/5 = 6/10.  A has two representations:  0.999...
>>>>> and 1.000....   The 3rd digit after the decimal point of
>>>>> representation 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>
>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>
>>>>> What density property is that?  (And how do you think it is broken?)
>>>>
>>>> I believe he means the property that between any two members of the
>>>> Real, or the Rationals, there will ALWAYS be another member of that
>>>> set between them. I.E., there is NOT a 'next' value from a given value.
>>>>
>>>
>>> Yes, I thought he might mean that.  I wouldn't call that "density"
>>> myself, as "density" has a different meaning.  Perhaps the
>>> "denseness" property?
>>>
>>> But if A=1, how does this break the property?  It doesn't - the
>>> property breaks if A != 1, so that would make wij's claim plain
>>> Wrong, like everything else he said.  :)   [no surprise, I guess.]
>>>
>>> Mike.
>>>
>>>> One value between x and y will be (x+y)/2
>>>>
>>>> The problem with thinking of 0.9999.... as something distinct from 1
>>>> is THAT breaks the density property, as there can be no number
>>>> bigger than 0.9999... and less than 1.0000
>>
>> It finally came back to me I think: sets with such order relations are
>> called "dense in themselves".
>
> As I've heard the term used that's slightly different.  It's from the
> field of topology, meaning that every point of the set is a limit point
> of the set.  So it's more to do with neighbourhoods (closeness) than
> order - but now I've remembered we also have "densely ordered" sets
> which is exactly what we want, sets having the "in between" property.
>
> But just to complicate things, a total ordering induces a topology for
> the space (the "order topology" based on open intervals) so for a
> totally ordered set like Q, both definitions make sense.  And Q is both
> dense in itself and densely ordered.  Same for R.
>
> Going even further OT, that makes me wonder whether for totally ordered
> sets, the two definitions amount to the same thing?  But a little
> thought tells me that for such sets:
>
>    densely ordered implies dense in itself
>
>    dense in itself does not imply densely ordered!
>
> so the definitions aren't equivalent.  [Umm, I guess we need to also
> insist the sets have at least two elements to eliminate pathelogical
> edge cases...]
I found an old pointer on my desktop:
https://en.wikipedia.org/wiki/Tarski%27s_axiomatization_of_the_reals
It's Axiom 2 of the Tarski axioms of the reals; it reads
If x < z, there exists a y such that x < y and y < z. In other words,
"<" is dense in R.
I recall that when I was writing that fluff piece mentioned above, I
knew the axiom but could not recall its "street name", so I went a
searching. At first, I tried Archimedean but that wasn't it. I finally
hit the Tarski page and there it was. But no interest name. It's funny
how bits and pieces from school days lurk some place in the head and
slowly get mushy.
--
Jeff Barnett

On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry wrote:
> On 13/12/2021 15:53, wij wrote:
> > On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
> >> On 13/12/2021 12:27, wij wrote:
> >>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> >>>> On 12/12/2021 18:15, Richard Damon wrote:
> >>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
> >>>>>> On 12/12/2021 08:48, wij wrote:
> >>>>>>> Example 1:
> >>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>>>>>
> >>>>>> ok. So A = 1
> >>>>>>
> >>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>>>>>
> >>>>>> Does not compute. 999... and 1000... are not numbers.
> >>>>>>
> >>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>>>>>
> >>>>>> No, of course not. (Does not compute)
> >>>>>>
> >>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
> >>>>>>
> >>>>>> Real numbers may have one or two decimal representations, a bit like
> >>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >>>>>> 1.000.... The 3rd digit after the decimal point of representation
> >>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
> >>>>>>
> >>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>>>>>
> >>>>>> What density property is that? (And how do you think it is broken?)
> >>>>>
> >>>>> I believe he means the property that between any two members of the
> >>>>> Real, or the Rationals, there will ALWAYS be another member of that set
> >>>>> between them. I.E., there is NOT a 'next' value from a given value.
> >>>>>
> >>>> Yes, I thought he might mean that. I wouldn't call that "density"
> >>>> myself, as "density" has a different meaning. Perhaps the "denseness"
> >>>> property?
> >>>>
> >>>
> >>> Amazing! What a phenomenon!
> >>>
> >>> google "density property".
> >>>
> >>> Density Property::=
> >>> For any two different numbers, there exists another different number in
> >>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> >>> i<k<j.
> >>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >>>
> >> So we were right!
> >>
> >> So if A=1, how does this break the density property?
> >
> > When there ever exists a time a n such that A(n)=1,
> Where does A(n) come from? I'll take it that you're using
>
> A(n) = 1 - (1/10)^n
>
> so A(n) is the sequence (0.9, 0.99, 0.999, ...)
>
> Well, there IS NO n SUCH THAT A(n) = 1

ah hum, what is the point?

> > "A(n)<A(n+1)<1"
> > then, the density property does not hold.
> 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.
If ∃m, A(m)=1 (x=A(m)=1) then, x=b => "x<b" is false (Density property is broken)

> > This leads to 0.999...≠1.
> But your argument above is nonsense, so it doesn't lead to that at all. :)
>
> Mike.

I am not sure what is in you mind.
Please provide a non-nonsense argument.

> >
> >>>> But if A=1, how does this break the property? It doesn't - the property
> >>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
> >>>> everything else he said. :) [no surprise, I guess.]
> >>>>
> >>>> Mike.
> >>>>> One value between x and y will be (x+y)/2
> >>>>>
> >>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
> >>>>> THAT breaks the density property, as there can be no number bigger than
> >>>>> 0.9999... and less than 1.0000
> >>>>>
> >>>>>> Mike.
> >>>>>
> >>>
> >>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
> >>>
> >>> Given two different numbser A(n), and 1, there always exists another different
> >>> number A(n+1) such that A(n)<A(n+1)<1
> >> Correct.
> >>
> >>>
> >>> When A(n)=1? Infinity?
> >>
> >> A(n) < 1 for all n. (n = oo is not in the range of n)
> >>
> >>
> >> Mike.
> >
> > Correct.
> >

On 13/12/2021 19:35, wij wrote:
> On Tuesday, 14 December 2021 at 01:51:57 UTC+8, Mike Terry wrote:
>> On 13/12/2021 15:53, wij wrote:
>>> On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
>>>> On 13/12/2021 12:27, wij wrote:
>>>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>>>>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>>>>> Example 1:
>>>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>>>>
>>>>>>>> ok. So A = 1
>>>>>>>>
>>>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>>>>
>>>>>>>> Does not compute. 999... and 1000... are not numbers.
>>>>>>>>
>>>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>>>>
>>>>>>>> No, of course not. (Does not compute)
>>>>>>>>
>>>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>>>>
>>>>>>>> Real numbers may have one or two decimal representations, a bit like
>>>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>>>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>>>>
>>>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>>>>
>>>>>>>> What density property is that? (And how do you think it is broken?)
>>>>>>>
>>>>>>> I believe he means the property that between any two members of the
>>>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>>>>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>>>>>
>>>>>> Yes, I thought he might mean that. I wouldn't call that "density"
>>>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
>>>>>> property?
>>>>>>
>>>>>
>>>>> Amazing! What a phenomenon!
>>>>>
>>>>> google "density property".
>>>>>
>>>>> Density Property::=
>>>>> For any two different numbers, there exists another different number in
>>>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
>>>>> i<k<j.
>>>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>>>
>>>> So we were right!
>>>>
>>>> So if A=1, how does this break the density property?
>>>
>>> When there ever exists a time a n such that A(n)=1,
>> Where does A(n) come from? I'll take it that you're using
>>
>> A(n) = 1 - (1/10)^n
>>
>> so A(n) is the sequence (0.9, 0.99, 0.999, ...)
>>
>> Well, there IS NO n SUCH THAT A(n) = 1
>
> ah hum, what is the point?
>

Above you said "When there ever exists a time a n such that A(n)=1,".

That is not a gramatical sentence, so I may have misunderstood what you
meant to say. There are two "a" words - the first (".. a time ..") must
be the indefinite article, but what is the second ("..time a n such..")?
It doesn't make sense so maybe you should clarify it. Also,
mathematical statements are not time-dependant, so what is the word
"time" doing in the sentence?

I guessed what you said as saying A(n) = 1 for some n, but there is no
such n. So I brought that to your attention. If you know and
understand that, just ignore my comment. :)

>>> "A(n)<A(n+1)<1"
>>> then, the density property does not hold.
>> 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.

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:
> >> On 13/12/2021 15:53, wij wrote:
> >>> On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
> >>>> On 13/12/2021 12:27, wij wrote:
> >>>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> >>>>>> On 12/12/2021 18:15, Richard Damon wrote:
> >>>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
> >>>>>>>> On 12/12/2021 08:48, wij wrote:
> >>>>>>>>> Example 1:
> >>>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>>>>>>>
> >>>>>>>> ok. So A = 1
> >>>>>>>>
> >>>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>>>>>>>
> >>>>>>>> Does not compute. 999... and 1000... are not numbers.
> >>>>>>>>
> >>>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>>>>>>>
> >>>>>>>> No, of course not. (Does not compute)
> >>>>>>>>
> >>>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
> >>>>>>>>
> >>>>>>>> Real numbers may have one or two decimal representations, a bit like
> >>>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >>>>>>>> 1.000.... The 3rd digit after the decimal point of representation
> >>>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
> >>>>>>>>
> >>>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>>>>>>>
> >>>>>>>> What density property is that? (And how do you think it is broken?)
> >>>>>>>
> >>>>>>> I believe he means the property that between any two members of the
> >>>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
> >>>>>>> between them. I.E., there is NOT a 'next' value from a given value.
> >>>>>>>
> >>>>>> Yes, I thought he might mean that. I wouldn't call that "density"
> >>>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
> >>>>>> property?
> >>>>>>
> >>>>>
> >>>>> Amazing! What a phenomenon!
> >>>>>
> >>>>> google "density property".
> >>>>>
> >>>>> Density Property::=
> >>>>> For any two different numbers, there exists another different number in
> >>>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> >>>>> i<k<j.
> >>>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >>>>>
> >>>> So we were right!
> >>>>
> >>>> So if A=1, how does this break the density property?
> >>>
> >>> When there ever exists a time a n such that A(n)=1,
> >> Where does A(n) come from? I'll take it that you're using
> >>
> >> A(n) = 1 - (1/10)^n
> >>
> >> so A(n) is the sequence (0.9, 0.99, 0.999, ...)
> >>
> >> Well, there IS NO n SUCH THAT A(n) = 1
> >
> > ah hum, what is the point?
> >
> Above you said "When there ever exists a time a n such that A(n)=1,".
>
> That is not a gramatical sentence, so I may have misunderstood what you
> meant to say. There are two "a" words - the first (".. a time ..") must
> be the indefinite article, but what is the second ("..time a n such..")?
> It doesn't make sense so maybe you should clarify it. Also,
> mathematical statements are not time-dependant, so what is the word
> "time" doing in the sentence?
>
> I guessed what you said as saying A(n) = 1 for some n, but there is no
> such n. So I brought that to your attention. If you know and
> understand that, just ignore my comment. :)
> >>> "A(n)<A(n+1)<1"
> >>> then, the density property does not hold.
> >> 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 12/13/21 10:53 AM, wij wrote:
> On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
>> On 13/12/2021 12:27, wij wrote:
>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>>> Example 1:
>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>>
>>>>>> ok. So A = 1
>>>>>>
>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>>
>>>>>> Does not compute. 999... and 1000... are not numbers.
>>>>>>
>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>>
>>>>>> No, of course not. (Does not compute)
>>>>>>
>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>>
>>>>>> Real numbers may have one or two decimal representations, a bit like
>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>>
>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>>
>>>>>> What density property is that? (And how do you think it is broken?)
>>>>>
>>>>> I believe he means the property that between any two members of the
>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>>>
>>>> Yes, I thought he might mean that. I wouldn't call that "density"
>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
>>>> property?
>>>>
>>>
>>> Amazing! What a phenomenon!
>>>
>>> google "density property".
>>>
>>> Density Property::=
>>> For any two different numbers, there exists another different number in
>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
>>> i<k<j.
>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>
>> So we were right!
>>
>> So if A=1, how does this break the density property?
>
> When there ever exists a time a n such that A(n)=1, "A(n)<A(n+1)<1"
> then, the density property does not hold.
> This leads to 0.999...≠1.

But the question wasn't if A(n) was 1, the question was if A, the limit
as n goes to infinity was 1. The limit of a sequence is normally NOT an
element of the sequence.

Note, as pointed below, 'infinity' is not a value in N, Q, or R.

>
>>>> But if A=1, how does this break the property? It doesn't - the property
>>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>>>> everything else he said. :) [no surprise, I guess.]
>>>>
>>>> Mike.
>>>>> One value between x and y will be (x+y)/2
>>>>>
>>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>>>> THAT breaks the density property, as there can be no number bigger than
>>>>> 0.9999... and less than 1.0000
>>>>>
>>>>>> Mike.
>>>>>
>>>
>>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>>>
>>> Given two different numbser A(n), and 1, there always exists another different
>>> number A(n+1) such that A(n)<A(n+1)<1
>> Correct.
>>
>>>
>>> When A(n)=1? Infinity?
>>
>> A(n) < 1 for all n. (n = oo is not in the range of n)
>>
>>
>> Mike.
>
> Correct.
>

On 12/13/21 10:25 AM, wij wrote:
> On Monday, 13 December 2021 at 20:53:29 UTC+8, richar...@gmail.com wrote:
>> On 12/13/21 7:27 AM, wij wrote:
>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>>> Example 1:
>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>>
>>>>>> ok. So A = 1
>>>>>>
>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>>
>>>>>> Does not compute. 999... and 1000... are not numbers.
>>>>>>
>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>>
>>>>>> No, of course not. (Does not compute)
>>>>>>
>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>>
>>>>>> Real numbers may have one or two decimal representations, a bit like
>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>>
>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>>
>>>>>> What density property is that? (And how do you think it is broken?)
>>>>>
>>>>> I believe he means the property that between any two members of the
>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>>>
>>>> Yes, I thought he might mean that. I wouldn't call that "density"
>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
>>>> property?
>>>>
>>>
>>> Amazing! What a phenomenon!
>>>
>>> google "density property".
>>>
>>> Density Property::=
>>> For any two different numbers, there exists another different number in
>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
>>> i<k<j.
>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>
>>>> But if A=1, how does this break the property? It doesn't - the property
>>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>>>> everything else he said. :) [no surprise, I guess.]
>>>>
>>>> Mike.
>>>>> One value between x and y will be (x+y)/2
>>>>>
>>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>>>> THAT breaks the density property, as there can be no number bigger than
>>>>> 0.9999... and less than 1.0000
>>>>>
>>>>>> Mike.
>>>>>
>>>
>>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>>>
>>> Given two different numbser A(n), and 1, there always exists another different
>>> number A(n+1) such that A(n)<A(n+1)<1
>>>
>>> When A(n)=1? Infinity?
>>>
>> There is no FINITE n where A(n) is equal to 1
>>
>
> Neither a FINITE n is in limit.
> What is the n in "lim(n->∞) A(n)=1"? Finite, infinite, or not a number?

Each n is a finite number.

The key is that the limit of a sequence doesn't need to be a member of
the sequence, and in fact, normally isn't.

>
>> Note, for the Reals, Naturals, etc., 'Infinity' isn't a value, only a
>> 'limiting case'
>>
>> Thus the limit(n->infinity) A(n) is 1, even though no individual A(n) is 1.
>>
>> This is a common property of limits.
>>
>
> "No individual A(n) is 1. But limit(n->infinity) A(n) is 1".
> So limit theory turns 'approaching' to 'equal' in term of the limit smoke.
> Where I can find evidence that A(∞)=1 but from the 'approaching is equal'
> theory is the problem.

Right, the terms approach the limit.

The limit is that value that terms get arbitraryily close to.

A(infinity) isn't a proper notation, as A is a sequnce with Natural
Number indexes, and infinity isn't a Natural Number.

One definition of 'The Limit' of a sequence is the number L, that for
any given arbirary positive value e, there is some N where all elements
of the seqence A(n), for all n > N, that |A(n) - L| < e

i.e, for any arbitrarily chosen precision, we can find a point in the
sequence where it stays inside that bound.

>
>>
>> Just like 0.9999... for any finite number of 9s isn't equal to 1, but
>> the limiting case with the endless 9s is.
>
> I am not talking about 'limiting case'. Limit theory is full of inconsistency.
> (Every one learned 'limit method' should have a sense of this. I do not what to dig into this shit deep)
> We should be interested in the case that 0.999... equal to 1 or not, not the "limiting case".

Maybe you should look at it again.

If you aren't going to use the right definition of Limit, and the range
of the Natural, Rational, and Real number, don't use those terms.

>
> And, here, right now, the density property in this thread.
> Does not 'density property' mean to hold infinitely?
> The problem is: The density property procedure can go on infinitely. Can not?
>

Not sure what you mean by 'infinitely' here, especially if you reject
the concept of a limit. It can be done an unbounded number of times.

Remember, when we are talking about counting with Natural numbers, there
is NO infinity. Infinity is just a limit we can approach.

On Tuesday, 14 December 2021 at 08:32:06 UTC+8, richar...@gmail.com wrote:
> On 12/13/21 10:53 AM, wij wrote:
> > On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
> >> On 13/12/2021 12:27, wij wrote:
> >>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> >>>> On 12/12/2021 18:15, Richard Damon wrote:
> >>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
> >>>>>> On 12/12/2021 08:48, wij wrote:
> >>>>>>> Example 1:
> >>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>>>>>
> >>>>>> ok. So A = 1
> >>>>>>
> >>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>>>>>
> >>>>>> Does not compute. 999... and 1000... are not numbers.
> >>>>>>
> >>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>>>>>
> >>>>>> No, of course not. (Does not compute)
> >>>>>>
> >>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
> >>>>>>
> >>>>>> Real numbers may have one or two decimal representations, a bit like
> >>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >>>>>> 1.000.... The 3rd digit after the decimal point of representation
> >>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
> >>>>>>
> >>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>>>>>
> >>>>>> What density property is that? (And how do you think it is broken?)
> >>>>>
> >>>>> I believe he means the property that between any two members of the
> >>>>> Real, or the Rationals, there will ALWAYS be another member of that set
> >>>>> between them. I.E., there is NOT a 'next' value from a given value.
> >>>>>
> >>>> Yes, I thought he might mean that. I wouldn't call that "density"
> >>>> myself, as "density" has a different meaning. Perhaps the "denseness"
> >>>> property?
> >>>>
> >>>
> >>> Amazing! What a phenomenon!
> >>>
> >>> google "density property".
> >>>
> >>> Density Property::=
> >>> For any two different numbers, there exists another different number in
> >>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> >>> i<k<j.
> >>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >>>
> >> So we were right!
> >>
> >> So if A=1, how does this break the density property?
> >
> > When there ever exists a time a n such that A(n)=1, "A(n)<A(n+1)<1"
> > then, the density property does not hold.
> > This leads to 0.999...≠1.
> But the question wasn't if A(n) was 1, the question was if A, the limit
> as n goes to infinity was 1. The limit of a sequence is normally NOT an
> element of the sequence.

The question was whether A(n) equals 1 or not when n approaches infinity and the
related density property.
You were still stuck in using 'limit theory' to miss the issue.

> Note, as pointed below, 'infinity' is not a value in N, Q, or R.
> >
> >>>> But if A=1, how does this break the property? It doesn't - the property
> >>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
> >>>> everything else he said. :) [no surprise, I guess.]
> >>>>
> >>>> Mike.
> >>>>> One value between x and y will be (x+y)/2
> >>>>>
> >>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
> >>>>> THAT breaks the density property, as there can be no number bigger than
> >>>>> 0.9999... and less than 1.0000
> >>>>>
> >>>>>> Mike.
> >>>>>
> >>>
> >>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
> >>>
> >>> Given two different numbser A(n), and 1, there always exists another different
> >>> number A(n+1) such that A(n)<A(n+1)<1
> >> Correct.
> >>
> >>>
> >>> When A(n)=1? Infinity?
> >>
> >> A(n) < 1 for all n. (n = oo is not in the range of n)
> >>
> >>
> >> Mike.
> >
> > Correct.
> >

On Tuesday, 14 December 2021 at 11:34:59 UTC+8, Richard Damon wrote:
> On 12/13/21 10:25 AM, wij wrote:
> > On Monday, 13 December 2021 at 20:53:29 UTC+8, richar...@gmail.com wrote:
> >> On 12/13/21 7:27 AM, wij wrote:
> >>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
> >>>> On 12/12/2021 18:15, Richard Damon wrote:
> >>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
> >>>>>> On 12/12/2021 08:48, wij wrote:
> >>>>>>> Example 1:
> >>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
> >>>>>>
> >>>>>> ok. So A = 1
> >>>>>>
> >>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
> >>>>>>
> >>>>>> Does not compute. 999... and 1000... are not numbers.
> >>>>>>
> >>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
> >>>>>>
> >>>>>> No, of course not. (Does not compute)
> >>>>>>
> >>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
> >>>>>>
> >>>>>> Real numbers may have one or two decimal representations, a bit like
> >>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
> >>>>>> 1.000.... The 3rd digit after the decimal point of representation
> >>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
> >>>>>>
> >>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
> >>>>>>
> >>>>>> What density property is that? (And how do you think it is broken?)
> >>>>>
> >>>>> I believe he means the property that between any two members of the
> >>>>> Real, or the Rationals, there will ALWAYS be another member of that set
> >>>>> between them. I.E., there is NOT a 'next' value from a given value.
> >>>>>
> >>>> Yes, I thought he might mean that. I wouldn't call that "density"
> >>>> myself, as "density" has a different meaning. Perhaps the "denseness"
> >>>> property?
> >>>>
> >>>
> >>> Amazing! What a phenomenon!
> >>>
> >>> google "density property".
> >>>
> >>> Density Property::=
> >>> For any two different numbers, there exists another different number in
> >>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
> >>> i<k<j.
> >>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> >>>
> >>>> But if A=1, how does this break the property? It doesn't - the property
> >>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
> >>>> everything else he said. :) [no surprise, I guess.]
> >>>>
> >>>> Mike.
> >>>>> One value between x and y will be (x+y)/2
> >>>>>
> >>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
> >>>>> THAT breaks the density property, as there can be no number bigger than
> >>>>> 0.9999... and less than 1.0000
> >>>>>
> >>>>>> Mike.
> >>>>>
> >>>
> >>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
> >>>
> >>> Given two different numbser A(n), and 1, there always exists another different
> >>> number A(n+1) such that A(n)<A(n+1)<1
> >>>
> >>> When A(n)=1? Infinity?
> >>>
> >> There is no FINITE n where A(n) is equal to 1
> >>
> >
> > Neither a FINITE n is in limit.
> > What is the n in "lim(n->∞) A(n)=1"? Finite, infinite, or not a number?
> Each n is a finite number.
>
> The key is that the limit of a sequence doesn't need to be a member of
> the sequence, and in fact, normally isn't.
> >
> >> Note, for the Reals, Naturals, etc., 'Infinity' isn't a value, only a
> >> 'limiting case'
> >>
> >> Thus the limit(n->infinity) A(n) is 1, even though no individual A(n) is 1.
> >>
> >> This is a common property of limits.
> >>
> >
> > "No individual A(n) is 1. But limit(n->infinity) A(n) is 1".
> > So limit theory turns 'approaching' to 'equal' in term of the limit smoke.
> > Where I can find evidence that A(∞)=1 but from the 'approaching is equal'
> > theory is the problem.
> Right, the terms approach the limit.
>
> The limit is that value that terms get arbitraryily close to.
>
> A(infinity) isn't a proper notation, as A is a sequnce with Natural
> Number indexes, and infinity isn't a Natural Number.
>
> One definition of 'The Limit' of a sequence is the number L, that for
> any given arbirary positive value e, there is some N where all elements
> of the seqence A(n), for all n > N, that |A(n) - L| < e
>
> i.e, for any arbitrarily chosen precision, we can find a point in the
> sequence where it stays inside that bound.
> >
> >>
> >> Just like 0.9999... for any finite number of 9s isn't equal to 1, but
> >> the limiting case with the endless 9s is.
> >
> > I am not talking about 'limiting case'. Limit theory is full of inconsistency.
> > (Every one learned 'limit method' should have a sense of this. I do not what to dig into this shit deep)
> > We should be interested in the case that 0.999... equal to 1 or not, not the "limiting case".
> Maybe you should look at it again.
>
> If you aren't going to use the right definition of Limit, and the range
> of the Natural, Rational, and Real number, don't use those terms.
> >
> > And, here, right now, the density property in this thread.
> > Does not 'density property' mean to hold infinitely?
> > The problem is: The density property procedure can go on infinitely. Can not?
> >
> Not sure what you mean by 'infinitely' here, especially if you reject
> the concept of a limit. It can be done an unbounded number of times.
>
> Remember, when we are talking about counting with Natural numbers, there
> is NO infinity. Infinity is just a limit we can approach.

Right. "Infinity is just a limit we can approach". So the following:
0.999... can never reach the limit 1.

Let f(n)= (2*n+1)!/((n!)^2*2^(3*k+1))
S= Σ(n=0,∞) f(n) = √2
S can never reach the limit √2 (albeit infinitely approaching, and, all the
instances of the sequence and the sum are rational).

This is the blind spot of Pythagoreans:
--- Infinitely approaching means equal. Number too small equals zero. ---

Yes, we should remember, "Infinity is just a limit we can approach".

On 12/14/21 5:10 AM, wij wrote:
> On Tuesday, 14 December 2021 at 11:34:59 UTC+8, Richard Damon wrote:
>> On 12/13/21 10:25 AM, wij wrote:
>>> On Monday, 13 December 2021 at 20:53:29 UTC+8, richar...@gmail.com wrote:
>>>> On 12/13/21 7:27 AM, wij wrote:
>>>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>>>>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>>>>> Example 1:
>>>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>>>>
>>>>>>>> ok. So A = 1
>>>>>>>>
>>>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>>>>
>>>>>>>> Does not compute. 999... and 1000... are not numbers.
>>>>>>>>
>>>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>>>>
>>>>>>>> No, of course not. (Does not compute)
>>>>>>>>
>>>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>>>>
>>>>>>>> Real numbers may have one or two decimal representations, a bit like
>>>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>>>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>>>>
>>>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>>>>
>>>>>>>> What density property is that? (And how do you think it is broken?)
>>>>>>>
>>>>>>> I believe he means the property that between any two members of the
>>>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>>>>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>>>>>
>>>>>> Yes, I thought he might mean that. I wouldn't call that "density"
>>>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
>>>>>> property?
>>>>>>
>>>>>
>>>>> Amazing! What a phenomenon!
>>>>>
>>>>> google "density property".
>>>>>
>>>>> Density Property::=
>>>>> For any two different numbers, there exists another different number in
>>>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
>>>>> i<k<j.
>>>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>>>
>>>>>> But if A=1, how does this break the property? It doesn't - the property
>>>>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>>>>>> everything else he said. :) [no surprise, I guess.]
>>>>>>
>>>>>> Mike.
>>>>>>> One value between x and y will be (x+y)/2
>>>>>>>
>>>>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>>>>>> THAT breaks the density property, as there can be no number bigger than
>>>>>>> 0.9999... and less than 1.0000
>>>>>>>
>>>>>>>> Mike.
>>>>>>>
>>>>>
>>>>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>>>>>
>>>>> Given two different numbser A(n), and 1, there always exists another different
>>>>> number A(n+1) such that A(n)<A(n+1)<1
>>>>>
>>>>> When A(n)=1? Infinity?
>>>>>
>>>> There is no FINITE n where A(n) is equal to 1
>>>>
>>>
>>> Neither a FINITE n is in limit.
>>> What is the n in "lim(n->∞) A(n)=1"? Finite, infinite, or not a number?
>> Each n is a finite number.
>>
>> The key is that the limit of a sequence doesn't need to be a member of
>> the sequence, and in fact, normally isn't.
>>>
>>>> Note, for the Reals, Naturals, etc., 'Infinity' isn't a value, only a
>>>> 'limiting case'
>>>>
>>>> Thus the limit(n->infinity) A(n) is 1, even though no individual A(n) is 1.
>>>>
>>>> This is a common property of limits.
>>>>
>>>
>>> "No individual A(n) is 1. But limit(n->infinity) A(n) is 1".
>>> So limit theory turns 'approaching' to 'equal' in term of the limit smoke.
>>> Where I can find evidence that A(∞)=1 but from the 'approaching is equal'
>>> theory is the problem.
>> Right, the terms approach the limit.
>>
>> The limit is that value that terms get arbitraryily close to.
>>
>> A(infinity) isn't a proper notation, as A is a sequnce with Natural
>> Number indexes, and infinity isn't a Natural Number.
>>
>> One definition of 'The Limit' of a sequence is the number L, that for
>> any given arbirary positive value e, there is some N where all elements
>> of the seqence A(n), for all n > N, that |A(n) - L| < e
>>
>> i.e, for any arbitrarily chosen precision, we can find a point in the
>> sequence where it stays inside that bound.
>>>
>>>>
>>>> Just like 0.9999... for any finite number of 9s isn't equal to 1, but
>>>> the limiting case with the endless 9s is.
>>>
>>> I am not talking about 'limiting case'. Limit theory is full of inconsistency.
>>> (Every one learned 'limit method' should have a sense of this. I do not what to dig into this shit deep)
>>> We should be interested in the case that 0.999... equal to 1 or not, not the "limiting case".
>> Maybe you should look at it again.
>>
>> If you aren't going to use the right definition of Limit, and the range
>> of the Natural, Rational, and Real number, don't use those terms.
>>>
>>> And, here, right now, the density property in this thread.
>>> Does not 'density property' mean to hold infinitely?
>>> The problem is: The density property procedure can go on infinitely. Can not?
>>>
>> Not sure what you mean by 'infinitely' here, especially if you reject
>> the concept of a limit. It can be done an unbounded number of times.
>>
>> Remember, when we are talking about counting with Natural numbers, there
>> is NO infinity. Infinity is just a limit we can approach.
>
> Right. "Infinity is just a limit we can approach". So the following:
> 0.999... can never reach the limit 1.

Wrong, by that logic we can't have a number like 0.9999.... or 0.3333...
because they are only that value 'in the limit' when we get to the
infinite number of digits.

No finite number of 9s in 0.99999 will be equal to 1, but IN THE LIMIT,
when we imagine that we reach that infinite end, it is.

>
> Let f(n)= (2*n+1)!/((n!)^2*2^(3*k+1))
> S= Σ(n=0,∞) f(n) = √2
> S can never reach the limit √2 (albeit infinitely approaching, and, all the
> instances of the sequence and the sum are rational).

Wrong, No S(k) = Σ(n=0,k) f(n) will equal √2, but IN THE LIMIT, S does.

>
> This is the blind spot of Pythagoreans:
> --- Infinitely approaching means equal. Number too small equals zero. ---
>
> Yes, we should remember, "Infinity is just a limit we can approach".

And thus, 'In the limit', we reach it.
Not for any finite step, but in the limit.
>

On 12/14/21 5:06 AM, wij wrote:
> On Tuesday, 14 December 2021 at 08:32:06 UTC+8, richar...@gmail.com wrote:
>> On 12/13/21 10:53 AM, wij wrote:
>>> On Monday, 13 December 2021 at 23:30:04 UTC+8, Mike Terry wrote:
>>>> On 13/12/2021 12:27, wij wrote:
>>>>> On Monday, 13 December 2021 at 02:38:48 UTC+8, Mike Terry wrote:
>>>>>> On 12/12/2021 18:15, Richard Damon wrote:
>>>>>>> On 12/12/21 10:24 AM, Mike Terry wrote:
>>>>>>>> On 12/12/2021 08:48, wij wrote:
>>>>>>>>> Example 1:
>>>>>>>>> Let A≡ Σ(n=1,∞) 9/10^n= 0.999...
>>>>>>>>
>>>>>>>> ok. So A = 1
>>>>>>>>
>>>>>>>>> = 999.../1000...= (3*3*(11...1))/(5*2)^n
>>>>>>>>
>>>>>>>> Does not compute. 999... and 1000... are not numbers.
>>>>>>>>
>>>>>>>>> If A=1, can we conclude that (3*3*(11..1)) equals to (5*2)^∞ ?
>>>>>>>>
>>>>>>>> No, of course not. (Does not compute)
>>>>>>>>
>>>>>>>>> If A=1, what is the 3rd digit after the decimal point of A?
>>>>>>>>
>>>>>>>> Real numbers may have one or two decimal representations, a bit like
>>>>>>>> +0 = -0, or 3/5 = 6/10. A has two representations: 0.999... and
>>>>>>>> 1.000.... The 3rd digit after the decimal point of representation
>>>>>>>> 0.999 is 9, while for the representation 1.000 it is 0.
>>>>>>>>
>>>>>>>>> If A=1, density property (of ℚ and ℝ) is broken (false).
>>>>>>>>
>>>>>>>> What density property is that? (And how do you think it is broken?)
>>>>>>>
>>>>>>> I believe he means the property that between any two members of the
>>>>>>> Real, or the Rationals, there will ALWAYS be another member of that set
>>>>>>> between them. I.E., there is NOT a 'next' value from a given value.
>>>>>>>
>>>>>> Yes, I thought he might mean that. I wouldn't call that "density"
>>>>>> myself, as "density" has a different meaning. Perhaps the "denseness"
>>>>>> property?
>>>>>>
>>>>>
>>>>> Amazing! What a phenomenon!
>>>>>
>>>>> google "density property".
>>>>>
>>>>> Density Property::=
>>>>> For any two different numbers, there exists another different number in
>>>>> between. Take interval [a,b] for instance: ∀i,j∈[a,b], i<j such that ∃k,
>>>>> i<k<j.
>>>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
>>>>>
>>>> So we were right!
>>>>
>>>> So if A=1, how does this break the density property?
>>>
>>> When there ever exists a time a n such that A(n)=1, "A(n)<A(n+1)<1"
>>> then, the density property does not hold.
>>> This leads to 0.999...≠1.
>> But the question wasn't if A(n) was 1, the question was if A, the limit
>> as n goes to infinity was 1. The limit of a sequence is normally NOT an
>> element of the sequence.
>
> The question was whether A(n) equals 1 or not when n approaches infinity and the
> related density property.
> You were still stuck in using 'limit theory' to miss the issue.

No, *A* was the INFINITE sum (not the finite partial sum on the way),
which can only be evaluate IN THE LIMIT, since infinity isn't a number
(in Natural, Rationals, or Reals).

Now, if you want to talk about a different field, lke the Surreals, then
the rules change.

>
>> Note, as pointed below, 'infinity' is not a value in N, Q, or R.
>>>
>>>>>> But if A=1, how does this break the property? It doesn't - the property
>>>>>> breaks if A != 1, so that would make wij's claim plain Wrong, like
>>>>>> everything else he said. :) [no surprise, I guess.]
>>>>>>
>>>>>> Mike.
>>>>>>> One value between x and y will be (x+y)/2
>>>>>>>
>>>>>>> The problem with thinking of 0.9999.... as something distinct from 1 is
>>>>>>> THAT breaks the density property, as there can be no number bigger than
>>>>>>> 0.9999... and less than 1.0000
>>>>>>>
>>>>>>>> Mike.
>>>>>>>
>>>>>
>>>>> Let A(0)=0, A(n)=(A(n-1)+1)/2, n∈ℕ
>>>>>
>>>>> Given two different numbser A(n), and 1, there always exists another different
>>>>> number A(n+1) such that A(n)<A(n+1)<1
>>>> Correct.
>>>>
>>>>>
>>>>> When A(n)=1? Infinity?
>>>>
>>>> A(n) < 1 for all n. (n = oo is not in the range of n)
>>>>
>>>>
>>>> Mike.
>>>
>>> Correct.
>>>