If 0.999... is rational, then:
0.999....= p/q (p,q∈ℕ)
<=> 0.999...*q=p
If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
since 0.999... is defined as infinite repeating, and q is finite,
but 0.999...*q is never finite.
Conclusion: Repeating decimals are irrational.

I am kind of reluctant to raise this issue again, but many problems would lead
to a basic issue: what is infinity. It is simple as it merely means "never end".
In this interpretation, 'time'/'steps' is meant existent in algorithms and lies hidden
in math. concept. E.g. Is infinity a natural numbers?
Think about that the Peano model does not reject infinity explicitly. I think
this is the root of many confusions (and errors in math.)

An article is published for this idea.
https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
The article is not aimed to explain the infinity idea (I added some to find out
adding too many such wordings would pollute the original intent)

wij <wyniijj2@gmail.com> writes:

> If 0.999... is rational, then:
> 0.999....= p/q (p,q∈ℕ)
> <=> 0.999...*q=p

Yes. For example 0.999... = 1/1.

> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> since 0.999... is defined as infinite repeating, and q is finite,
> but 0.999...*q is never finite.

0.999...*q is finite for all q in N.

> I am kind of reluctant to raise this issue again,

Really? Where would we be without someone denying (or not knowing) that
.... denotes a limit every few weeks?

--
Ben.

On Monday, 8 August 2022 at 21:16:39 UTC+8, Ben Bacarisse wrote:
> wij <wyni...@gmail.com> writes:
>
> > If 0.999... is rational, then:
> > 0.999....= p/q (p,q∈ℕ)
> > <=> 0.999...*q=p
> Yes. For example 0.999... = 1/1.

I don't see valid logical derivation.
By definition "repeating decimal" (0.999...) never ends.
"0.999..." and "1/1" is apparently different. So, what is the derivation?
We have gone through this a lot. Nothing from the limit theory is valid (this
is a very fundamental arithmetic problem, any 'higher' level theory will suffer
circular argument problem).

> > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > since 0.999... is defined as infinite repeating, and q is finite,
> > but 0.999...*q is never finite.
> 0.999...*q is finite for all q in N.

ditto. I don't see valid logical derivation.

> > I am kind of reluctant to raise this issue again,
> Really? Where would we be without someone denying (or not knowing) that
> ... denotes a limit every few weeks?
>
> --
> Ben.

I would say it's Emperor's clothes. Innocent child can see it.
You can keep teaching a false thing.

wij <wyniijj2@gmail.com> writes:

> On Monday, 8 August 2022 at 21:16:39 UTC+8, Ben Bacarisse wrote:
>> wij <wyni...@gmail.com> writes:
>>
>> > If 0.999... is rational, then:
>> > 0.999....= p/q (p,q∈ℕ)
>> > <=> 0.999...*q=p
>> Yes. For example 0.999... = 1/1.
>
> I don't see valid logical derivation.

That's to be expected. I don't think you know what 0.999... means.

> By definition "repeating decimal" (0.999...) never ends.

Yes, it represents a limit. (Technically, it is a never ending sequence
of partial sums, whose limit is the sum.)

> "0.999..." and "1/1" is apparently different.

Yes, and "2/2" and "1+0" and "001" are all apprently different too. But
all denote the same number: 1.

> So, what is the derivation?

0.999... means

Sum n=1 to oo [9/10^n]
= limit n->oo [Sum i=1,n [9/10^i]]
= 1

(Do you know how limits are calculated?)

> We have gone through this a lot.

Yes.

> Nothing from the limit theory is valid (this
> is a very fundamental arithmetic problem, any 'higher' level theory will suffer
> circular argument problem).

Then the notation 0.999... is invalid since it denotes a limit. Don't
write limits if you don't consider them valid.

Please don't play the classic crank game where you use conventional
notation to mean something vague that is not what the notation normally
means. That's so boring:

"The sun in made of cheese!", you say. And after a few rounds it turns
out that dairy products are invalid, so "cheese" must mean something
else.

> I would say it's Emperor's clothes.

Curious that. I am clothed in (a) a definition of what X.YYY... means;
(b) a consistent theory that defines the resulting limits.

You don't even have a definition of what the notation you are misusing
means. One of us certainly naked!

--
Ben.

On 8/8/2022 5:27 AM, wij wrote:
> If 0.999... is rational, then:
> 0.999....= p/q (p,q∈ℕ)
> <=> 0.999...*q=p
> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> since 0.999... is defined as infinite repeating, and q is finite,
> but 0.999...*q is never finite.
> Conclusion: Repeating decimals are irrational.

No. Repeating decimals are rational.

On Monday, August 8, 2022 at 2:37:59 PM UTC+1, wyni...@gmail.com wrote:
> On Monday, 8 August 2022 at 21:16:39 UTC+8, Ben Bacarisse wrote:
> > wij <wyni...@gmail.com> writes:
> >
> > > If 0.999... is rational, then:
> > > 0.999....= p/q (p,q∈ℕ)
> > > <=> 0.999...*q=p
> > Yes. For example 0.999... = 1/1.
> I don't see valid logical derivation.
> By definition "repeating decimal" (0.999...) never ends.
> "0.999..." and "1/1" is apparently different. So, what is the derivation?
> We have gone through this a lot. Nothing from the limit theory is valid (this
> is a very fundamental arithmetic problem, any 'higher' level theory will suffer
> circular argument problem).
> > > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > > since 0.999... is defined as infinite repeating, and q is finite,
> > > but 0.999...*q is never finite.
> > 0.999...*q is finite for all q in N.
> ditto. I don't see valid logical derivation.

I think you may be confusing "finite", meaning the size of the number, with "infinite repeating" where the calculation to determine the number does not stop.

0.999... is clearly more than 0, and clearly less than 2. 0 and 2 are both finite, so 0.999... is also finite.

Likewise, taking q=3 as an example, 3 * 0.999... > 0 and 3 * 0.999... < 6, so 3 * 0.999... is finite.

On 8/8/22 8:27 AM, wij wrote:
> If 0.999... is rational, then:
> 0.999....= p/q (p,q∈ℕ)
> <=> 0.999...*q=p
> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> since 0.999... is defined as infinite repeating, and q is finite,
> but 0.999...*q is never finite.
> Conclusion: Repeating decimals are irrational.

Wrong, 9/9 can generate the pattern 0.9999...

This can be proven by 9 = 3+3+3 and the assocative property thus 9/9 =
(3+3+3)/9 = 3/9 + 3/9 + 3/9 = 0.3333... + 0.3333... + 0.3333... =
0.9999....

>
> I am kind of reluctant to raise this issue again, but many problems would lead
> to a basic issue: what is infinity. It is simple as it merely means "never end".
> In this interpretation, 'time'/'steps' is meant existent in algorithms and lies hidden
> in math. concept. E.g. Is infinity a natural numbers?
> Think about that the Peano model does not reject infinity explicitly. I think
> this is the root of many confusions (and errors in math.)
>
> An article is published for this idea.
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> The article is not aimed to explain the infinity idea (I added some to find out
> adding too many such wordings would pollute the original intent)

On Monday, 8 August 2022 at 20:27:54 UTC+8, wij wrote:
> If 0.999... is rational, then:
> 0.999....= p/q (p,q∈ℕ)
> <=> 0.999...*q=p
> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> since 0.999... is defined as infinite repeating, and q is finite,
> but 0.999...*q is never finite.
> Conclusion: Repeating decimals are irrational.
>
> I am kind of reluctant to raise this issue again, but many problems would lead
> to a basic issue: what is infinity. It is simple as it merely means "never end".
> In this interpretation, 'time'/'steps' is meant existent in algorithms and lies hidden
> in math. concept. E.g. Is infinity a natural numbers?
> Think about that the Peano model does not reject infinity explicitly. I think
> this is the root of many confusions (and errors in math.)
>
> An article is published for this idea.
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> The article is not aimed to explain the infinity idea (I added some to find out
> adding too many such wordings would pollute the original intent)

To add more material came up to me (not well ordered):

----------------------------
There are quite a number of proofs of "repeating decimals are irrational".
The basic is the correct equation of 1/3 and its decimal form from long
division (kids understand this 'infinity' with no problem) should be:

1/3= 0.333... + nonzero_remainder.

----------------------------
To translate the 0.999... problem to limit:

Let A= lim(n->∞) 1-1/2^n = 0.999...
B= lim(n->∞) 1-1/10^n = 0.999...

Assume A=B
<=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
<=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
<=> lim(n->∞) 1 = lim(n->∞) 1/5^n
<=> 1=0

[Note] I just demonstrate an instance. The limit theory can evolve as it does
(e.g. one-sided limit... There are many slightly different versions of
interpretation of limit as it evolves). Readers might find different
authors use different rules.
Limit is a technic to find its 'limit', it cannot form a logically
consistent theory for real number, e.g. the result of limit in general
must be verified, e.g. numerically, one cannot absolutely trust the
result of limit arithmetic. And at final, lim(x->c) f(c)= L does not
'deduce' f(c)=L (In text book, probably just reads "lim(x->c) f(c)= L, SO
WRITTEN as f(c)=L"). Limit theory only says the limit of 0.999... is 1,
the theory does not say 0.999...=1. There is no equality concept in the
ε-δ theory.
If one resorts to Dedekind-cut-like theories (I did not really read it),
from the knowledge that all the combinations of discrete symbols cannot
represent all the real numbers, I can conclude what those theories
claim are false, let alone I suspect there should be circular arguments
there, because many terms there must be well defined as a fundamental
theory, are undefined (prove me wrong).

The limit example above demonstrated "0.999..." cannot denote a specific number,
which also means "repeating decimal" cannot specify a unique number (A!=B).
Using limit is invalid for me (for this question) but the result is correct,
see the provided reference (I found a typo there).

-----------------------
Simple arithmetic (this should also be a valid way 2.718... is calculated):
(0.999....)^n approaches 1/e
(1.000...1)^n approaches e (or defined as e)
A possible rebuttal might be that the (1-1/n) in lim(n->∞) (1-1/n)^n is an invalid
number (approximated like 0.999...), or it is a 'concept' etc...
But if it is not a number, the whole equation is broken.

-----------------------
A[0]=0
A[n]=(A[n-1]+1)/2

The density property says (implicitly) n can enumerate infinitely (otherwise, it
won't be a rule) and A[∞] never be 1. A[n] infinitely approaches 1 in form
like 0.999.... This is like in the case of the interval [0,1), infinite numbers
of 0.999...s are located near the open end of [0,1).
Can we infinitely refine the scale of a ruler and the last scale never touches
the scale of 1? I think, yes, something like the √2 story, otherwise all numbers
can be 'proved' rational.

wij <wyniijj2@gmail.com> writes:

> To add more material came up to me (not well ordered):
>
> ----------------------------
> There are quite a number of proofs of "repeating decimals are
> irrational".

I've not seen any. Are any of them published (you know what I mean
here -- properly published) so I can read them clearly presented?

> The basic is the correct equation of 1/3 and its decimal form from long
> division (kids understand this 'infinity' with no problem) should be:
>
> 1/3= 0.333... + nonzero_remainder.

Kids are not very good at mathematics. They tend to generalise without
justification. There is no r =/= 0 such that 1/3 = 0.333... + r.

This is because 0.333... means

lim(n->oo) Sum k=1,n (3/10^k)

which limit is 1/3. Proof of X by "kids think X" is not a sound
argument.

> ----------------------------
> To translate the 0.999... problem to limit:
>
> Let A= lim(n->∞) 1-1/2^n = 0.999...
> B= lim(n->∞) 1-1/10^n = 0.999...

Or, more simply, A=B=1.

> Assume A=B

Or, if you like, one can prove that A=B.

> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n

You can't do basic algebra. For pity's s, just pick up a maths book! I
presume you think you are multiplying both sizes by 2, yes? If so...

2 * lim(n->oo) 1/2^n = lim(n->oo) 2/2^n = lim(n->oo) 2/2^(n-1) = 0

> <=> 1=0

so no.

> Simple arithmetic (this should also be a valid way 2.718... is calculated):
> (0.999....)^n approaches 1/e

No it doesn't. Basic algebra again.

> (1.000...1)^n approaches e (or defined as e

Now this one in unclear. Usually, the ... in the middle like that
denote some constant number of repetitions. If that's what you mean,
then no. For any fixed x=1.000...1, x^n diverges.

> A[0]=0
> A[n]=(A[n-1]+1)/2
>
> The density property says (implicitly) n can enumerate infinitely
> (otherwise, it won't be a rule) and A[∞] never be 1.

This is literal nonsense. Inductive definitions like that for A define
A for all N (and no more). oo is not in N.

> A[n] infinitely approaches 1 in form like 0.999....

Yes, and ∀n A[n] < 1.

Where do you get this stuff? Is there some sort of Info Wars for maths
out there that I've missed?

--
Ben.

On 09/08/2022 13:07, wij wrote:
[Wij still:]
>> I am kind of reluctant to raise this issue again, [...].

Not reluctant enough, apparently!

> There are quite a number of proofs of "repeating decimals are irrational".

There are no /correct/ proofs of that.

> The basic is the correct equation of 1/3 and its decimal form from long
> division (kids understand this 'infinity' with no problem) should be:
> 1/3= 0.333... + nonzero_remainder.

"Kids" perhaps understand "1/3 == 0.333..." where "..." in their
minds means "and so on". They don't, in general, have an understanding
of limits, infinitesimals, non-standard analysis and other relevant
material. If you're going to perpetuate your "correct equation", then
you need at least to understand what you and/or others mean by "...".
It is normally taken, in this context, to mean the limit, in which case
there is no "non-zero remainder". If you want it to mean something else,
then what? You can't legitimately mix "Wij numbers" [ill-defined, thus
far] with the usual [standard] "real" numbers.

> To translate the 0.999... problem to limit:
> Let A= lim(n->∞) 1-1/2^n = 0.999...
> B= lim(n->∞) 1-1/10^n = 0.999...
> Assume A=B
> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n

OK so far; this is simply "0 == 0".

> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n

Illegitimate. You seem to have multiplied the equation
on the previous line by "2^n", but that is not a constant, and
indeed is increasing with "n" [which is not "free"]. See

https://en.wikipedia.org/wiki/Free_variables_and_bound_variables

[...]
> If one resorts to Dedekind-cut-like theories (I did not really read it),
> from the knowledge that all the combinations of discrete symbols cannot
> represent all the real numbers, I can conclude what those theories
> claim are false, let alone I suspect there should be circular arguments
> there, because many terms there must be well defined as a fundamental
> theory, are undefined (prove me wrong).

I try very hard to allow for English not being your first
language, but I cannot make sense of the above [or what you wrote
later]. But it does suggest that before making claims about numbers,
limits and so on, you ought at least to try to understand the basics
of analysis. It would help you a /lot/ if you looked up things like
"accumulation point":

https://en.wikipedia.org/wiki/Accumulation_point

and followed some of the material there, about limit points of sets,
boundary points, closed/open sets, neighbourhoods and similar.
Nothing on that page about Dedekind!

--
Andy Walker, Nottingham.
Andy's music pages: www.cuboid.me.uk/andy/Music
Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Ravel

On Mon, 8 Aug 2022 05:27:52 -0700 (PDT)
wij <wyniijj2@gmail.com> wrote:

> If 0.999... is rational, then:
> 0.999....= p/q (p,q∈ℕ)
> <=> 0.999...*q=p
> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> since 0.999... is defined as infinite repeating, and q is finite,
> but 0.999...*q is never finite.
> Conclusion: Repeating decimals are irrational.
>
> I am kind of reluctant to raise this issue again, but many problems
> would lead to a basic issue: what is infinity. It is simple as it
> merely means "never end". In this interpretation, 'time'/'steps' is
> meant existent in algorithms and lies hidden in math. concept. E.g.
> Is infinity a natural numbers? Think about that the Peano model does
> not reject infinity explicitly. I think this is the root of many
> confusions (and errors in math.)
>
> An article is published for this idea.
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> The article is not aimed to explain the infinity idea (I added some
> to find out adding too many such wordings would pollute the original
> intent)

Repeating decimals in base 10 will terminate in some other base thus
are rational numbers and not irrational numbers.

/Flibble

On Tuesday, 9 August 2022 at 23:53:06 UTC+8, Andy Walker wrote:
> On 09/08/2022 13:07, wij wrote:
> [Wij still:]
> >> I am kind of reluctant to raise this issue again, [...].
>
> Not reluctant enough, apparently!
> > There are quite a number of proofs of "repeating decimals are irrational".
> There are no /correct/ proofs of that.
> > The basic is the correct equation of 1/3 and its decimal form from long
> > division (kids understand this 'infinity' with no problem) should be:
> > 1/3= 0.333... + nonzero_remainder.
> "Kids" perhaps understand "1/3 == 0.333..." where "..." in their
> minds means "and so on". They don't, in general, have an understanding
> of limits, infinitesimals, non-standard analysis and other relevant
> material. If you're going to perpetuate your "correct equation", then
> you need at least to understand what you and/or others mean by "...".
> It is normally taken, in this context, to mean the limit, in which case
> there is no "non-zero remainder". If you want it to mean something else,
> then what? You can't legitimately mix "Wij numbers" [ill-defined, thus
> far] with the usual [standard] "real" numbers.
> > To translate the 0.999... problem to limit:
> > Let A= lim(n->∞) 1-1/2^n = 0.999...
> > B= lim(n->∞) 1-1/10^n = 0.999...
> > Assume A=B
> > <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> > <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> OK so far; this is simply "0 == 0".
> > <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
> Illegitimate. You seem to have multiplied the equation
> on the previous line by "2^n", but that is not a constant, and
> indeed is increasing with "n" [which is not "free"]. See
>
> https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
>

I purposely omitted comment to this SO SIMPLE derivation.
https://tutorial.math.lamar.edu/classes/calci/limitsproperties.aspx
You and Ben really showed you don't understand what you think understand.
When things come to 'research level', you 'teachers' are all failed.
Ben is worse, like to play clever talks as if it seems a clever proof.

> [...]
> > If one resorts to Dedekind-cut-like theories (I did not really read it),
> > from the knowledge that all the combinations of discrete symbols cannot
> > represent all the real numbers, I can conclude what those theories
> > claim are false, let alone I suspect there should be circular arguments
> > there, because many terms there must be well defined as a fundamental
> > theory, are undefined (prove me wrong).
> I try very hard to allow for English not being your first
> language, but I cannot make sense of the above [or what you wrote
> later]. But it does suggest that before making claims about numbers,
> limits and so on, you ought at least to try to understand the basics
> of analysis. It would help you a /lot/ if you looked up things like
> "accumulation point":
>
> https://en.wikipedia.org/wiki/Accumulation_point
>
> and followed some of the material there, about limit points of sets,
> boundary points, closed/open sets, neighbourhoods and similar.
> Nothing on that page about Dedekind!
>
> --
> Andy Walker, Nottingham.
> Andy's music pages: www.cuboid.me.uk/andy/Music
> Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Ravel

Keep reciting your 'trained' dogma to your students.

On Wednesday, 10 August 2022 at 01:09:36 UTC+8, Mr Flibble wrote:
> On Mon, 8 Aug 2022 05:27:52 -0700 (PDT)
> wij <wyni...@gmail.com> wrote:
>
> > If 0.999... is rational, then:
> > 0.999....= p/q (p,q∈ℕ)
> > <=> 0.999...*q=p
> > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > since 0.999... is defined as infinite repeating, and q is finite,
> > but 0.999...*q is never finite.
> > Conclusion: Repeating decimals are irrational.
> >
> > I am kind of reluctant to raise this issue again, but many problems
> > would lead to a basic issue: what is infinity. It is simple as it
> > merely means "never end". In this interpretation, 'time'/'steps' is
> > meant existent in algorithms and lies hidden in math. concept. E.g.
> > Is infinity a natural numbers? Think about that the Peano model does
> > not reject infinity explicitly. I think this is the root of many
> > confusions (and errors in math.)
> >
> > An article is published for this idea.
> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> > The article is not aimed to explain the infinity idea (I added some
> > to find out adding too many such wordings would pollute the original
> > intent)
> Repeating decimals in base 10 will terminate in some other base thus
> are rational numbers and not irrational numbers.
>
> /Flibble

Rational number means the ratio of two integer numbers p/q.

On Tue, 9 Aug 2022 10:42:48 -0700 (PDT)
wij <wyniijj2@gmail.com> wrote:

> On Wednesday, 10 August 2022 at 01:09:36 UTC+8, Mr Flibble wrote:
> > On Mon, 8 Aug 2022 05:27:52 -0700 (PDT)
> > wij <wyni...@gmail.com> wrote:
> >
> > > If 0.999... is rational, then:
> > > 0.999....= p/q (p,q∈ℕ)
> > > <=> 0.999...*q=p
> > > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > > since 0.999... is defined as infinite repeating, and q is finite,
> > > but 0.999...*q is never finite.
> > > Conclusion: Repeating decimals are irrational.
> > >
> > > I am kind of reluctant to raise this issue again, but many
> > > problems would lead to a basic issue: what is infinity. It is
> > > simple as it merely means "never end". In this interpretation,
> > > 'time'/'steps' is meant existent in algorithms and lies hidden in
> > > math. concept. E.g. Is infinity a natural numbers? Think about
> > > that the Peano model does not reject infinity explicitly. I think
> > > this is the root of many confusions (and errors in math.)
> > >
> > > An article is published for this idea.
> > > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> > > The article is not aimed to explain the infinity idea (I added
> > > some to find out adding too many such wordings would pollute the
> > > original intent)
> > Repeating decimals in base 10 will terminate in some other base
> > thus are rational numbers and not irrational numbers.
> >
> > /Flibble
>
> Rational number means the ratio of two integer numbers p/q.

I know what rational numbers are: not sure why you felt the need to
state the bleeding obvious. Numbers are either rational or irrational.

/Flibble

On Wednesday, 10 August 2022 at 06:35:56 UTC+8, Mr Flibble wrote:
> On Tue, 9 Aug 2022 10:42:48 -0700 (PDT)
> wij <wyni...@gmail.com> wrote:
>
> > On Wednesday, 10 August 2022 at 01:09:36 UTC+8, Mr Flibble wrote:
> > > On Mon, 8 Aug 2022 05:27:52 -0700 (PDT)
> > > wij <wyni...@gmail.com> wrote:
> > >
> > > > If 0.999... is rational, then:
> > > > 0.999....= p/q (p,q∈ℕ)
> > > > <=> 0.999...*q=p
> > > > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > > > since 0.999... is defined as infinite repeating, and q is finite,
> > > > but 0.999...*q is never finite.
> > > > Conclusion: Repeating decimals are irrational.
> > > >
> > > > I am kind of reluctant to raise this issue again, but many
> > > > problems would lead to a basic issue: what is infinity. It is
> > > > simple as it merely means "never end". In this interpretation,
> > > > 'time'/'steps' is meant existent in algorithms and lies hidden in
> > > > math. concept. E.g. Is infinity a natural numbers? Think about
> > > > that the Peano model does not reject infinity explicitly. I think
> > > > this is the root of many confusions (and errors in math.)
> > > >
> > > > An article is published for this idea.
> > > > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> > > > The article is not aimed to explain the infinity idea (I added
> > > > some to find out adding too many such wordings would pollute the
> > > > original intent)
> > > Repeating decimals in base 10 will terminate in some other base
> > > thus are rational numbers and not irrational numbers.
> > >
> > > /Flibble
> >
> > Rational number means the ratio of two integer numbers p/q.
> I know what rational numbers are: not sure why you felt the need to
> state the bleeding obvious. Numbers are either rational or irrational.
>
> /Flibble

Save craps, just prove what you say.
I can input 0.999... to my computer to any desired length and expect (0.999....)^n
can approach 1/e to any precision I desire.
I can input 1.000..1 to my computer to any desired length and expect (1.000...1)^n
can approach e to any precision I desire.

Note that lim(n->∞) (1+k/n)^n is a strictly increasing or decreasing function.
This is my early "exponential amplifier" proof.
Try your computer to see if it agree with your "bleeding obvious".
(I assume you cannot see there are infinite number of 0.999... are at the open
end of interval [0,1), and 1.00..1 also equals to 1)

On Tuesday, 9 August 2022 at 17:15:14 UTC+2, Ben Bacarisse wrote:
> This is because 0.333... means
>
> lim(n->oo) Sum k=1,n (3/10^k)

I notice that you think yourself very skilled in telling other people what stuff means.
So I am curious whether you know what a limit means.

Are you using the epsilon-delta definition, or the infinitesimal definition?

On Monday, 8 August 2022 at 14:27:54 UTC+2, wyni...@gmail.com wrote:
> If 0.999... is rational, then:
> 0.999....= p/q (p,q∈ℕ)
> <=> 0.999...*q=p
> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> since 0.999... is defined as infinite repeating, and q is finite,
> but 0.999...*q is never finite.
> Conclusion: Repeating decimals are irrational.
>
> I am kind of reluctant to raise this issue again, but many problems would lead
> to a basic issue: what is infinity. It is simple as it merely means "never end".
> In this interpretation, 'time'/'steps' is meant existent in algorithms and lies hidden
> in math. concept. E.g. Is infinity a natural numbers?
> Think about that the Peano model does not reject infinity explicitly. I think
> this is the root of many confusions (and errors in math.)
>
> An article is published for this idea.
> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> The article is not aimed to explain the infinity idea (I added some to find out
> adding too many such wordings would pollute the original intent)

You are raising all the usual objections to these issues and you are going to be brushed away by all the dogmatists who know the “correct way” to think about the problem.

Asking what infinity is doesn’t give you any useful answers either.
Go the other way - start with infinitesimals and the hyperreal number line…

https://www.khanacademy.org/college-careers-more/bjc/2015-challenge/2015-math/v/infinitesimals-non-standards-analysis

On Monday, 8 August 2022 at 15:16:39 UTC+2, Ben Bacarisse wrote:
> Yes. For example 0.999... = 1/1.

Ok. So it needs saying so I will say it. You are an ass.

Few posts ago (on another thread) you were giving me a lecture about functions needing domains/context.

Whether 0.999... = 1/1 is a theorem or not absolutely depends on the number system. The domain of your symbols.

It is a theorem in R.
It isn’t a theorem in *R

On Tuesday, 9 August 2022 at 02:30:19 UTC+2, richar...@gmail.com wrote:
> On 8/8/22 8:27 AM, wij wrote:
> > If 0.999... is rational, then:
> > 0.999....= p/q (p,q∈ℕ)
> > <=> 0.999...*q=p
> > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > since 0.999... is defined as infinite repeating, and q is finite,
> > but 0.999...*q is never finite.
> > Conclusion: Repeating decimals are irrational.
> Wrong, 9/9 can generate the pattern 0.9999...
>
> This can be proven by 9 = 3+3+3 and the assocative property thus 9/9 =
> (3+3+3)/9 = 3/9 + 3/9 + 3/9 = 0.3333... + 0.3333... + 0.3333... =
> 0.9999....

Ooooh! The associative property! I know how to abuse that too.

S = 1 + 2 + 3 + 4 + 5 ...
S = 1 + (2+3+4) + (5+6+7) + (8+9+10)...
S-1 = 9 + 18 + 28 ...
S-1 = 9(1 + 2 + 3 ...)
S-1 = 9S
-8S = 1
S = -1/8
1+2+3+4+5... = -1/8

On 8/11/22 4:50 AM, Skep Dick wrote:
> On Monday, 8 August 2022 at 15:16:39 UTC+2, Ben Bacarisse wrote:
>> Yes. For example 0.999... = 1/1.
>
> Ok. So it needs saying so I will say it. You are an ass.
>
> Few posts ago (on another thread) you were giving me a lecture about functions needing domains/context.
>
> Whether 0.999... = 1/1 is a theorem or not absolutely depends on the number system. The domain of your symbols.
>
> It is a theorem in R.
> It isn’t a theorem in *R
>
>

Which is why you need to be clear what domain you are talking about,
especially if you are not talking about the domain that people will assume.

That just shows that you believe it is ok to be deceptive by not being
clear.

The number, 0.9999... if not otherwise indicated by context, is a Real
Number in the Real Number system. To later say you are talking about
some other system *R, is just showing you are being deceptive.

Virtually ANY statement can be proven or disproven if you allow it to be
moved into an arbitrary logic system crafted for that purpose, so
context is vital for communication.

On 8/11/22 5:28 AM, Skep Dick wrote:
> On Tuesday, 9 August 2022 at 02:30:19 UTC+2, richar...@gmail.com wrote:
>> On 8/8/22 8:27 AM, wij wrote:
>>> If 0.999... is rational, then:
>>> 0.999....= p/q (p,q∈ℕ)
>>> <=> 0.999...*q=p
>>> If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
>>> since 0.999... is defined as infinite repeating, and q is finite,
>>> but 0.999...*q is never finite.
>>> Conclusion: Repeating decimals are irrational.
>> Wrong, 9/9 can generate the pattern 0.9999...
>>
>> This can be proven by 9 = 3+3+3 and the assocative property thus 9/9 =
>> (3+3+3)/9 = 3/9 + 3/9 + 3/9 = 0.3333... + 0.3333... + 0.3333... =
>> 0.9999....
>
> Ooooh! The associative property! I know how to abuse that too.
>
> S = 1 + 2 + 3 + 4 + 5 ...
> S = 1 + (2+3+4) + (5+6+7) + (8+9+10)...
> S-1 = 9 + 18 + 28 ...
> S-1 = 9(1 + 2 + 3 ...)
> S-1 = 9S
> -8S = 1
> S = -1/8 > 1+2+3+4+5... = -1/8
>
>
>

Except that you did it with non-convergent sums, I didn't it with a
finite value.

The value of the sum 1 + 2 + 3 + 4 + 5 + ... doesn't HAVE a value, so
you can't apply associativity.

The value 1 / 3 as 0.3333... IS a value (since you propose that
0.9999... is something that can be evaluated)

The key here is the assocative property applies to the domain of the REALS.

1/3 is a real

1 + 2 + 3 + 4 + 5 .... is not, so you can't apply it there.

It is a well established property of infinite sums that many of the
conventional operations only "work" if the sum is in fact, convergent.

The thing you run into in your proof is that the your first line we get
the value of "infinity" for S.

The problem is infinity-1 is the same as infinity (at least for some
versions of infinity), so everything breaks down. The number systems
that handle the new values, like infinity, have this annoying problem of
losing characteristics that we like to assume we have.

On Thursday, 11 August 2022 at 13:55:09 UTC+2, richar...@gmail.com wrote:
> Except that you did it with non-convergent sums, I didn't it with a
> finite value.
So what? You did it with an infinite representation.

If you are going to substitute a finite representation with an infinite one it's on you to prove that they are equivalent representations of the object.

> 1/3 is a real
Why do you say that?

1/3 ∈ ℚ
1/3 ∈ *R
1/3 ∈ R

It's whatever you want it to be.

> 1 + 2 + 3 + 4 + 5 .... is not, so you can't apply it there.
That's a lot of talk what I can and can't do. Where's this rulebook of yours?

You sure as hell haven't produces anything formal yet!

> It is a well established property of infinite sums that many of the
> conventional operations only "work" if the sum is in fact, convergent.
> The thing you run into in your proof is that the your first line we get
> the value of "infinity" for S.

Which is a perfectly valid thing to do in some number systems.
Why are you discriminating against those number systems?

> The problem is infinity-1 is the same as infinity (at least for some
> versions of infinity), so everything breaks down.
Not in my model. Maybe your model is incomplete?

> that handle the new values, like infinity, have this annoying problem of
> losing characteristics that we like to assume we have.
That's a really easy problem to solve! Stop making stupid assumptions.

On Thursday, 11 August 2022 at 16:45:05 UTC+8, Skep Dick wrote:
> On Monday, 8 August 2022 at 14:27:54 UTC+2, wyni...@gmail.com wrote:
> > If 0.999... is rational, then:
> > 0.999....= p/q (p,q∈ℕ)
> > <=> 0.999...*q=p
> > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > since 0.999... is defined as infinite repeating, and q is finite,
> > but 0.999...*q is never finite.
> > Conclusion: Repeating decimals are irrational.
> >
> > I am kind of reluctant to raise this issue again, but many problems would lead
> > to a basic issue: what is infinity. It is simple as it merely means "never end".
> > In this interpretation, 'time'/'steps' is meant existent in algorithms and lies hidden
> > in math. concept. E.g. Is infinity a natural numbers?
> > Think about that the Peano model does not reject infinity explicitly. I think
> > this is the root of many confusions (and errors in math.)
> >
> > An article is published for this idea.
> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download
> > The article is not aimed to explain the infinity idea (I added some to find out
> > adding too many such wordings would pollute the original intent)
> You are raising all the usual objections to these issues and you are going to be brushed away by all the dogmatists who know the “correct way” to think about the problem.
>
> Asking what infinity is doesn’t give you any useful answers either.
> Go the other way - start with infinitesimals and the hyperreal number line…
>
> https://www.khanacademy.org/college-careers-more/bjc/2015-challenge/2015-math/v/infinitesimals-non-standards-analysis

The video seems blocked to where I live. I had skimmed a little what so called
"non-standard analysis", I did not like the impression read to me (too artificial).

On Thursday, 11 August 2022 at 17:29:00 UTC+8, Skep Dick wrote:
> On Tuesday, 9 August 2022 at 02:30:19 UTC+2, richar...@gmail.com wrote:
> > On 8/8/22 8:27 AM, wij wrote:
> > > If 0.999... is rational, then:
> > > 0.999....= p/q (p,q∈ℕ)
> > > <=> 0.999...*q=p
> > > If 0.999...∈ℕ, there exist q∈ℕ such that 0.999...*q∈ℕ
> > > since 0.999... is defined as infinite repeating, and q is finite,
> > > but 0.999...*q is never finite.
> > > Conclusion: Repeating decimals are irrational.
> > Wrong, 9/9 can generate the pattern 0.9999...
> >
> > This can be proven by 9 = 3+3+3 and the assocative property thus 9/9 =
> > (3+3+3)/9 = 3/9 + 3/9 + 3/9 = 0.3333... + 0.3333... + 0.3333... =
> > 0.9999....
> Ooooh! The associative property! I know how to abuse that too.
>
> S = 1 + 2 + 3 + 4 + 5 ...
> S = 1 + (2+3+4) + (5+6+7) + (8+9+10)...
> S-1 = 9 + 18 + 28 ...
> S-1 = 9(1 + 2 + 3 ...)
> S-1 = 9S
> -8S = 1
> S = -1/8
> 1+2+3+4+5... = -1/8

I think you should provide a solution to really solve the issue.

My idea of infinity is simple. '∞' denotes a unique number like the i (unit of
imaginary number).

Definition of ∞:
1. ∀n∈ℕ, n<∞
2. The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞

The meaning of ∞ in 'thinking' is merely (a process/procedure) 'never end'.

I think I solved the basic 'paradoxes' of infinite series. The basics is that
the addend of an infinite series cannot be rearranged.
Everybody seems to agree this point, but really as did?, or I formalized the idea.
Snippet from https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

+-----------------+
| Infinite Series |
+-----------------+
Series::= S= Σ(n=0,k) a(n)= a(0)+ a(1)+ a(2) +... +a(k)
a(n) is called the general term, addend, summand. n is referred as the
index. Series S is the sum from the first term a(0) to the last term a(k).
The sum of those first terms (n<k) is called the partial sum.
"a(0)+...+a(k)" is called expanded form.

Infinite Series::= If the series S refers to infinite terms (n=∞), S is called
an infinite series. Note that there are infinite addend. The sum cannot be
completed by enumeration (∞ means unreachable, by definition).

In the concept that number-is-an-expression-of-computation, infinite series is
a number, no such concern of converge/diverge (statement when number converges
is a number, diverges is not, is self-controdictory). The computaion rule of
infinite series is based on the expanded form and concepts mentioned above.
Noteworthy difference is that the interpretation of "..." in the expanded form
is a "fixed/unique" number of terms, i.e. "∞+1≠∞" (not the notion of Cantor's
infinite correspondence).

Arithmetic of expanded form:
Ex1: Let S= Σ(n=0,∞) a^n = 1+a+a^2+...+a^∞)
S= 1+a*(1+a+a^2+...+a^∞)- a*a^∞
<=> S= 1+a*S-a^(∞+1)
<=> S(1-a)=1-a^(∞+1)
<=> S= (1-a^(∞+1))/(1-a)

Ex2: Let S= Σ(n=1,∞) n = 1+2+3+...+n
S= 1+2+3+...+n // (1)
S= n+...+3+2+1 // (2)
2S= n*(n+1) // (1)+(2)
<=> S= n*(n+1)/2

∴ Basically, formula for 'finite' series is applicable to infinite series.
(note that mathematical inducion cannot prove such formulas because by
definition, ∞ is unreachable by counting.)

Rule: Handling of the expanded form of infinite series must list the last
addend. Otherwise, the expanded form is ill-formed (obscure semantics and
information being lost cannot conduct valid deduction).

Ex.1 (the last addend is omitted):
A=1+2+3+4+5+...
=(1+2)+(3+4)+5+...
=3+7+5+... // ill-formed, obscure semantics.

Last addend listed:
A=1+2+3+4+5+...+∞ // well-formed, the exanded form of Σ(n=1,∞) {n}

Ex.2:
S=1+2+4+8+... // ill-formed
<=> S=1+2(1+2+4+8+...)
<=> S=1+2S
<=> S=-1

Last addend listed:
S=1+2+4+8+...+2^∞
<=> S=1+2(1+2+4+...+2^(∞-1))
<=> S=1+2S-2^(∞+1)
<=> S=2^(∞+1)-1 // Lots of similar "magic calculation" deriving the result
// S=-1 can be found in youtube. (the term containing the
// last addend ∞ is ignored)

Ex.3:
"f(n)= Σ(k=0,n) 1/k! => f(∞)=e(The base of natural logarithm)"?
We know for sure ∀n∈ℕ, f(n)∈ℚ. To get the result f(n)=e (f(n)∉ℚ), the only
current option is n=∞. But the issue whether or not f(∞)=e (exact equal by
definition) can only be decided via definition, e.g. e≡f(∞). Otherwise, we
can only say f(∞)≈e. (In considering the definition of the equal sign '=',
other forms of e are likely not mutually replaceable with f(∞))

Ex.4: x= Σ(n=1,∞) 1/n²
A common expression is x= Σ(n=1,∞) 1/n²= π²/6, therefore, π=√(6*x)
The issue here is: Lots of π can be derived from various infinite serieses.
But, according to the definition of '=', the result of mutual substitution
may become inconsistent.
For now, the uncontroversial definition of π is the ratio of the
circumference of a circle to its diameter (no computable definition), it is
more correct to use '≈'.
Therefore, Σ(n=1,∞) 1/n² ≈ π²/6 is what it is.

[Note1] "..." in expression is normally indeterminant, of vague semantic.
"0.999..." is also indeterminant before the "..." is eliminated, the
number "0.999..." represents is uncertain, must be removed to ensure
what the number is.
Ex1: Let x=0.999...
10*x= 9+x // This is the result of x after interpreted, not necessarily
// the result followed from "x=0.999..."
// This equation must be given to define x (eliminate the
// ambiguous "...")
Ex2: Let x=√(2+√(2+√(2+...))). Then, possible interpretation of x are:
x=√(2+x)
x=√(2+√(2+x))
x=√(2+√(2+√(2+x)))
...

Ex3: "0.999..." usual 'repeating decimal' cannot denote a unique number.
Let A= Σ(n=1,∞) 1/2^n = 0.999...
B= Σ(n=1,∞) 9/10^n = 0.999...

Let A=B
<=> 1-1/2^∞= 1-1/10^∞ // converted from the formula of geometric series
<=> 1/2^∞= 1/10^∞
<=> 10^∞= 2^∞
<=> 5^∞=1
<=> false

[Note2] Expanded form is prone to magic tricks, perhaps owing to conceptional
generalization of visual illusion too easy to form. It is an error
because the regrouping of the expanded form hides the fact that the
original way of computation is reformulated.
Ex: S can be the sum of any sequence of natural numbers.
S= Σ(n=1,∞) n= 1+2+3+... =1+1+1+1+...= (1+1)+(1+1+1)+...
= Σ(n=1,∞) n+1 // S is modified

Axiom: Σ(n=0,∞) a(n)= a(0)+ Σ(n=1,∞) a(n)
= a(∞)+ Σ(n=0,∞-1) a(n)
Theorem1: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)
Theorem2: Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))
Proof: Omitted (Can be derived from the expanded form)

Ex1: Σ 2*n =Σ (n+n) =Σ n + Σ n
If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural
numbers, the notion that the whole is greater than the part is conflicted
by this rule (many paradoxical and current text book arithmetic have the
same issue using Theorem2 like in Ex3).
But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?
An idea that using C-language's for loop expression might solve this
problem (or, at least, better than the traditional Σ notation):
for(n=0;;++n) n; or f(n=0;;n+=2) n;
Benefit of such a notation is 1.the symbol '∞' can be omitted 2. the
meaning is more concrete, reducing mathematical imagination of 'Σ'.

Temporary Conclusion: The essence of an infinite series may be a number whose
computation never terminates because of infinite number of non-zero
addends), or could be imagined as a 'running' number (density property
requires the existence of such an 'irrational' number).
------------------------- End of Quote

On Thursday, 11 August 2022 at 19:43:10 UTC+8, richar...@gmail.com wrote:
> On 8/11/22 4:50 AM, Skep Dick wrote:
> > On Monday, 8 August 2022 at 15:16:39 UTC+2, Ben Bacarisse wrote:
> >> Yes. For example 0.999... = 1/1.
> >
> > Ok. So it needs saying so I will say it. You are an ass.
> >
> > Few posts ago (on another thread) you were giving me a lecture about functions needing domains/context.
> >
> > Whether 0.999... = 1/1 is a theorem or not absolutely depends on the number system. The domain of your symbols.
> >
> > It is a theorem in R.
> > It isn’t a theorem in *R
> >
> >
> Which is why you need to be clear what domain you are talking about,
> especially if you are not talking about the domain that people will assume.
>
> That just shows that you believe it is ok to be deceptive by not being
> clear.
>
>
> The number, 0.9999... if not otherwise indicated by context, is a Real
> Number in the Real Number system. To later say you are talking about
> some other system *R, is just showing you are being deceptive.
>
> Virtually ANY statement can be proven or disproven if you allow it to be
> moved into an arbitrary logic system crafted for that purpose, so
> context is vital for communication.

Definition man, allow me to call you such (another meaning of definition is
that your understanding stops there, you can't go anything beyond).
Your R is only valid in 'standard + academic + examination' reinforced environments.
'Real number' is actually quite physical and global, in ancient time or present.
If a theory of 'Real number' does not fit 'what is observed', it gets modified.
There may be a standard R, like in physics there is a standard model, it is just
a standardized resolution (AND, still in revision process) among many.

The real number 0.999... is a number roughly
A[0]=0
A[n]=(A[n-1]+1)/2

When n goes infinitely large. And it could be 'visualized' in thought logically.
If possible, it could also be the experimental reality, like the scale marks.