Discussion:
Does the number of nines increase?
(too old to reply)
WM
2024-06-25 20:18:23 UTC
Permalink
Let the infinite sequence 0.999... be multiplied by 10. Does the number of
nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?

Regards, WM
Jim Burns
2024-06-25 22:11:07 UTC
Permalink
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1

tl;dr
No.
Post by WM
Does the number of nines grow
when in 0.999 the decimal point is shifted
by one or more position?
By only one or any finite number of positions,
no.

Nuance:
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.

The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓

for each j ∈ ℕᴬ⤾⁺¹₀ᐣ⤓
⟨0…j⟩ ̊< ⟨0…j⁺¹⟩ ̊≤ ℕ ̊= ℵ₀
⟨j……⟩ ̊≤ ⟨j⁺¹……⟩ ̊≤ ⟨j……⟩ ̊= ℕ ̊= ℵ₀

tl;dr
No.
WM
2024-06-26 07:15:49 UTC
Permalink
Post by Jim Burns
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
Is the set of natural indices complete such that no natural number can be
added?
Post by Jim Burns
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓
And they are fixed. Therefore your answer is correct:No. Therefore
9.999... has one 9 less after the decimal point than 0.999... .

Regards, WM
Jim Burns
2024-06-26 14:20:28 UTC
Permalink
Post by WM
Post by Jim Burns
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
Is the set of natural indices complete
such that
no natural number can be added?
You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'
if you draw a distinction,
if being understood is something you want.


The well.ordered inductive natural.number.indices
are complete
such that
no natural.number.index is not
a natural.number.index.

The well.ordered inductive natural.number.indices
are complete
such that
no natural.number.index is without
its natural.number.index.successor
and
such that
no nonempty natural.number.index.set is without
its natural.number.index.first.member.

The well.ordered inductive natural.number.indices
are complete
such that
no natural.number.index.pair is without
its natural.number.index.sum.
Post by WM
Post by Jim Burns
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓
And they are fixed.
They are certainly well.ordered and inductive.

If, by 'fixed', you mean that
each is not anything other than itself,
then yes.
They are fixedly well.ordered and inductive.
Post by WM
Therefore your answer is correct:No.
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
No.
"Move" the decimal point.
0.9⃒9999... and 09.9⃒9999...
0.99⃒999... and 09.99⃒999...
0.999⃒99... and 09.999⃒99...
0.9999⃒9... and 09.9999⃒9...
0.99999⃒... and 09.99999⃒...
...

Inductive:
for each 9 there is its successor.9
There isn't one 9 fewer.

Therefore,
once again,
'humongous' and 'infinite' are different.
Chris M. Thomasson
2024-06-26 18:49:24 UTC
Permalink
Post by Jim Burns
Post by WM
Post by Jim Burns
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
Is the set of natural indices complete
such that
no natural number can be added?
You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'
if you draw a distinction,
if being understood is something you want.
The well.ordered inductive natural.number.indices
are complete
such that
no natural.number.index is not
a natural.number.index.
The well.ordered inductive natural.number.indices
are complete
such that
no natural.number.index is without
its natural.number.index.successor
and
such that
no nonempty natural.number.index.set is without
its natural.number.index.first.member.
The well.ordered inductive natural.number.indices
are complete
such that
no natural.number.index.pair is without
its natural.number.index.sum.
Post by WM
Post by Jim Burns
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓
And they are fixed.
They are certainly well.ordered and inductive.
If, by 'fixed', you mean that
each is not anything other than itself,
then yes.
They are fixedly well.ordered and inductive.
Post by WM
Therefore your answer is correct:No.
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
No.
"Move" the decimal point.
0.9⃒9999...  and  09.9⃒9999...
0.99⃒999...  and  09.99⃒999...
0.999⃒99...  and  09.999⃒99...
0.9999⃒9...  and  09.9999⃒9...
0.99999⃒...  and  09.99999⃒...
...
for each 9  there is its successor.9
There isn't one 9 fewer.
I still wonder why WM thinks there is one 9 "fewer"... Strange one!
Post by Jim Burns
Therefore,
once again,
'humongous' and 'infinite' are different.
Indeed! :^)
Jim Burns
2024-06-26 21:55:27 UTC
Permalink
Post by Chris M. Thomasson
Post by Jim Burns
for each 9  there is its successor.9
There isn't one 9 fewer.
I still wonder why WM thinks
there is one 9 "fewer"... Strange one!
My theory is that
WM thinks an infiniteᵂᴹ number is
very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ AKA humongous.

My own experience with physics problems is
part of what leads me to think he thinks that.
Many's the time I've scratched out 'negligible' terms.
Doing that is treating 'teensy' like 'infinitesimal' and
'humongous' like 'infinite'.

Sometimes, doing that works well enough.
Especially in physics, Kingdom of the Well.Behaved Function.

Sometimes, infiniteⁿᵒᵗᐧᵂᴹ is different.
Sometimes, Bob disappears.
Sometimes, a ball is sliced into finitely.many pieces and
re.assembled as two balls each as large as the one original.
WM's response (with terminology corrected) is
"But that's not what humongous is like!"
To which the answer comes back,
yes, you're right. It's NOT like humongous.
Post by Chris M. Thomasson
Post by Jim Burns
Therefore,
once again,
'humongous' and 'infinite' are different.
Indeed! :^)
WM
2024-06-27 12:15:30 UTC
Permalink
Post by Jim Burns
WM thinks an infiniteᵂᴹ number is
very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
No, I assume that sets are complete. Therefore ℕ_0 as a proper superset
of ℕ has one elements more than ℕ. Infinity does not make them equal.

Regards, WM
FromTheRafters
2024-06-27 12:33:31 UTC
Permalink
Post by Jim Burns
WM thinks an infiniteᵂᴹ number is
very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
No, I assume that sets are complete. Therefore ℕ_0 as a proper superset of ℕ
has one elements more than ℕ. Infinity does not make them equal.
Yes it does, cardinal arithmetic works differently in the two realms of
finite and infinite.
WM
2024-06-27 19:47:58 UTC
Permalink
Post by FromTheRafters
Therefore ℕ_0 as a proper superset of ℕ
has one elements more than ℕ. Infinity does not make them equal.
Yes it does, cardinal arithmetic works differently in the two realms of
finite and infinite.
Cardinality is nonsense. Set differences are established by their
elements. ℕ_0 differs from ℕ by one element. Do you want to
contradict? Don't make a fool of yourself!

Regards, WM
Richard Damon
2024-06-28 00:03:22 UTC
Permalink
Post by WM
Post by FromTheRafters
Therefore ℕ_0 as a proper superset of ℕ has one elements more than ℕ.
Infinity does not make them equal.
Yes it does, cardinal arithmetic works differently in the two realms
of finite and infinite.
Cardinality is nonsense. Set differences are established by their
elements. ℕ_0 differs from ℕ by one element. Do you want to contradict?
Don't make a fool of yourself!
Regards, WM
As was saidd, Cardinatality of infinte sets work differently than that
of finite sets.

If you can't handle that difference, you can't deal with the truly
infinite sets, only your sort-of looks a little bit like infinite sets
of your FISONs, which are just arbitrarily large, not infinite.
WM
2024-06-28 13:41:57 UTC
Permalink
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than that
of finite sets.
Here is no cardinality asked for.

A non-terminating digit sequence does not determine a real number.

The limit of a strictly monotonic sequence is not among its terms.

The same distinction has to be observed with series. There must not be a
difference in the mathematical contents whether the partial sums are
written separately like

3, 3.1, 3.14, 3.141, 3.1415, ... (11.1)

or are written in one line with interruptions

(((((3.)1)4)1)5)... (11.2)

or without interruptions

3.1415... . (11.3)

Regards, WM
Chris M. Thomasson
2024-06-28 22:23:29 UTC
Permalink
Post by WM
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than that
of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
Huh? Sure it can. What is:

.(428571)

? It is a non-terminating digit sequence in base ten. It's rational, and
real... The numerator and denominator are natural numbers.

;^)

[...]
WM
2024-06-29 17:01:10 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than that
of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
.(428571)
That is a formula determining an infinite digit sequence
Post by Chris M. Thomasson
? It is a non-terminating digit sequence in base ten.
No. Like "0.111..." it is a formula.

Formulas determine sequences. The other way round is not possible.

Regards, WM
Chris M. Thomasson
2024-06-29 19:11:18 UTC
Permalink
Post by WM
Post by WM
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than
that of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
.(428571) >
That is a formula determining an infinite digit sequence
Try 3/7 represented in base 10 decimal notation ;^)
Post by WM
? It is a non-terminating digit sequence in base ten.
No. Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Here is a simple recursive formula that creates infinite 9's during
iteration:

r[0] = .9
r[n] = r[n - 1] + 10^(-n) * .9 = .(9) = 1 (limit)

let's expand:

r[0] = 0
r[1] = .9 + 10^(-1) * .9 = .99
r[2] = .99 + 10^(-2) * .9 = .999
r[3] = .999 + 10^(-3) * .9 = .9999

... on and on ...

See?
Chris M. Thomasson
2024-06-30 01:06:46 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by WM
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than
that of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
.(428571) >
That is a formula determining an infinite digit sequence
Try 3/7 represented in base 10 decimal notation ;^)
Post by WM
? It is a non-terminating digit sequence in base ten.
No. Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Here is a simple recursive formula that creates infinite 9's during
r[0] = .9
r[n] = r[n - 1] + 10^(-n) * .9 = .(9) = 1 (limit)
r[0] = 0
God damn it! r[0] should be equal to .9 in this expansion! Fucking typos!
Post by Chris M. Thomasson
r[1] = .9 + 10^(-1) * .9 = .99
r[2] = .99 + 10^(-2) * .9 = .999
r[3] = .999 + 10^(-3) * .9 = .9999
... on and on ...
See?
WM
2024-06-30 14:38:48 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by WM
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than
that of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
.(428571) >
That is a formula determining an infinite digit sequence
Try 3/7 represented in base 10 decimal notation ;^)
That is a formula too.
Post by Chris M. Thomasson
Post by WM
? It is a non-terminating digit sequence in base ten.
No. Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Here is a simple recursive formula that creates infinite 9's during
Formulas determine sequences.

Regards, WM
joes
2024-06-29 21:18:32 UTC
Permalink
Post by WM
Post by WM
Post by Richard Damon
As was saidd, Cardinatality of infinte sets work differently than
that of finite sets.
Here is no cardinality asked for.
A non-terminating digit sequence does not determine a real number.
Huh? Sure it can. What is: .(428571)?
That is a formula determining an infinite digit sequence
Which number does it describe?
Post by WM
It is a non-terminating digit sequence in base ten.
No. Like "0.111..." it is a formula.
Formulas determine sequences. The other way round is not possible.
Then all real numbers are formulas.
--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.
joes
2024-06-28 08:38:14 UTC
Permalink
Post by WM
WM thinks an infiniteᵂᴹ number is very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
No, I assume that sets are complete. Therefore ℕ_0 as a proper superset
of ℕ has one elements more than ℕ. Infinity does not make them equal.
What does „complete” mean?
With which numbers do you describe the sizes of N and N_0?
WM
2024-06-28 13:55:34 UTC
Permalink
Post by joes
Post by WM
Post by Jim Burns
WM thinks an infiniteᵂᴹ number is
very.very.very.large.but.finiteⁿᵒᵗᐧᵂᴹ
No, I assume that sets are complete. Therefore ℕ_0 as a proper superset
of ℕ has one elements more than ℕ. Infinity does not make them equal.
What does „complete” mean?
It means that no natural number can be added to
{0, 1, 2, 3, ..., ω}
It means that the subtraction of the complete set leaves
{0, 1, 2, 3, ..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.

Regards, WM
joes
2024-06-28 17:23:43 UTC
Permalink
Post by joes
Post by WM
WM thinks an infinite number is
very.large.but.finite
No, I assume that sets are complete. Therefore ℕ_0 as a proper
superset of ℕ has one elements more than ℕ. Infinity does not make
them equal.
Infinity does not have a predecessor like finite numbers.
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
It means that the subtraction of the complete set leaves {0, 1, 2, 3,
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
There is not, since there are infinitely many of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
FromTheRafters
2024-06-28 17:46:41 UTC
Permalink
Post by joes
Post by joes
Post by WM
WM thinks an infinite number is
very.large.but.finite
No, I assume that sets are complete. Therefore ℕ_0 as a proper
superset of ℕ has one elements more than ℕ. Infinity does not make
them equal.
Infinity does not have a predecessor like finite numbers.
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
It means that the subtraction of the complete set leaves {0, 1, 2, 3,
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
There is not, since there are infinitely many of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
There's a number of them, however, how many there are is not a number.
WM
2024-06-28 19:58:17 UTC
Permalink
Post by FromTheRafters
Post by joes
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
There's a number of them, however, how many there are is not a number.
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different numbers of
nines is stupid.

Regards, WM
FromTheRafters
2024-06-28 21:01:49 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by joes
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
There's a number of them, however, how many there are is not a number.
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not changed
by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different numbers of
nines is stupid.
The number of nines is not a number.
Ross Finlayson
2024-06-29 01:05:43 UTC
Permalink
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by joes
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
There's a number of them, however, how many there are is not a number.
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different
numbers of nines is stupid.
The number of nines is not a number.
I thought .999... = 1, ....

You know, or it "goes" to.
Ross Finlayson
2024-06-29 01:57:48 UTC
Permalink
Post by Ross Finlayson
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by joes
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
There's a number of them, however, how many there are is not a number.
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different
numbers of nines is stupid.
The number of nines is not a number.
I thought .999... = 1, ....
You know, or it "goes" to.
When I saw this thread subject the first thing that came to
mind was "casting out 9's", which is according to this fact
that for integers in base b, that b-1 or roots of b-1 can
be provided a divisibility test by recursively summing the
digits and testing when convenient the recursion for divisibility.

Here though it's about the radix, often a name for the base,
eg. "the hexadecimal radix", yet here mostly about the
modular unit, it's the decimal point, the positional
placeholder between > 1 and < 1, the radix point.

The idea of, "10.1", then, vis-a-vis, 1, gets into,
for example, "100.01", and so on, or for writing
"1|1" instead of "." (the radix, the modular radix).
or, "... times 1 plus 1 divided by ...".

Now, this would be inconvenient in a usual sense,
to have sort of an integer part and non-integer part,
connected with addition, instead of the most usual
sort of field, where the placeholder is what moves,
when multiplying or dividing by the radix.

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

The Wiki page doesn't really address this other quite
usual notion of the definition of "radix".

It links to "non-standard positional number system"
which is usually just about calendar and clock
arithmetic and still quite all in the modular,
yet the definition of "decimal point" makes
for that the placeholder at the end of the integer
part and beginning of the non-integer part,
is also called the "radix".

On the Wiki, the entry for "decimal point" points
to "decimal separator".

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

It has a mention of "radix point".

https://en.wikipedia.org/wiki/Decimal_separator#Radix_point


The radix _point_ is part of the character of the radix,
it sort of is the "unit radix".

So, imagine a fixed-width machine fixed-point number,
where the radix point is in the middle. Then the least
(non-zero) value, for example, looks like

0000.0001

https://en.wikipedia.org/wiki/Fixed-point_arithmetic

In fixed point arithmetic, the _increment_ of the
underlying machine number, usually is considered
to represent this "dialing the knob", if you recall
the discussion this year here about models of arithmetic
that fulfill having these finite means, then, in
the unbounded of those.


So anyways, having equal densities of 0's and 1's
can be said to be so for infinite sequences of 0's
and 1's, _among all of them_, and folding them
in half about the radix point _keeps this also so_,
and running out in the unbounded that "restricted
sequence element interchange" runs out, _does run out_.

Thusly, in a greater setting, such natural statements
like "half the integers are even" are quite so.

You're welcome to say "they have the same cardinality
the integers and even integers", and not welcome to
say "the density of the even integers being 1/2 is not so".

Then, as a course-of-passage _goes_ from 0 to 1,
through each point between and in order, like
time-ordering what results modular clock arithmetic,
it's most totally usual in all systems parameterized by "t".
FromTheRafters
2024-06-29 09:36:56 UTC
Permalink
Post by Ross Finlayson
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by joes
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
There's a number of them, however, how many there are is not a number.
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different
numbers of nines is stupid.
The number of nines is not a number.
I thought .999... = 1, ....
You know, or it "goes" to.
Yes, but the number of nines in the sequence after the radix point is
countably infinite rather than finite, hence NaN in this context.
WM
2024-06-29 17:10:52 UTC
Permalink
Post by FromTheRafters
Post by Ross Finlayson
You know, or it "goes" to.
Yes, but the number of nines in the sequence after the radix point is
countably infinite rather than finite, hence NaN in this context.
All nines are from the sequence 0.9, 0.09, 0.009, ... None of the ℵo
nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1. ℵo nines
fail.

Regards, WM
Chris M. Thomasson
2024-06-29 19:29:08 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by Ross Finlayson
You know, or it "goes" to.
Yes, but the number of nines in the sequence after the radix point is
countably infinite rather than finite, hence NaN in this context.
All nines are from the sequence 0.9, 0.09, 0.009, ... None of the ℵo
nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1. ℵo nines
fail.
a = 1/9 = .(1) = (.111...)

b = a * 9 = 1
FromTheRafters
2024-06-30 09:55:32 UTC
Permalink
Post by FromTheRafters
Post by Ross Finlayson
You know, or it "goes" to.
Yes, but the number of nines in the sequence after the radix point is
countably infinite rather than finite, hence NaN in this context.
All nines are from the sequence 0.9, 0.09, 0.009, ... None of the ℵo nines
makes its partial sum 0,9, 0.99, 0.999, ... equal to 1
Yes it does, the sequence of partial sums approaches/converges to one.
This *complete* ordered field of reals guarantees cauchy sequence
convergence. See how real numbers are defined.
Ross Finlayson
2024-06-30 14:48:04 UTC
Permalink
Post by FromTheRafters
Post by WM
Post by FromTheRafters
Post by Ross Finlayson
You know, or it "goes" to.
Yes, but the number of nines in the sequence after the radix point is
countably infinite rather than finite, hence NaN in this context.
All nines are from the sequence 0.9, 0.09, 0.009, ... None of the ℵo
nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1
Yes it does, the sequence of partial sums approaches/converges to one.
This *complete* ordered field of reals guarantees cauchy sequence
convergence. See how real numbers are defined.
It's axiomatic, and about the usual open topology.

There are others, ....
WM
2024-06-29 17:08:02 UTC
Permalink
Post by Ross Finlayson
I thought .999... = 1, ....
That is wrong. All nines are from the sequence 0.9, 0.09, 0.009, ... None
of the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1.

Regards, WM
Jim Burns
2024-06-29 18:01:12 UTC
Permalink
Post by Ross Finlayson
Post by FromTheRafters
The number of nines is not a number.
I thought .999... = 1, ....
1 is near almost.all (all.but.finitely.many) of
0.9 0.99 0.999 0.9999 0.99999 ...
for any sense > 0 of 'near'

For that reason,
we assign 1 to 0.999...

The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
Post by Ross Finlayson
You know, or it "goes" to.
<RF>
Post by Ross Finlayson
This recalls the "First of Zen Koans" bit again,
Two Buddhist priests observe a flag in the wind.
The first says, "the wind, moves, the flag".
The second says, "ah, that flag, moves, in the wind."
A third says "it is your mind that moves".
</RF>
Date: Fri, 28 Jun 2024 21:38:40 -0700

The sequence
0.9 0.99 0.999 0.9999 0.99999 ...
does not move.

It is your mind that moves,
imagining the next, and the next, and the next.

Remember that
0.9 0.99 0.999 0.9999 0.99999 ...
doesn't "go to" even Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
not inside our 13.8×10⁹.year.old universe.

However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.

It's the same for infinitely.many 9s in that
we can't go to them, but
we can reason about them.
But
'infinite' is different from 'humongous' and
different conclusions get concluded.
WM
2024-06-29 18:14:57 UTC
Permalink
Post by Chris M. Thomasson
However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.
We can reason about ℵo nines, all missing the limit.
Post by Chris M. Thomasson
It's the same for infinitely.many 9s in that
we can't go to them, but
we can reason about them.
But most matheologians don't understand that the sequence
0.9, 0.09, 0.009, ... contains ℵo nines without containing the limit 0.
Post by Chris M. Thomasson
But
'infinite' is different from 'humongous' and
different conclusions get concluded.
If you think straight, then only one conclusion follows: 0.999... < 1.
And Bob does not disappear.

Regards, WM
Chris M. Thomasson
2024-06-29 19:30:16 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.
We can reason about ℵo nines, all missing the limit.
Post by Chris M. Thomasson
It's the same for infinitely.many 9s  in that
we can't go to them,  but
we can reason about them.
But most matheologians don't understand that the sequence
0.9, 0.09, 0.009, ... contains ℵo nines without containing the limit 0.
Post by Chris M. Thomasson
But
'infinite' is different from 'humongous' and
different conclusions get concluded.
If you think straight, then only one conclusion follows: 0.999... < 1.
And Bob does not disappear.
(0.999...) = 1 in base ten. Got it?
Jim Burns
2024-06-30 10:28:59 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
However,
we can reason about Avogadroᴬᵛᵒᵍᵃᵈʳᵒ 9s
by finite not.first.false claim.sequence
without going to them.
Each non.empty.set of 9s holds a first.in.set 9
Each 9 has a first.after 9 and a last.before 9,
except the first 9, which only has a first.after 9
Post by WM
We can reason about ℵo nines,
all missing the limit.
1 is near almost.all (all.but.finitely.many) of
0.9 0.99 0.999 0.9999 0.99999 ...
for any sense > 0 of 'near'

For that reason,
we assign 1 to 0.999...

The values of infinite.length decimals
are assigned by a method different from how
the values of finite.length decimals are assigned.
Post by WM
We can reason about ℵo nines,
all missing the limit.
None of
0.9 0.99 0.999 0.9999 0.99999 ...
is near almost.all of
0.9 0.99 0.999 0.9999 0.99999 ...
for any sense > 0 of 'near'
but
the values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999... = 1
0.999... ≠ 0.9
0.999... ≠ 0.99
0.999... ≠ 0.999
0.999... ≠ 0.9999
0.999... ≠ 0.99999
...
Post by WM
Post by Chris M. Thomasson
It's the same for infinitely.many 9s  in that
we can't go to them,  but
we can reason about them.
But most matheologians don't understand that
the sequence 0.9, 0.09, 0.009, ... contains
ℵo nines without containing the limit 0.
1 is near almost.all (all.but.finitely.many) of
0.9 0.99 0.999 0.9999 0.99999 ...
for any sense > 0 of 'near'
0.999... = 1
0.999... ≠ 0.9
0.999... ≠ 0.99
0.999... ≠ 0.999
0.999... ≠ 0.9999
0.999... ≠ 0.99999
...
Post by WM
Post by Chris M. Thomasson
But
'infinite' is different from 'humongous' and
different conclusions get concluded.
If you think straight,
0.999... < 1.
The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999... ≠ 0.9 < 1
0.999... ≠ 0.99 < 1
0.999... ≠ 0.999 < 1
0.999... ≠ 0.9999 < 1
0.999... ≠ 0.99999 < 1
...

0.999... = 1

----
Post by WM
And Bob does not disappear.
The cardinal ℵ₀ of the set ℕ of
all cardinals.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
is a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)

/ Assume otherwise
| Assume ℵ₀⁺¹ > ℵ₀
| There are at.least.ℵ₀⁺¹ cardinals.growable.by.1
| However, ℵ₀ is how.many cardinals.growable.by.1
\ Contradiction.

ℵ₀ = |ℕ| is not.growable.by.1
|ℕ| = |ℕ∪{Bob]|
exists f: ℕ∪{Bob} → ℕ : bijection
Bob ∉ f(ℕ∪{Bob})
And Bob disappears.
WM
2024-06-30 14:51:42 UTC
Permalink
Post by Jim Burns
Post by WM
If you think straight,
0.999... < 1.
The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
0.999... ≠ 0.9 < 1
0.999... ≠ 0.99 < 1
0.999... ≠ 0.999 < 1
0.999... ≠ 0.9999 < 1
0.999... ≠ 0.99999 < 1
...
If you use only definable length, then always ℵo terms are missing.

All finite indices guarantee finite length.
0.999... = 0.999... < 1
Post by Jim Burns
0.999... = 1
That is wrong.
If you insert parentheses, nothing changes

..((((0,9)9)9)9)... = 0,999...

0.9 + 0.09 + 0.009 + ... contains ℵo terms with ℵo nines, all together
smaller than 1.

Regards, WM

WM
2024-06-29 16:58:36 UTC
Permalink
Post by FromTheRafters
Post by WM
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not changed
by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different numbers of
nines is stupid.
The number of nines is not a number.
The set of natural indices is a set.
Shifting it to the left shifts every index and every indexed nine to the
left.

Regards, WM
joes
2024-06-29 20:29:11 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Which it hasn’t.
Post by WM
Post by FromTheRafters
Post by WM
Or this is not the case.
But assuming completeness and simultaneously claiming different
numbers of nines is stupid.
They are the same.
Post by WM
Post by FromTheRafters
The number of nines is not a number.
The set of natural indices is a set.
Shifting it to the left shifts every index and every indexed nine to the
left.
So what? Every index is increased by 1, blah blah PA.
--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.
FromTheRafters
2024-06-30 10:02:48 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different numbers of
nines is stupid.
The number of nines is not a number.
The set of natural indices is a set.
Shifting it to the left shifts every index and every indexed nine to the
left.
120 volts is a good amount of voltage for US household use.

That still doesn't affect the fact that aleph_null is not a real
number.
WM
2024-06-28 18:25:57 UTC
Permalink
Post by joes
Post by joes
Post by WM
WM thinks an infinite number is
very.large.but.finite
No, I assume that sets are complete. Therefore ℕ_0 as a proper
superset of ℕ has one elements more than ℕ. Infinity does not make
them equal.
Infinity does not have a predecessor like finite numbers.
What is immediately before ω?
Post by joes
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
What is immediately before ω?
Post by joes
It means that the subtraction of the complete set leaves {0, 1, 2, 3,
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
There is not, since there are infinitely many of them.
Linearity excludes more than one at a position. Immediately before ω
there is at most one natural number.
Post by joes
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many
of them there are.
|N| + 1 = |N_0|

Regards, WM
joes
2024-06-29 20:46:53 UTC
Permalink
Post by WM
Post by joes
Post by joes
Post by WM
WM thinks an infinite number is very.large.but.finite
No, I assume that sets are complete. Therefore ℕ_0 as a proper
superset of ℕ has one elements more than ℕ. Infinity does not make
them equal.
It does.
Post by WM
What is immediately before ω?
Post by joes
Infinity does not have a predecessor like finite numbers.
Post by joes
What does „complete” mean?
It means that no natural number can be added to {0, 1, 2, 3, ..., ω}
Duh, the set of all natural numbers N contains all of them.
This is misleadingly notated, implying a predecessor. You mean ω u ℕ,
or {ℕ}∪ℕ = {{0, 1, 2, …}, 0, 1, 2, …}. [Neo layout FTW]
Post by WM
Post by joes
It means that the subtraction of the complete set leaves {0, 1, 2, 3,
..., ω} \ ℕ = {0, ω}.
It means that in {0, 1, 2, 3, ..., ω} before ω there is a natural
number.
There is not, since there are infinitely many of them.
Linearity excludes more than one at a position. Immediately before ω
there is at most one natural number.
You are confusing the infinitely long, linearly scaled natural number
line with the ordinal number line, which can be drawn with 0, ω, ω*2(?),
ω^2, …(?) at fixed/constant intervals. They do not live in the same
world. You want to shoehorn ω and the rest somewhere onto the infinite
N, continuing in steps of 1. This already fails in finding the non-
existent far end. ω is not connected, as you say, to 0 by a finite
number of steps. It is an augmentation to the whole set, giving a new
anchor. Likewise ω*2 = ω+ω is obviously infinitely far away from ω.
Post by WM
Post by joes
Post by joes
With which numbers do you describe the sizes of N and N_0?
Most of them are dark and cannot be used as individuals.
Not their elements. I was asking for their number, how many of them
there are.
|N| + 1 = |N_0|
And that is…? You gave only a relation.
--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.
WM
2024-06-27 12:03:14 UTC
Permalink
Post by Jim Burns
Post by WM
Is the set of natural indices complete
such that
no natural number can be added?
You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'
Why? There is nothing different. The set of numbers and of indices is ℕ.
Post by Jim Burns
The well.ordered inductive natural.number.indices
are complete
Therefore the indices of the nines of 0.999... are the complete set ℕ.
When they are shifted to 9.999..., none is added. One is missing at the
right of the decimal point.
Post by Jim Burns
Post by WM
Post by Jim Burns
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
If, by 'fixed', you mean that
each is not anything other than itself,
then yes.
Yes.
Post by Jim Burns
Post by WM
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
No.
You believe in the magic of matheology. Try to think. "Countable" is an
unsharp notion, too unsharp to measure the fact that {1, 2, 3, ...} has
one number less than {0, 1, 2, 3, ...} although this is obvious: ℕ is a
proper subsets of ℕ_0.
Post by Jim Burns
"Move" the decimal point.
0.9⃒9999... and 09.9⃒9999...
0.99⃒999... and 09.99⃒999...
0.999⃒99... and 09.999⃒99...
0.9999⃒9... and 09.9999⃒9...
0.99999⃒... and 09.99999⃒...
...
Not readable what you try here.
Post by Jim Burns
There isn't one 9 fewer.
As long as you are claiming that elements can be lost by exchanging, your
arguments are not relevant in any respect.

Regards, WM
Jim Burns
2024-06-27 16:30:29 UTC
Permalink
Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
The well.ordered inductive natural.number.indices
are complete
Therefore the indices of the nines of 0.999...
are the complete set ℕ.
Each subset of ℕ holds a first, excepting {}, completely.
Each number in ℕ has a successor and a predecessor,
excepting 0, which has only a successor, completely.

For each number in ℕ
there is a decimal place in 0.999... and a 9 is in it,
completely.
Post by WM
When they are shifted to 9.999..., none is added.
One is missing at the right of the decimal point.
0 is in {0,1,2,...} and not.in {1,2,3,...}
Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
No.
⎛ for each j in {0,1,2,...}: j⁺¹ is in {1,2,3,...}
⎜ j ↦ j⁺¹ is 1.to.1
⎜ {0,1,2,...} ̊≤ {1,2,3,...}

⎜ {0,1,2,...} ⊇ {1,2,3,...}
⎜ {0,1,2,...} ̊≥ {1,2,3,...}

⎝ {0,1,2,...} ̊= {1,2,3,...}

Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Post by WM
Post by Jim Burns
Post by WM
Therefore
9.999... has one 9 less [one 9 fewer]
after the decimal point than 0.999... .
No.
You believe in the magic of matheology.
I believe j ↦ j⁺¹ is 1.to.1
Post by WM
Try to think.
"Countable" is an unsharp notion,
Consider instead the notion '1.to.1 function'
Post by WM
"Countable" is an unsharp notion,
too unsharp to measure the fact that
{1, 2, 3, ...} has one number less than
{0, 1, 2, 3, ...}
ℕ is a proper subsets of ℕ_0.
Consider the set ℕ of
all cardinalities which can grow by 1

One more than a cardinality which can grow by 1
is also a cardinality which can grow by 1
and is also in ℕ

If ℕ has a cardinality which can grow by 1
then there are more than ℕ.many cardinalities in ℕ
and ℕ is larger than ℕ

Because ℕ is not larger than ℕ
the cardinality of ℕ cannot grow by 1
but
there are sets which are
ℕ with 1 element inserted or deleted.
Those sets have the same cardinality as ℕ
The cardinality didn't grow by 1
Post by WM
Post by Jim Burns
Post by WM
Is the set of natural indices complete
such that
no natural number can be added?
You'd do better at being understood if
you said what distinguishes 'natural number'
from 'natural index'
if you draw a distinction,
if being understood is something you want.
Why? There is nothing different.
The set of numbers and of indices is ℕ.
https://en.wiktionary.org/wiki/if

You're welcome.
WM
2024-06-27 18:37:33 UTC
Permalink
Post by Jim Burns
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
The set ℕ of indices of the nines in 0.999... cannot grow by a natural
number.
Post by Jim Burns
Post by WM
Try to think.
"Countable" is an unsharp notion,
Consider instead the notion '1.to.1 function'
That is as unsharp. It shows equinumerosity of primes and rationals.
Post by Jim Burns
Because ℕ is not larger than ℕ
the cardinality of ℕ cannot grow by 1
but
there are sets which are
ℕ with 1 element inserted or deleted.
Those sets have the same cardinality as ℕ
The cardinality didn't grow by 1
As I said cardinality is useless to check equinumerosity.
Post by Jim Burns
Post by WM
The set of numbers and of indices is ℕ.
Therefore in 0,999... there are precisely as many nines as in 9.999....
Therefore behind the decimal point there are less after shifting it.

Regards, WM
Chris M. Thomasson
2024-06-27 19:57:53 UTC
Permalink
Post by WM
Post by Jim Burns
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
The set ℕ of indices of the nines in 0.999... cannot grow by a natural
number.
Post by Jim Burns
Post by WM
Try to think.
"Countable" is an unsharp notion,
Consider instead the notion '1.to.1 function'
That is as unsharp. It shows equinumerosity of primes and rationals.
Post by Jim Burns
Because ℕ is not larger than ℕ
the cardinality of ℕ cannot grow by 1
but
there are sets which are
ℕ with 1 element inserted or deleted.
Those sets have the same cardinality as ℕ
The cardinality didn't grow by 1
As I said cardinality is useless to check equinumerosity.
Post by Jim Burns
Post by WM
The set of numbers and of indices is ℕ.
Therefore in 0,999... there are precisely as many nines as in 9.999....
Therefore behind the decimal point there are less after shifting it.
Oh boy, you are a special one... ;^o
Jim Burns
2024-06-28 04:31:42 UTC
Permalink
Post by WM
Post by Jim Burns
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Sets can have elements inserted,
which makes them different sets.
The effect on size of inserting 1
is not the same for all sets.

Larger than all the
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
are the
infiniteⁿᵒᵗᐧᵂᴹ.sizes.which.cannot.grow.by.1

For each
finiteⁿᵒᵗᐧᵂᴹ.size.which.can.grow.by.1
in the set ℕ of all
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
there is a larger
finiteⁿᵒᵗᐧᵂᴹ.size.which.can.grow.by.1

It would lead to contradictions for
for the size of the set ℕ of all
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
to be smaller than one of the
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
_in itself_

So,
the size of the set ℕ of all
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
isn't any of the
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1

Only
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
can grow by 1, which
the size of the set ℕ of all
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
isn't any of.

The size of the set ℕ of all
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
cannot grow by 1
It is an
infiniteⁿᵒᵗᐧᵂᴹ.size.which.cannot.grow.by.1

The effect on size of inserting 1
is not the same for all sets.
WM
2024-06-28 13:50:06 UTC
Permalink
Post by Jim Burns
Post by WM
Post by Jim Burns
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Sets can have elements inserted,
which makes them different sets.
The effect on size of inserting 1
is not the same for all sets.
It is changing the infinite set but not its cardinality. Therefore
cardinality is useless for my proof.
Post by Jim Burns
The size of the set ℕ of all
finiteⁿᵒᵗᐧᵂᴹ.sizes.which.can.grow.by.1
cannot grow by 1
It is an
infiniteⁿᵒᵗᐧᵂᴹ.size.which.cannot.grow.by.1
Your discussion is off topic.
Post by Jim Burns
The effect on size of inserting 1
is not the same for all sets.
The effect of removing a nine from 0.999... is changing its value.
When in 0.999... the decimal point is shifted, the number of nines remains
constant. That has nothing to do with cardinalities.

Regards, WM
joes
2024-06-28 17:16:56 UTC
Permalink
Post by WM
Cardinalities which can grow by 1 are finite. Cardinalities which
cannot grow by 1 are infinite.
Cardinalities are useless. Sets can grow by 1 element.
So what is the size of N\{2}?
Post by WM
Sets can have elements inserted, which makes them different sets.
The effect on size of inserting 1 is not the same for all sets.
It is changing the infinite set but not its cardinality. Therefore
cardinality is useless for my proof.
What are you replacing it with?
Post by WM
The size of the set ℕ of all finite.sizes.which.can.grow.by.1
cannot grow by 1 It is an infinite.size.which.cannot.grow.by.1
Your discussion is off topic.
The size of N, containing all finite numbers, is itself infinite.
Post by WM
The effect on size of inserting 1 is not the same for all sets.
The effect of removing a nine from 0.999... is changing its value.
When in 0.999... the decimal point is shifted, the number of nines
remains constant. That has nothing to do with cardinalities.
You can’t remove from the right, since there is no end. You can only
remove from the left (dividing by 10), or cut it off, replacing all
following places with zeroes. If you removed one from the middle (i.e.
not the first one), there are still infinitely many.
WM
2024-06-28 18:19:11 UTC
Permalink
Post by joes
Post by WM
Cardinalities which can grow by 1 are finite. Cardinalities which
cannot grow by 1 are infinite.
Cardinalities are useless. Sets can grow by 1 element.
So what is the size of N\{2}?
It is |ℕ| - 1. That was easy.
Post by joes
Post by WM
Sets can have elements inserted, which makes them different sets.
The effect on size of inserting 1 is not the same for all sets.
It is changing the infinite set but not its cardinality. Therefore
cardinality is useless for my proof.
What are you replacing it with?
Number of elements.
Post by joes
Post by WM
The size of the set ℕ of all finite.sizes.which.can.grow.by.1
cannot grow by 1 It is an infinite.size.which.cannot.grow.by.1
Your discussion is off topic.
The size of N, containing all finite numbers, is itself infinite.
Yes, it is |ℕ|.
Post by joes
Post by WM
The effect on size of inserting 1 is not the same for all sets.
The effect of removing a nine from 0.999... is changing its value.
When in 0.999... the decimal point is shifted, the number of nines
remains constant. That has nothing to do with cardinalities.
You can’t remove from the right, since there is no end.
If all are a complete set, then we can shift all by one position to the
left.
Post by joes
You can only
remove from the left (dividing by 10)
Removing from the right is done by multiplying by 10.

Regards, WM
Chris M. Thomasson
2024-06-29 00:23:30 UTC
Permalink
On 6/28/2024 11:19 AM, WM wrote:
[...]
Post by WM
Removing from the right is done by multiplying by 10.
Huh?

.(9) * 10 = 9.(9) = 10

There is no removing a nine from the pool of infinite nines where it
somehow makes the pool contain less than infinite supply of nines is
already has.

WTF is wrong with you!

You are a fraud?
joes
2024-06-29 20:57:34 UTC
Permalink
Post by WM
Post by joes
Post by WM
Cardinalities which can grow by 1 are finite. Cardinalities which
cannot grow by 1 are infinite.
Cardinalities are useless. Sets can grow by 1 element.
So what is the size of N\{2}?
It is |ℕ| - 1. That was easy.
Why is that the same as ℕ\{1}? They are not supersets of each other.
Post by WM
Post by joes
Post by WM
Sets can have elements inserted, which makes them different sets.
The effect on size of inserting 1 is not the same for all sets.
It is changing the infinite set but not its cardinality. Therefore
cardinality is useless for my proof.
What are you replacing it with?
Number of elements.
Which you define how? Starting maybe with |ℕ|. What numbers do you use?
Post by WM
Post by joes
Post by WM
The size of the set ℕ of all finite sizes which can grow by 1 cannot
grow by 1. It is an infinite size which cannot grow by 1.
The size of N, containing all finite numbers, is itself infinite.
Yes, it is |ℕ|.
Which is as a concrete number…? [not OT?]
Post by WM
Post by joes
Post by WM
The effect of removing a nine from 0.999... is changing its value.
Only when replacing with another digit.
Post by WM
Post by joes
Post by WM
When in 0.999... the decimal point is shifted, the number of nines
remains constant. That has nothing to do with cardinalities.
Yes it does; that is how WE define the number of nines.
Post by WM
Post by joes
You can’t remove from the right, since there is no end.
If all are a complete set, then we can shift all by one position to the
left.
Or equivalently, add 9.
It is as a rope coming from the machine: there’s always more.
Post by WM
Post by joes
You can only remove from the left (dividing by 10)
Removing from the right is done by multiplying by 10.
Where do the superfluous digits go: 0.(9)9 ?
--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.
FromTheRafters
2024-06-30 10:13:47 UTC
Permalink
Post by joes
Post by WM
Post by joes
Cardinalities which can grow by 1 are finite. Cardinalities which
cannot grow by 1 are infinite.
Cardinalities are useless. Sets can grow by 1 element.
So what is the size of N\{2}?
It is |ℕ| - 1. That was easy.
Why is that the same as ℕ\{1}? They are not supersets of each other.
Two sets with the same cardinality are not necessarily the same. Two
sets can be unequal to each other but still be equal in cardinality.

Each set above is created by filtering one element out of consideration
in forming the new set.
Jim Burns
2024-06-29 13:24:24 UTC
Permalink
Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Sets can have elements inserted,
which makes them different sets.
The effect on size of inserting 1
is not the same for all sets.
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.

Not all cardinals are growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
A set with a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
can change without its cardinality changing.

⎛ ...not that setsⁿᵒᵗᐧᵂᴹ changeⁿᵒᵗᐧᵂᴹ
⎜ Adjunct X∪{Y} isn't X
⎜( Sets can grow by 1 element.
⎜ means
⎝( For each X,Y X∪{Y} exists

The cardinal ℵ₀ for
all cardinals.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
is a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)

| Assume otherwise.
| Assume ℵ₀ is growable.by.1
|
| If
| the cardinal ℵ₀ of
| all cardinals.growable.by.1
| is growable.by.1
| then
| the cardinal ℵ₀ᐡᴮᵒᵇ for
| all cardinals.growable.by.1
| and Bob too
| is growable.by.1 too
| and
| there are more.than.ℵ₀.many
| cardinals.growable.by.1
|
| However,
| there are ℵ₀.many
| cardinalities growable.by.1
| Contradiction.

Therefore,
the cardinal ℵ₀ for
all cardinals.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
is a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)

Not all cardinals are growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
A set with a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
can change without its cardinality changing.
Post by WM
Therefore
cardinality is useless for my proof.
if we stop _saying_
( cardinal of {k……} ̊≤ cardinal of {k⁺¹……}
k ↦ k⁺¹ doesn't stop _being_
1.to.1 from {k……} to {k⁺¹……}

For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
Ross Finlayson
2024-06-29 14:33:46 UTC
Permalink
Post by Jim Burns
Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
Cardinalities which can grow by 1 are finite.
Cardinalities which cannot grow by 1 are infinite.
Cardinalities are useless.
Sets can grow by 1 element.
Sets can have elements inserted,
which makes them different sets.
The effect on size of inserting 1
is not the same for all sets.
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.
Not all cardinals are growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
A set with a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
can change without its cardinality changing.
⎛ ...not that setsⁿᵒᵗᐧᵂᴹ changeⁿᵒᵗᐧᵂᴹ
⎜ Adjunct X∪{Y} isn't X
⎜( Sets can grow by 1 element.
⎜ means
⎝( For each X,Y X∪{Y} exists
The cardinal ℵ₀ for
all cardinals.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
is a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
| Assume otherwise.
| Assume ℵ₀ is growable.by.1
|
| If
| the cardinal ℵ₀ of
| all cardinals.growable.by.1
| is growable.by.1
| then
| the cardinal ℵ₀ᐡᴮᵒᵇ for
| all cardinals.growable.by.1
| and Bob too
| is growable.by.1 too
| and
| there are more.than.ℵ₀.many
| cardinals.growable.by.1
|
| However,
| there are ℵ₀.many
| cardinalities growable.by.1
| Contradiction.
Therefore,
the cardinal ℵ₀ for
all cardinals.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
is a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
Not all cardinals are growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
A set with a cardinal.not.growable.by.1 (infiniteⁿᵒᵗᐧᵂᴹ)
can change without its cardinality changing.
Post by WM
Therefore cardinality is useless for my proof.
if we stop _saying_
( cardinal of {k……} ̊≤ cardinal of {k⁺¹……}
k ↦ k⁺¹ doesn't stop _being_
1.to.1 from {k……} to {k⁺¹……}
For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
Yet, 'density', is its own thing, ....

Then you either got two 'not.resolves' or none,
was the idea, ....

(This was from the other day when I remarked on
the essential lack of discernability, or discernibility,
of two propositions which confound each other,
thus requiring a deconstructive account, where
the joke went 'pick one' and the joke was
'ha, I put them together, it's both or none'.)

Somehow I went all through school and calculus and
econometrics and statistics and with real analysis,
and transfinite cardinals were never introduced, ....

Then later I was interested in what I learned was
foundations most particularly about the continuous
and discrete, and I learned about the Galilean notion
of that countably infinite sets are equivalent, while
then there's duBois-Reymond with the diagonal argument,
that nested intervals is the usual old infinite-divisibility,
while there's then Cantor and the m-w proof and in the
language of set theory one of the consequences of
comprehension or the powerset result, then as with
regards to that making an inequality in cardinals,
that the real numbers were a rather usual construction
axiomatically with axiomatizing least-upper-bound
and axiomatizing measure 1.0, yet, I also learned
about the 'density' asymptotically or Schnirelmann,
about Katz' OUTPACING and similar notions, then as
well that the notion of continuous domain was largely
un-defined yet as with regards to Jordan measure,
Lebesgue measure, and Shannon/Nyquist supersampling,
that there's at least three.

The "extra-ordinary" then or Mirimanoff, who is to
be thanked for the introduction of both the "ordinary"
and "extra-ordinary" in what resulted ordinary regular
set theory, Zermelo-Fraenkel set theory most usually
in a model theory, though with of course "naive set-builder"
having been ubiquitous since forever, really gets into
the logical foundations, then for why dually-self-infraconsistency
is a thing and that the more one studies foundations
classically and modernly, that it all results sort of one thing,
a platonist's ideal strong mathematical platonism, of
a universe of mathematical objects like a "Hilbert's
Infinite Living, Working Museum of Mathematics, Now Open".
WM
2024-06-29 17:18:26 UTC
Permalink
Post by Jim Burns
Post by WM
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.
Cardinality is irrelevant.
Post by Jim Burns
Post by WM
Therefore
cardinality is useless for my proof.
For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
Cardinality does not present a difficulty. It is simply unable to
distinguish |ℕ| and |ℕ_0|.

All nines of 0.999... are from the sequence 0.9, 0.09, 0.009, ... None of
the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1.
ℵo nines fail to make 0.999... = 1.

Regards, WM

Regards, WM
Chris M. Thomasson
2024-06-29 19:24:39 UTC
Permalink
Post by WM
Post by Jim Burns
Post by WM
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.
Cardinality is irrelevant.
Post by Jim Burns
Post by WM
Therefore cardinality is useless for my proof.
For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
Cardinality does not present a difficulty. It is simply unable to
distinguish |ℕ| and |ℕ_0|.
All nines of 0.999... are from the sequence 0.9, 0.09, 0.009, ... None
The sum of the sequence:

.9 + .09 + .009 + ... = .(9) = 1 in base ten

There are infinite 9's...
Post by WM
of the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1.
ℵo nines fail to make 0.999... = 1.
Regards, WM
Regards, WM
joes
2024-06-29 20:25:10 UTC
Permalink
Post by WM
Post by WM
It is changing the infinite set but not its cardinality.
A set with a cardinal.growable.by.1 (finite)
cannot change without its cardinality changing.
Cardinality is irrelevant.
Then how do you measure infinite sets?
Post by WM
Post by WM
Therefore cardinality is useless for my proof.
For any difficulty which cardinality presents,
not saying 'cardinal' not.resolves the difficulty.
Cardinality does not present a difficulty. It is simply unable to
distinguish |ℕ| and |ℕ_0|.
Working as designed. How do you distinguish them?
Post by WM
All nines of 0.999... are from the sequence 0.9, 0.09, 0.009, ... None
of the ℵo nines makes its partial sum 0,9, 0.99, 0.999, ... equal to 1.
ℵo nines fail to make 0.999... = 1.
That’s one way to construct it. Of course none of the infinite series
individually reaches the limit (otherwise the following ones :-P would go
above it?). Together they do. It is reached after exactly Aleph0 steps.
--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.
Jim Burns
2024-06-29 22:48:23 UTC
Permalink
Post by WM
Post by Jim Burns
Post by WM
It is changing the infinite set
but not its cardinality.
A set with a cardinal.growable.by.1 (finiteⁿᵒᵗᐧᵂᴹ)
cannot change without its cardinality changing.
Cardinality is irrelevant.
Post by Jim Burns
Post by WM
Therefore cardinality is useless for my proof.
For any difficulty which cardinality presents,
not.saying 'cardinal' not.resolves the difficulty.
Cardinality does not present a difficulty.
It is simply unable to distinguish |ℕ| and |ℕ_0|.
A not.at.all.drug.impaired use of the word 'irrelevant'
declaring that cardinality is irrelevantᵂᴹ to
cardinals |ℕ₁| and |ℕ₀|

Cardinality does not distinguish cardinals |ℕ₁| and |ℕ₀|
because
j ↦ j⁺¹ is 1.to.1 from ℕ₀ to ℕ₁
and
k ↤ k is 1.to.1 to ℕ₀ from ℕ₁

Without 'cardinality'
you (WM) still have your (WM's) difficulty,
that
j ↦ j⁺¹ is 1.to.1 from ℕ₀ to ℕ₁
and
k ↤ k is 1.to.1 to ℕ₀ from ℕ₁
Post by WM
All nines of 0.999... are from
the sequence 0.9, 0.09, 0.009, ...
None of the ℵo nines makes
its partial sum 0,9, 0.99, 0.999, ... equal to 1.
Each non.empty.set of 9s holds a first.in.set 9
Each 9 has a first.after 9 and a last.before 9,
except the first 9, which only has a first.after 9

None of those is the last 9
None of those equal 1
Post by WM
ℵo nines fail to make 0.999... = 1.
1 is near almost.all (all.but.finitely.many) of
0.9 0.99 0.999 0.9999 0.99999 ...
for any sense > 0 of 'near'

For that reason,
we assign 1 to 0.999...

The values of infinite.length decimals
are assigned by a different method from how
the values of finite.length decimals are assigned.
Chris M. Thomasson
2024-06-26 18:46:27 UTC
Permalink
Post by WM
Post by Jim Burns
Post by WM
Let the infinite sequence 0.999...
be multiplied by 10.
Does the number of nines grow?
Cardinalities which can grow by 1 are finite.
The number of nines in 0.999... is
larger than each finite cardinality.
It does not equal any finite cardinality.
It cannot grow by 1
tl;dr
No.
Is the set of natural indices complete such that no natural number can
be added?
Post by Jim Burns
There are _only_ positions in 0.999... which
are separated by some finite number,
even though there are infinitely.many of them.
The positions in 0.999... correspond to
numbers in well.ordered inductive ℕᴬ⤾⁺¹₀ᐣ⤓
And they are fixed. Therefore your answer is correct:No. Therefore
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10

There are still infinite nines... :^)

However, just keep in mind that .(9) represents 1 in base 10 therefore:

9.(9) is 10

So, .(9) * 10 = 1 * 10 = 10

and: .(9) * 9.(9) = 1 * 10 = 10

(.999...) * (9.999...) = 10

Got it?

:^)
WM
2024-06-27 12:09:17 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
Wrong.

10*0.999...999 = 9.999...990

==> 10*0.999...999 < 9.999...999

==> 9*0.999...999 < 9

==> 0.999... < 1
Post by Chris M. Thomasson
There are still infinite nines... :^)
If the set ℕ is actually complete, then the set ℕ_0 has one element
more.

Regards, WM
FromTheRafters
2024-06-27 12:38:37 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
Post by WM
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
Wrong.
10*0.999...999 = 9.999...990
==> 10*0.999...999 < 9.999...999
==> 9*0.999...999 < 9
==> 0.999... < 1
Post by Chris M. Thomasson
There are still infinite nines... :^)
If the set ℕ is actually complete, then the set ℕ_0 has one element more.
If by 'complete' you mean something entirely different from what it
means to most of the rest of the posters here. The set of naturals is
'complete' in your sense because it is defined such that it contains
*all* of its elements. All sets are 'complete' in that sense.
WM
2024-06-27 19:52:54 UTC
Permalink
Post by FromTheRafters
Post by WM
If the set ℕ is actually complete, then the set ℕ_0 has one element more.
If by 'complete' you mean something entirely different from what it
means to most of the rest of the posters here.
No, it is a fact that can be proved by 0 and cannot be disproved.
Post by FromTheRafters
The set of naturals is
'complete' in your sense because it is defined such that it contains
*all* of its elements. All sets are 'complete' in that sense.
But you believe that the set of indices in 0.999... is not complete?

Regards, WM
Chris M. Thomasson
2024-06-27 20:20:55 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by WM
If the set ℕ is actually complete, then the set ℕ_0 has one element more.
If by 'complete' you mean something entirely different from what it
means to most of the rest of the posters here.
No, it is a fact that can be proved by 0 and cannot be disproved.
Post by FromTheRafters
The set of naturals is 'complete' in your sense because it is defined
such that it contains *all* of its elements. All sets are 'complete'
in that sense.
But you believe that the set of indices in 0.999... is not complete?
.999... = .(9) = 1

Got it?
Post by WM
Regards, WM
Chris M. Thomasson
2024-06-27 19:56:14 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
Post by WM
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
Wrong.
10*0.999...999 = 9.999...990
Huh? Where is that last zero coming from? Your very strange darkness?
Anyway:

Are you suggesting a prefix and a postfix, a header and a footer setup?

[prolog][infinity][epilog]

Fwiw, this can be modeled in a difference between two n-ary points.

take two 4d points:

p0 = (1, 2, 3, 4)
p1 = (-1, -.42, 2, .001)
dif = p1 - p0
mid = p0 + dif / 2

Anyway, back to a way base 10 can represent infinitely repeating decimals:

0.999... = 1
0.(9) = 1
9.(9) = 10

10 * (.999...) = 10 * 1 = 10
9.(9) * .(9) = 10 * 1 = 10

Take .(1122) where 101/900 is good enough for that precision, 4 digits.
Now take:

.(4269123) = 474347/1111111

You must be familiar with:

355/113
Post by WM
==> 10*0.999...999 < 9.999...999
==> 9*0.999...999 < 9
==> 0.999... < 1
Post by Chris M. Thomasson
There are still infinite nines... :^)
If the set ℕ is actually complete, then the set ℕ_0 has one element more.
You cannot add another natural number to N because its already there. N
is all of them. So, N + 1 = N
WM
2024-06-27 20:06:47 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
Wrong.
10*0.999...999 = 9.999...990
Where is that last zero coming from?
In 0.999... there is a constant set of indices indexing all nines. If a
point of fly droppings becomes visible at another place this does not
change the set of indices of nines. Hence also an arbitrary shift of the
decimal point by multiplication does not change the number of nines.

Regards, WM
Chris M. Thomasson
2024-06-27 20:20:11 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
Post by WM
Post by Chris M. Thomasson
Post by WM
9.999... has one 9 less after the decimal point than 0.999... .
NO! Think of: .(9) * 10 = 9.(9) = 10
Wrong.
10*0.999...999 = 9.999...990
Where is that last zero coming from?
In 0.999... there is a constant set of indices indexing all nines. If a
point of fly droppings becomes visible at another place this does not
change the set of indices of nines. Hence also an arbitrary shift of the
decimal point by multiplication does not change the number of nines.
.(9) = 1

There is always an infinite number of nines in Say:

9.(9) = 10

This is the way base 10 can work.

You cannot magically add one!
WM
2024-06-28 13:59:02 UTC
Permalink
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been produced
by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been produced
by adding 9 to 0.999...?

Regards, WM
joes
2024-06-28 17:19:26 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”. They are the same number 10.
WM
2024-06-28 18:21:45 UTC
Permalink
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.

Regards, WM
Chris M. Thomasson
2024-06-28 22:25:18 UTC
Permalink
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?

;^o
Ross Finlayson
2024-06-29 01:04:29 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten? You know, add a zero?
Chris M. Thomasson
2024-06-29 02:05:59 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten?  You know, add a zero?
(.999...) * (9.999...) = (9.999...)

(9.999...) * 10 = (99.999...)

(99.999...) * 10 = (999.999...)

Where are the missing nines? ;^)
Ross Finlayson
2024-06-29 02:43:38 UTC
Permalink
Post by Chris M. Thomasson
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten? You know, add a zero?
(.999...) * (9.999...) = (9.999...)
(9.999...) * 10 = (99.999...)
(99.999...) * 10 = (999.999...)
Where are the missing nines? ;^)
Oh, so you add 9's, ....
Chris M. Thomasson
2024-06-29 02:49:16 UTC
Permalink
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten?  You know, add a zero?
(.999...) * (9.999...) = (9.999...)
(9.999...) * 10 = (99.999...)
(99.999...) * 10 = (999.999...)
Where are the missing nines? ;^)
Oh, so you add 9's, ....
There are infinite nines here, just moving the decimal point wrt base 10.
Ross Finlayson
2024-06-29 04:38:40 UTC
Permalink
Post by Chris M. Thomasson
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten? You know, add a zero?
(.999...) * (9.999...) = (9.999...)
(9.999...) * 10 = (99.999...)
(99.999...) * 10 = (999.999...)
Where are the missing nines? ;^)
Oh, so you add 9's, ....
There are infinite nines here, just moving the decimal point wrt base 10.
This recalls the "First of Zen Koans" bit again,

Two Buddhist priests observe a flag in the wind.

The first says, "the wind, moves, the flag".
The second says, "ah, that flag, moves, in the wind."

A third says "it is your mind that moves".



And like Max in the land of the Wild Things
they made him their king.
Chris M. Thomasson
2024-06-29 03:04:38 UTC
Permalink
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten?  You know, add a zero?
(.999...) * (9.999...) = (9.999...)
(9.999...) * 10 = (99.999...)
(99.999...) * 10 = (999.999...)
Where are the missing nines? ;^)
Oh, so you add 9's, ....
r[0] = .9
r[n] = r[n] + 10^(-n) * .9

Should show infinite nines .999... wrt a summation.

// expand:

r[0] = .9
r[1] = .9 + 10^(-1) * .9 = .99
r[2] = .99 + 10^(-2) * .9 = .999
...

Heck, even r[0] can be: .0 + 10^(-0) * .9

So:

r[0] = .0 + 10^(-0) * .9 = .9
r[1] = .9 + 10^(-1) * .9 = .99
r[2] = .99 + 10^(-2) * .9 = .999
r[3] = .999 + 10^(-3) * .9 = .9999
...

:^)
Chris M. Thomasson
2024-06-29 04:49:33 UTC
Permalink
On 6/28/2024 8:04 PM, Chris M. Thomasson wrote:
[...]
Post by Chris M. Thomasson
r[0] = .9
r[n] = r[n] + 10^(-n) * .9
[...]

Oh my, should that be: ? lol.. wow, sorry, brain fart big time:

r[n] = r[n - 1] + 10^(-n) * .9

;^o

r[0] = .9
r[1] = r[n - 1] + 10^(-n) * .9

as:

r[1] = r[1 - 1] + 10^(-1) * .9

where:


r[1] = r[0] + 10^(-1) * .9


finally:


r[1] = .9 + 10^(-1) * .9 = .99

God damn it!



Sorry everybody! Shit happens.
FromTheRafters
2024-06-29 09:39:53 UTC
Permalink
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten? You know, add a zero?
(.999...) * (9.999...) = (9.999...)
(9.999...) * 10 = (99.999...)
(99.999...) * 10 = (999.999...)
Where are the missing nines? ;^)
Oh, so you add 9's, ....
A musical term.
Ross Finlayson
2024-06-29 15:00:50 UTC
Permalink
Post by FromTheRafters
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by Ross Finlayson
Post by Chris M. Thomasson
Post by WM
Post by joes
Post by WM
Post by Chris M. Thomasson
9.(9) = 10
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by multiplying 0.999... by 10?
Is in 9.999... one 9 more than in 0.999... when 9.999... has been
produced by adding 9 to 0.999...?
No, they both have infinitely many 9s. It does not matter how they
were „produced”.
Then the set of indices is not constant. Try again.
Post by joes
They are the same number 10.
Wrong. Try again.
Did you get dropped on your head at a really young age or something?
;^o
How many times do you multiply it by ten? You know, add a zero?
(.999...) * (9.999...) = (9.999...)
(9.999...) * 10 = (99.999...)
(99.999...) * 10 = (999.999...)
Where are the missing nines? ;^)
Oh, so you add 9's, ....
A musical term.


"Decriptive differential dynamics: stop-derivative and motion"
FromTheRafters
2024-06-25 23:17:09 UTC
Permalink
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the number of
nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the decimal
point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
Ross Finlayson
2024-06-26 00:31:25 UTC
Permalink
Post by FromTheRafters
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
What does Simon Stevin say?

The number of nines left of the radix grows, ....

Though, one might aver it's the "count" of nines,
it's also its number.

Counting and numbering are two different things,
though they're often conflated, not to be confused.

Numbers "have" a number and "make" a count.
FromTheRafters
2024-06-26 10:36:54 UTC
Permalink
Post by Ross Finlayson
Post by FromTheRafters
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
What does Simon Stevin say?
The number of nines left of the radix grows, ....
Though, one might aver it's the "count" of nines,
it's also its number.
Counting and numbering are two different things,
though they're often conflated, not to be confused.
Numbers "have" a number and "make" a count.
...000099.999...

There are countably many nines on either side of the radix point. This
is a 'representation' of a number which has a countably infinite
sequence of symbols left *and* right of the radix. Left of the radix
point it has countably infinitely many zeroes and finitely many
non-zero symbols. Right of the radix it has countably infinitely many
symbols which in this case is all nines and possibly finitely many
non-zero and non-nine leading symbols.

...000100.000...

Same number but no nines.
WM
2024-06-27 10:26:39 UTC
Permalink
Post by FromTheRafters
Post by Ross Finlayson
Post by FromTheRafters
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
What does Simon Stevin say?
The number of nines left of the radix grows, ....
And the number right of the radix point decreases.
Post by FromTheRafters
Post by Ross Finlayson
Though, one might aver it's the "count" of nines,
it's also its number.
Counting and numbering are two different things,
though they're often conflated, not to be confused.
Numbers "have" a number and "make" a count.
...000099.999...
There are countably many nines on either side of the radix point.
"Countably" is an unsharp notion because there are countably many prime
numbers and countably many algebraic numbers but obviously there are
algebraic numbers which are not prime numbers but all prime numbers are
algebraic numbers. So this notion too unsharp to measure the fact that {1,
2, 3, ...} has one number less than {0, 1, 2, 3, ...} although this is an
obvious fact.

Regards, WM
FromTheRafters
2024-06-27 12:41:17 UTC
Permalink
Post by WM
Post by Ross Finlayson
Post by FromTheRafters
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
What does Simon Stevin say?
The number of nines left of the radix grows, ....
And the number right of the radix point decreases.
No, it does not!
WM
2024-06-27 19:55:41 UTC
Permalink
Post by FromTheRafters
Post by WM
Post by Ross Finlayson
The number of nines left of the radix grows, ....
And the number right of the radix point decreases.
No, it does not!
The complete set of indices, when attached to the nines, can grow when
only a point of fly droppings is moved?

Regards, WM
joes
2024-06-28 08:35:38 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by WM
Post by Ross Finlayson
The number of nines left of the radix grows, ....
And the number right of the radix point decreases.
No, it does not!
The complete set of indices, when attached to the nines, can grow when
only a point of fly droppings is moved?
It does not grow. It stays the same. You can subtract an arbitrary finite
number and it doesn’t change.
WM
2024-06-28 13:51:48 UTC
Permalink
Post by joes
Post by WM
Post by FromTheRafters
Post by WM
Post by Ross Finlayson
The number of nines left of the radix grows, ....
And the number right of the radix point decreases.
No, it does not!
The complete set of indices, when attached to the nines, can grow when
only a point of fly droppings is moved?
It does not grow. It stays the same.
Therefore 10*0.999...999 = 9.999...990.

Regards, WM
joes
2024-06-28 17:27:40 UTC
Permalink
Post by WM
Post by joes
Post by WM
The complete set of indices, when attached to the nines, can grow when
only a point of fly droppings is moved?
It does not grow. It stays the same.
Therefore 10*0.999...999 = 9.999...990.
It stays infinite. What does your notation mean? You cannot add digits
„after” infinitely many; there is no after. You cannot count to or past
infinity with naturals, as ω is not itself in N.
WM
2024-06-28 18:28:54 UTC
Permalink
Post by joes
Post by WM
Post by joes
Post by WM
The complete set of indices, when attached to the nines, can grow when
only a point of fly droppings is moved?
It does not grow. It stays the same.
Therefore 10*0.999...999 = 9.999...990.
It stays infinite. What does your notation mean?
The set of indices is constant. It cannot grow. Shifting to the left
happens for all of them.
Post by joes
You cannot add digits
„after” infinitely many; there is no after.
You don't know it. That does not make it vanish.
Post by joes
You cannot count to or past
infinity with naturals,
because they are dark.
Post by joes
as ω is not itself in N.
But it is the first transfinite number.

Regards, WM
WM
2024-06-26 07:18:01 UTC
Permalink
Post by FromTheRafters
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the number of
nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the decimal
point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
You must understand that no natural number can be added to the complete
set. Therefore no index can be added here. 9.999... has as many nines as
0.999... . After the decimal point, there is a difference. In completed
infinity.

Regards, WM
FromTheRafters
2024-06-26 10:40:51 UTC
Permalink
Post by FromTheRafters
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the number of
nines grow?
No, both sequences are infinite.
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal expansion
representation?
You must understand that no natural number can be added to the complete set.
Of course not, a set contains *all* of its elements.
Therefore no index can be added here. 9.999... has as many nines as 0.999...
.
9.999... is not a set.
After the decimal point, there is a difference. In completed infinity.
You are delusional.
WM
2024-06-27 11:49:00 UTC
Permalink
Post by FromTheRafters
Post by FromTheRafters
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal expansion
representation?
You must understand that no natural number can be added to the complete set.
Of course not, a set contains *all* of its elements.
And all are indices of the nines in 0.999... .
Post by FromTheRafters
Therefore no index can be added here. 9.999... has as many nines as 0.999...
.
9.999... is not a set.
But the indices indexing the nines in 0.999... are the set ℕ. After
shifting the decimal point, there are just these indices at the right of
the decimal point. But one of them does not index a nine.
Post by FromTheRafters
After the decimal point, there is a difference. In completed infinity.
You are
using logic. The set of indices remains unchanges, the row of nines do not
acquire another one.

Regards, WM
FromTheRafters
2024-06-27 12:47:10 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by FromTheRafters
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
You must understand that no natural number can be added to the complete set.
Of course not, a set contains *all* of its elements.
And all are indices of the nines in 0.999... .
Post by FromTheRafters
Therefore no index can be added here. 9.999... has as many nines as
0.999... .
9.999... is not a set.
But the indices indexing the nines in 0.999... are the set ℕ. After shifting
the decimal point, there are just these indices at the right of the decimal
point. But one of them does not index a nine.
Post by FromTheRafters
After the decimal point, there is a difference. In completed infinity.
You are
using logic. The set of indices remains unchanges, the row of nines do not
acquire another one.
See Cardinal Arithmetic again, paying close attention to the
transfinite.

https://en.wikipedia.org/wiki/Cardinal_number
Chris M. Thomasson
2024-06-27 20:16:32 UTC
Permalink
Post by WM
Post by FromTheRafters
Post by FromTheRafters
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
You must understand that no natural number can be added to the complete set.
Of course not, a set contains *all* of its elements.
And all are indices of the nines in 0.999... .
r[0] = .9
r[n] = r[0] + 10^(-n) * .9

lets expand it for a couple of iterations:

r[0] = .9
r[1] = r[0] + 10^(-1) * .9 = .99
r[2] = r[1] + 10^(-2) * .9 = .999
r[3] = r[2] + 10^(-3) * .9 = .9999
...

r[n] limit is one in base 10...

Got it?

Want a bunch of 5's?

r[0] = .5
r[1] = r[0] + 10^(-1) * .5 = .55
r[2] = r[1] + 10^(-2) * .5 = .555
r[3] = r[2] + 10^(-3) * .5 = .5555

;^)
Post by WM
Post by FromTheRafters
Therefore no index can be added  here. 9.999... has as many nines as
0.999... .
9.999... is not a set.
But the indices indexing the nines in 0.999... are the set ℕ. After
shifting the decimal point, there are just these indices at the right of
the decimal point. But one of them does not index a nine.
Post by FromTheRafters
After the decimal point, there is a difference. In completed infinity.
You are
using logic. The set of indices remains unchanges, the row of nines do
not acquire another one.
Regards, WM
Chris M. Thomasson
2024-06-28 22:28:53 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Post by FromTheRafters
Post by FromTheRafters
Post by WM
Corollary-question: Does the number of nines grow when in 0.999
the decimal point is shifted by one or more position?
What do you think multiplying by ten does to a continued decimal
expansion representation?
You must understand that no natural number can be added to the complete set.
Of course not, a set contains *all* of its elements.
And all are indices of the nines in 0.999... .
r[0] = .9
r[n] = r[0] + 10^(-n) * .9
^^^^^^^^^^^^^^

Ahhh shit! Damn typos! That recursion should be:

r[n] = r[n - 1] + 10^(-n) * .9

Sorry everybody!
[...]
Chris M. Thomasson
2024-06-25 23:38:15 UTC
Permalink
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the number
of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Huh? .999 * 10 = 9.99

The number of nines is still three.

Want to make it grow by one more nine? Try:

.999 / 10 + .9 = .9999
Chris M. Thomasson
2024-06-25 23:39:08 UTC
Permalink
Post by Chris M. Thomasson
.999 / 10 + .9
Or simply 9999/10000 ;^)
Chris M. Thomasson
2024-06-25 23:44:08 UTC
Permalink
Post by Chris M. Thomasson
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
OOPS! I totally missed you .(9) notation. 1 * 10 = 10.
Post by Chris M. Thomasson
Post by WM
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Huh? .999 * 10 = 9.99
The number of nines is still three.
.999 / 10 + .9 = .9999
WM
2024-06-26 07:19:39 UTC
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Post by Chris M. Thomasson
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the number
of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Huh? .999 * 10 = 9.99
The number of nines is still three.
Hence after the decimal point it differs.

Regards, WM
Chris M. Thomasson
2024-06-26 18:48:08 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
Post by WM
Let the infinite sequence 0.999... be multiplied by 10. Does the
number of nines grow?
Corollary-question: Does the number of nines grow when in 0.999 the
decimal point is shifted by one or more position?
Huh? .999 * 10 = 9.99
The number of nines is still three.
Hence after the decimal point it differs.
I missed the part where you were using .999... = .(9)

Sorry about that, anyway,

(.999...) * (9.999...) = 9.(9) = 10

See?
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