Re: Does the number of nines increase?

On 6/27/2024 8:03 AM, WM wrote:

Le 26/06/2024 à 16:20, Jim Burns a écrit :

On 6/26/2024 3:15 AM, WM wrote:

Le 26/06/2024 à 00:11, Jim Burns a écrit :

On 6/25/2024 4:18 PM, WM wrote:

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.

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,...}

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.

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

Try to think.
"Countable" is an unsharp notion,

Consider instead the notion '1.to.1 function'

"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.

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

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.