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Subject: Re: Does the number of nines =?UTF-8?Q?increase=3F?=
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Date: Sun, 30 Jun 24 14:51:42 +0000
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From: WM <wolfgang.mueckenheim@tha.de>
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Le 30/06/2024 à 12:29, Jim Burns a écrit :
> On 6/29/2024 2:14 PM, WM wrote:

>> If you think straight,
>> then only one conclusion follows:
>> 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
> 
> 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