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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: Does the number of nines increase?
Date: Fri, 28 Jun 2024 20:04:38 -0700
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On 6/28/2024 7:43 PM, Ross Finlayson wrote:
> On 06/28/2024 07:05 PM, Chris M. Thomasson wrote:
>> On 6/28/2024 6:04 PM, Ross Finlayson wrote:
>>> On 06/28/2024 03:25 PM, Chris M. Thomasson wrote:
>>>> On 6/28/2024 11:21 AM, WM wrote:
>>>>> Le 28/06/2024 à 19:19, joes a écrit :
>>>>>> Am Fri, 28 Jun 2024 13:59:02 +0000 schrieb WM:
>>>>>>> Le 27/06/2024 à 22:20, "Chris M. Thomasson" a écrit :
>>>>>>>
>>>>>>>> There is always an infinite number of nines in Say:
>>>>>>>> 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.
>>>>>
>>>>>> 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
....

:^)