Sujet : Re: Does the number of nines increase?
De : chris.m.thomasson.1 (at) *nospam* gmail.com (Chris M. Thomasson)
Groupes : sci.mathDate : 28. Jun 2024, 23:28:53
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v5ndf6$3hout$1@dont-email.me>
References : 1 2 3 4 5 6
User-Agent : Mozilla Thunderbird
On 6/27/2024 1:16 PM, Chris M. Thomasson wrote:
On 6/27/2024 4:49 AM, WM wrote:
Le 26/06/2024 à 12:40, FromTheRafters a écrit :
WM has brought this to us :
>
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.
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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!
lets expand it for a couple of iterations:
[...]
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