Re: Does the number of nines increase?

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?

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What do you think multiplying by ten does to a continued decimal expansion representation?

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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
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
;^)

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Therefore no index can be added  here. 9.999... has as many nines as 0.999... .

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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.

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After the decimal point, there is a difference. In completed infinity.

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You are

 using logic. The set of indices remains unchanges, the row of nines do not acquire another one.
 Regards, WM