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On 06/28/2024 02:01 PM, FromTheRafters wrote:Yes, but the number of nines in the sequence after the radix point is countably infinite rather than finite, hence NaN in this context.WM pretended :>Le 28/06/2024 à 19:46, FromTheRafters a écrit :>joes was thinking very hard :>Am Fri, 28 Jun 2024 13:55:34 +0000 schrieb WM:>Duh, the set of all natural numbers N contains all of them.What does „complete” mean?It means that no natural number can be added to {0, 1, 2, 3, ..., ω}>>Not their elements. I was asking for their number, how manyWith which numbers do you describe the sizes of N and N_0?Most of them are dark and cannot be used as individuals.
of them there are.
There's a number of them, however, how many there are is not a number.
Either there is a complete and fixed set ℕ with a fixed number |ℕ| of
elements.
If so, then 0.999...*10 = 9,999... where the number of nines has not
changed by more than 0.
Or this is not the case.
But assuming completeness and simultaneously claiming different
numbers of nines is stupid.
The number of nines is not a number.
I thought .999... = 1, ....
>
You know, or it "goes" to.
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