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On 6/27/2024 5:09 AM, WM wrote:In 0.999... there is a constant set of indices indexing all nines. If a point of fly droppings becomes visible at another place this does not change the set of indices of nines. Hence also an arbitrary shift of the decimal point by multiplication does not change the number of nines.Le 26/06/2024 à 20:46, "Chris M. Thomasson" a écrit :Where is that last zero coming from?On 6/26/2024 12:15 AM, WM wrote:Wrong.9.999... has one 9 less after the decimal point than 0.999... .>
NO! Think of: .(9) * 10 = 9.(9) = 10
10*0.999...999 = 9.999...990
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