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Re: It is a very bad idea and nothing less than stupid to define 1/3 = 0.333...
Posted:
Oct 4, 2017 3:15 PM


On Wednesday, 4 October 2017 15:06:33 UTC4, Markus Klyver wrote: > Den tisdag 3 oktober 2017 kl. 22:04:52 UTC+2 skrev John Gabriel: > > On Tuesday, 3 October 2017 15:40:54 UTC4, burs...@gmail.com wrote: > > > Nope, doesn't make any sense at all. Just plain > > > crazy. Why would you write a formula for a partial > > > sum as a number: > > > > > > If ... means nth place, then this here: > > > > > > 0.999... > > > > > > Would mean: > > > > > > 110^(n) > > > > > > So it would not be a number, but an expression with > > > a varying place holder n. Making this exxpression > > > here also dependent on n: > > > > > > 0.999... = 1 > > > > > > But fact is that the following expression: > > > > > > 0.999... > > > > > > Has the meaning: > > > > > > lim n>oo (110^(n)) > > > > No. 0.999... = \sum_{k=1}^{\infty} 9/(10^k) > > > > > > > lim_{n \to \infty} \sum_{k=1}^{n} 9/(10^k) has the value: > > > > > > 1 > > > > > > Am Dienstag, 3. Oktober 2017 21:15:40 UTC+2 schrieb netzweltler: > > > > The meaning of "..." is absolutely clear in this context and we both know that there is a nth decimal place for each n ? N in 0.999... > > \sum_{k=1}^{\infty} 9/(10^k) is defined as lim_{n \to \infty} \sum_{k=1}^{n} 9/(10^k). Hence, 0.999... = 1.
"Nah Uh" is not a proof you silly twerp. I am not going to repeat myself to you again. You have been told over and over again.



