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Topic: It is a very bad idea and nothing less than stupid to define 1/3
= 0.333...

Replies: 5   Last Post: Oct 4, 2017 3:15 PM

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genmailus@gmail.com

Posts: 323
Registered: 8/18/16
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
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On Wednesday, 4 October 2017 15:06:33 UTC-4, 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 UTC-4, 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 n-th place, then this here:
> > >
> > > 0.999...
> > >
> > > Would mean:
> > >
> > > 1-10^(-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 (1-10^(-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.



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