0

So 1=0.999... can be proved by arguing that the only real number can be put between 1 and 0.999... is zero due to the fact that the only infinitesimal in real number is zero.

We can also write the statement as: $$ (\exists c \in \Bbb R)(1-0.999\cdots=c) \Rightarrow (c=0) $$

Also, I saw another proof in wiki by expressing 0.999.. in an infinite geometric series and calculate its limit to be 1. i.e. $$ 0.999\ldots = \lim_{n\to\infty}0.\underbrace{ 99\ldots9 }_{n} = \lim_{n\to\infty}\sum_{k = 1}^n\frac{9}{10^k} = \lim_{n\to\infty}\left(1-\frac{1}{10^n}\right) = 1-\lim_{n\to\infty}\frac{1}{10^n} = 1 \, - \, 0 = 1. $$

However, the epsilon-delta definition of limit can only ensure that the difference between the summation and its limit to be smaller than smaller than a positive real number $ \epsilon $ . Under such definition, the infinite geometric series above has a limit of 1 but it is actually never reaching 1. The best we can ensure that the infinite geometric series is $ \epsilon $ away from 1.

So my question is:

Under rigorous definition of limit by the epsilon-delta definition and real number system, is it valid to prove 0.999...=1 by infinite geometric series?

Supplement:

So some of you mentioned that the limit of such infinite geometric series is 1. According Epsilon-delta definition of limit, limit is

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon > 0)(\exists \ \delta > 0) (\forall x \in D)(0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon) $$

So we can only find an $ x $ such that $ |f(x) - L| < \varepsilon $. They never equal.The limit of f(x) equals to L doesn't mean f(x) itself equals to L when there exist an $ x $ such that $ 0 < |x - c | < \delta $ .

Ken T
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    Possible duplicate http://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1 – user326159 Mar 11 '17 at 11:47
  • No, my question is actually about the limit under epsilon-delta definition. – Ken T Mar 11 '17 at 11:51
  • "can only ensure that the difference between the summation and its limit to be smaller..." What do you mean by "the summation"? $\sum_{k=1}^\infty \frac{9}{10^k}$ is *by definition* equal to $\lim_{n\to\infty} \sum_{k=1}^n \frac{9}{10^k}$, and this limit is $1$. – littleO Mar 11 '17 at 12:14
  • Possible duplicate of [Is it true that $0.999999999\ldots = 1$?](http://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1) –  Mar 11 '17 at 14:29
  • @littleO I disagree. Please see my supplement to see if my understanding is correct. – Ken T Mar 11 '17 at 14:38
  • The series doesn't reach the limit exactly the same way the dots at the end of $0.999...$ never reach the end. – N. S. Mar 11 '17 at 14:46
  • @N.S. yes, this is what I mean! Everyone, please try to understand my argument. – Ken T Mar 23 '17 at 11:49

1 Answers1

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Not only is it valid, but given that the limit of the series is actually the very definition of what $0.999\ldots$ means, any rigorous proof of $0.999\ldots=1$ is ultimately either itself a prove of that limit, or in some way depends on the proof of that limit.

celtschk
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  • This has not answered my question. This simply asks me to believe. – Ken T Mar 23 '17 at 11:50
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    @KenT: It answers the question because the definition is *the only thing that gives the symbols a meaning*. Without using the definition, something like $0.9999\ldots$ *is not even a number.* It's just a meaningless combination of symbols, until you apply the definition of what the symbols mean. And the definition is exactly the series, plain and simple. – celtschk Mar 28 '17 at 09:25