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This question already has an answer here:

Surprisingly, I never came across a repeating decimal, which did not include the last digit, so I'm wondering if this would even be a valid notation?

$$0.\overline{0}1$$

So the following statement would be true?

$$1 - 0.\overline{9} = 0.\overline{0}1$$

Thanks for clarification.

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marked as duplicate by MJD, user147263, Shailesh, Em., user228113 Feb 2 '16 at 0:58

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No, it's not valid; if we specify any digits after the repeated sequence, the sequence has to repeat a specific, finite number of times. Also, note that $0.\overline{9} = 1$, so $1 - 0.\overline{9} = 0$.

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  • $\begingroup$ So, what would the notation of "indefinitely small" or rather "indefinitely close to $0$" be, if the notation analogous to $0.\overline{9}$, meaning "indefinitely close to $1$", is invalid? $\endgroup$ – qwertz Feb 1 '16 at 19:36
  • $\begingroup$ @Coodey We'd use something more rigorous than the $\overline{n}$ notation; the exact convention depends on the context. $\endgroup$ – DylanSp Feb 1 '16 at 19:39

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