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Does .99999… = 1?

After reading all the kind answers for this previous question question of mine, I wonder... How do we get a fraction whose decimal expansion is the simple $0.\overline{9}$?

I don't mean to look like kidding or joking (of course, one can teach math with fun so it becomes more interesting), but this series has really raised a flag here, because $\frac{9}{9}$ won't solve this case, although it solves for all other digits (e.g. $0.\overline{8}=\frac{8}{9}$ and so on).

Thanks! Beco.

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marked as duplicate by Qiaochu Yuan Mar 29 '11 at 4:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Since $0.\overline 9=1$, $\tfrac99$ does solve this case. – Mariano Suárez-Alvarez Mar 29 '11 at 4:29
I don't understand what you mean by "create." Is your question answered by the discussion at ? – Qiaochu Yuan Mar 29 '11 at 4:29
$0.9999... = 1$. You can create it with $1/1=1$. This is a duplicate of many questions, or if that information is not enough, please see – fdart17 Mar 29 '11 at 4:31
@Qiaochu: The OP just recently asked how, given a repeating decimal, get a fraction that has that decimal expansion. The title was not very felicitous, though. – Arturo Magidin Mar 29 '11 at 4:33
Hum... It looks like it is one of those corners of mathematics. If some mod feel it should close this question, please do so. Thanks your patience. – Dr Beco Mar 29 '11 at 4:34
up vote 5 down vote accepted

The number $0.9999\cdots$ is in fact equal to $1$, which is why you get $\frac{9}{9}$. See this previous question.

To see it is equal to $1$, you can use any number of ideas:

  1. The hand-wavy but convincing one: Let $x=0.999\cdots$. Then $10x = 9.999\cdots = 9 + x$. So $9x = 9$, hence $x=1$.

  2. The formal one. The decimal expansion describes an infinite series. Here we have that $$ x = \sum_{n=1}^{\infty}\frac{9}{10^n}.$$ This is a geometric series with common ration $\frac{1}{10}$ and initial term $\frac{9}{10}$, so $$x = \sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{\quad\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\quad\frac{9}{10}\quad} = 1.$$

In general, a number whose decimal expansion terminates (has a "tail of 0s") always has two decimal expansions, one with a tail of 9s. So: $$\begin{align*} 1.0000\cdots &= 0.9999\cdots\\ 2.480000\cdots &= 2.4799999\cdots\\ 1938.01936180000\cdots &= 1938.019361799999\cdots \end{align*}$$ etc.

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Thanks again Arturo. – Dr Beco Mar 29 '11 at 4:35
Sorry I can't accept the answer right now. Only in 6 minutes according with the site. I'll do it soon. – Dr Beco Mar 29 '11 at 4:36

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