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