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It is known that $0.\overline 9=1$.

I already have a couple of proofs.

Proof 1: Let $x=0.\bar{9}\Rightarrow 10x=9.\bar{9} \Rightarrow 9x=9\Rightarrow x=1$

Proof 2: Let $x=0.\bar{9}=0.9+0.09+0.009+...=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...=\frac{\frac{9}{10}}{\frac{9}{10}}=1$

Are there are other interesting proofs you can think of or remember?

F. A. Mala
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    Related:https://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1 – Wolgwang Jul 19 '21 at 07:59
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    One way which I remember being pretty convincing for me as a kid was adding three copies of the decimal expansion for $\frac13$ to get $1 = \frac33 = 0.\overline{9}.$ If you want the most fundamental/formal way then you can do it with limits, but that's basically equivalent to the second proof where you use an infinite series – Stephen Donovan Jul 19 '21 at 07:59
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    Whenever this comes up I always have to ask "what does $0.999\ldots$ mean?" Once you actually define it, it's usually clear (or easy to show) that it is equal to $1$. – pancini Jul 19 '21 at 08:00
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    Proof 2 is not a proof, it is the definition : $0.\overline 9= \sum_{k=1}^{\infty} \frac{9}{10^k}.$ This series is convergent and has the value $=1.$ – Fred Jul 19 '21 at 08:00
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    For every $\varepsilon >0$ there's some $n$ such that $1-0.\overline{9} < 1-0.\underbrace{9\ldots9}_{n}< \varepsilon$. – AlvinL Jul 19 '21 at 08:03
  • @Fred Is a proof! The definition is $\sum_{k=1}^\infty \frac{9}{10^k}$. The proof is the simplification of this sum into a closed form. – Theo Bendit Jul 19 '21 at 08:03

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