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I have stumbled across the proof that $0.\overline{9}=1$. The proof is as follows.

Let $x=0.\overline{9}$

$10\cdot x = 9.\overline{9}$

$10\cdot x = 9 + 0.\overline{9}$

Now that $0.\overline{9}=x$, $10\cdot x = 9 + x$.

We get $9x=9$ and $\therefore x=1$.

With this proof, we know that $0.\overline{9}=1$. Is there any scenario in math where using $0.\overline{9}$ instead of $1$ offers an easier solution to a problem?

Jaden Lee
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    Rounding is not required. $0.\overline{9} = 1$ without the need for rounding. – badjohn Jun 28 '20 at 09:21
  • @badjohn I do already know that $0.\overline{9}=1$. What I was asking was that is there any time when either solving a problem or proving something that using the $0.\overline{9}$ instead of $1$ is makes the math easier to solve. Perhaps I am not understanding your point. Do you mean that because $0.\overline{9}=1$, there is no such case when using $0.\overline{9}$ is more useful than $1$? – Jaden Lee Jun 28 '20 at 19:15
  • I wa referring to "rounding" in the title when there is no rounding in the body of the question. – badjohn Jun 28 '20 at 19:22

1 Answers1

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I don't know of any case where using this representation offers an easier solution to a given problem. But in one case it is useful is Cantor's diagonal argument for the proof of uncountablity of the reals using decimal number system. There you want to make sure that while listing out the reals, that you always consider one representation of such recurring $9$s vs $0$s consistently. Otherwise someone might claim that oh well! the number you constructed is not in our list, but an alternate representation might have it and other more subtle issues.So yeah! without this basic fact, proving uncountability of real becomes harder by Cantor's beautiful agrument.

Bhaswat
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  • I know that Cantor's diagonal argument can be used to prove that the set of real numbers between 0 to 1 is greater than the natural number set. However, the argument only requires binary numbers (0 and 1) to prove that its set is greater than the natural number set. How does using $0.\overline{9}$ instead of $1$ help support the uncountability of the real set? – Jaden Lee Jun 28 '20 at 08:44
  • it is the same! In binary numbers $ .01111111111111111111.... = .1$. You can read more about it in https://math.stackexchange.com/questions/2694917/disproving-cantors-diagonal-argument – Bhaswat Jun 28 '20 at 08:47