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There are tens of posts already on this site about whether $0.\overline{9} = 1$.

This is something that intrigues me, and I have a question about this, including a "proof" which I have found myself.

Question:

This comment says that

you need a terminating decimal to get something less than 1.

If so, does it mean that a non-terminating decimal (e.g. $0.\overline{9}$) is $\ge 1$?

So is $\frac{1}{3}$ ($0.\overline{3}$) also $\ge 1$? It is non-terminating, but you can subtract $\frac{1}{3}$ from $1$ to get $\frac{2}{3} = 0.\overline{6}$, which is another non-terminating decimal. How do those mechanics work?


Theorem: $0.99999... = 1(.00000... = 1)$

Proof:

\begin{align} \frac{1}{9} &= 0.11111... \\ \frac{2}{9} &= 0.22222... \\ \frac{3}{9} &= 0.33333... \\ \frac{4}{9} &= 0.44444... \\ \frac{5}{9} &= 0.55555... \\ \frac{6}{9} &= 0.66666... \\ \frac{7}{9} &= 0.77777... \\ \frac{8}{9} &= 0.88888... \\ \therefore \frac{9}{9} &= 0.99999... \\ &= 1 \end{align}

Is the above proof correct? I came up with it myself before I decided to ask this question, but I do not know if it is mathematically valid.

Aryan Beezadhur
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    For me, it's ok – Tito Eliatron Sep 30 '20 at 09:14
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    I would accept the proof. – Peter Sep 30 '20 at 09:22
  • I doubt that this is a "proof". Because how you are evaluating $\frac{1}{9} = 0.11111...$? You can do same process for $\frac{9}{9}$. – C.F.G Sep 30 '20 at 09:23
  • In what way you compute $\dfrac 9 9$ ? According to elementary-school teaching will be: how many times $9$ "is in" $9$ ? Answer: $1$. There is no way to produce $0.999 \ldots$ this way. – Mauro ALLEGRANZA Sep 30 '20 at 09:24
  • @MauroALLEGRANZA I just found this proof that I made in the post you link also, but [as a comment](https://math.stackexchange.com/q/11/717872#comment7407644_11) – Aryan Beezadhur Sep 30 '20 at 09:25
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    in math, calculator is not a precise device. – C.F.G Sep 30 '20 at 09:26
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    Of couse it is better to use the geometric series to prove it. Nevertheless, I would consider the proof to be valid. – Peter Sep 30 '20 at 09:28
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    @Aryan Beezadhur, you can safely say that because it is true, but calculator doesn't prove it. – Ennar Sep 30 '20 at 09:28
  • @Ennar but we can prove that $$\frac{1}{3} = 0.\overline{3}$$ since $0.9 / 3 = 0.3$ and $0.1$ remains, which will become $0.09 / 3$ and so on – Aryan Beezadhur Sep 30 '20 at 09:29
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    I think the statement of your question is likely to divide people, because really the whole issue $0.\bar{9} = 1$ is about offering an intuitive explanation of why the real numbers have certain properties and why the series definition of an infinite decimal is the right one. This doesn't usually involve any very rigorous proofs - to *prove* $0.\bar{9} = 1$ in the formal sense, you do little bit of real analysis. But of course that's not very intuitively convincing! – Izaak van Dongen Sep 30 '20 at 09:30
  • @Peter, please clarify. To me it looks like this says $1/9 = 0.\bar 1 \implies 9/9 = 0.\bar 9$. However, this just shifts the problem to something else. – Ennar Sep 30 '20 at 09:30
  • @AryanBeezadhur: How? it is just a process. i.e. first divide $10$ to $3$ then add a zero to its reminder then divide to $3$ so on. – C.F.G Sep 30 '20 at 09:31
  • @Aryan Beezadhur, of course you can prove it, but you didn't write a proof. – Ennar Sep 30 '20 at 09:31
  • The proof uses that the sequence $0.1,0.11,0.111,\cdots $ tends to $\frac{1}{9}$. If we multiply it with $9$ , the sequence is $0.9,0.99,0.999,\cdots $ which then tends to $1$. It might be a bit sloppy to write $0.\bar 1\cdot 9=0.\bar 9=1$ for a purist. – Peter Sep 30 '20 at 09:33
  • I mean the statement is true but that's not a proof –  Sep 30 '20 at 11:37

2 Answers2

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The non-terminating notation (either $0.9999\cdots$ or $0.\overline9$) is a disguised limit, namely

$$\lim_{n\to\infty}\sum_{k=1}^n\frac 9{10^k}$$ or $$\lim_{n\to\infty}\left(1-10^{-n}\right).$$

This limit equals $1$.


To relate this to your proof, we indeed have

$$0.\overline1=\lim_{n\to\infty}\sum_{k=1}^n\frac 1{10^k}=\frac19,$$

then

$$9\cdot 0.\overline1=9\lim_{n\to\infty}\sum_{k=1}^n\frac 1{10^k}=\lim_{n\to\infty}\sum_{k=1}^n\frac 9{10^k}=0.\overline 9$$ and $$\frac99=1.$$

But honestly, I see no benefit taking this indirect route through $9\cdot0.\overline 1$, and for completeness, you should explain why $\dfrac19=0.\overline1$, and why $9\cdot0.\overline1=0.\overline9$ like I did (or another way).

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What you wrote is simply $$\frac 19 = 0.\bar 1 \implies 1 = \frac 99 = 0.\bar 9$$ which is correct.

However, you didn't write anything to prove that $\frac 19 = 0.\bar 1$, so I wouldn't count this as a proof.

Any proof really needs to use that decimal number system is using geometric series to represent numbers that they converge to. You can use formula for the sum of geometric series, like Yves Daoust did, or you can use the same technique that's used in the proof of the formula directly:

$$x = 0.9999\ldots \implies 10 x = 9.9999\ldots = 9 + x \implies x = 1.$$

The above is just a notational shortcut for the following: \begin{align} 10\cdot\lim_{n\to\infty} \left(\frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^n}\right) &= \lim_{n\to\infty} 10\left(\frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^n}\right)\\ &= 9 + \lim_{n\to\infty} \left(\frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^{n-1}}\right). \end{align}

Ennar
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