Is this a valid proof that $0.\overline{9} = 1$?

Question

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?

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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.

Answer 1

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 $1$.

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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).

Answer 2

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}