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

terminatingdecimal 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.