First, let’s determine that

\[\newcommand\naturals{\mathbb{N}} \newcommand\rationals{\mathbb{Q}} \newcommand\reals{\mathbb{R}} \newcommand\xs{x_{1}x_{2}\dots x_n} \newcommand\nines{\overbrace{99\dots 9}^n} \newcommand\pq{\frac{p}{q}} \newcommand\qp{\frac{q}{p}} 0.(9) = 1\]

This may seem counter-intuitive, but demonstrably true. If \(0.(9) \neq 1\), then \(\exists n \in \reals: 0.(9) < n < 1\). To put it another way, there must be a number greater than 0.(9) and lesser than 1, equal to neither. Thinking about it makes it obvious this is not true.

Slightly more formally,

\[\begin{align*} 1 - 0.9 &= \frac{1}{10} \\ 1 - 0.99 &= \frac{1}{100} \\ 1 - 0.999 &= \frac{1}{10^{-3}} \\ \dots \\ 1 - 0.\nines &= \frac{1}{10^n} \\ \end{align*}\]

If \(0.(9) \neq 1\), the following must hold:

\[\forall n \in \naturals, \exists x \in \reals: x < \frac{1}{10^n}\]

It’s clear that the only such number is 0, making 0.(9) and 1 equal.

Let \(n \in [1,9]\). Is there \(\pq \in \rationals: \pq = 0.(n)\)?

For \(n = 9\), we’ve established \(p = 1, q = 1\). For the other values of \(n\) we can observe that

\[\begin{align*} 0.(1) \times 9 &= 0.(9) = 1 \\ 0.(2) \times 9/2 &= 0.(9) = 1 \\ 0.(3) \times 9/3 &= 0.(9) \\ \dots \\ 0.(8) \times 9/8 &= 0.(9) \\ 0.(9) \times 1 &= 0.(9) \end{align*}\]

So,

\[\forall n \in [1,9], \frac{n}{9} = 0.(n)\]

In general, let’s demonstrate that

\[\forall n \in \naturals, n > 0, \forall 0.(x_{1}x_{2}\dots x_n), \exists \pq \in \rationals: \pq = 0.(\xs)\]

Let \(p = \xs, q =\,\nines\). It’s clear that

\[\begin{align*} 0.(\xs) \times \qp &= 0.(\xs) \times \frac{\nines}{\xs} \\ &= \left(\frac{\xs}{10^n} + \frac{\xs}{10^{2n}} + \dots\right) \times \frac{\nines}{\xs} \\ &= \left(\frac{1}{10^n} + \frac{1}{10^{2n}} + \dots\right) \times\,\nines \\ &= 0.(9) \\ &= 1 \end{align*}\]

Finally, \(0.(\xs) \times \qp = 1 \implies \pq = 0.(\xs)\).

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