title: Typesetting math category: Features mathjax: true

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