From 84376189ffeabdd3b96bc1986f2e1b688bdd41ad Mon Sep 17 00:00:00 2001
From: Egor Tensin <Egor.Tensin@gmail.com>
Date: Mon, 7 Nov 2022 15:58:12 +0100
Subject: mathjax: replace with "Recurring decimals" from /blog/

It was always kinda lame there; belongs here IMO.
---
 _posts/2021-04-08-mathjax.md | 86 ++++++++++++++++++++++++++++++++++++++++----
 1 file changed, 80 insertions(+), 6 deletions(-)

(limited to '_posts')

diff --git a/_posts/2021-04-08-mathjax.md b/_posts/2021-04-08-mathjax.md
index d5f6e95..1c86957 100644
--- a/_posts/2021-04-08-mathjax.md
+++ b/_posts/2021-04-08-mathjax.md
@@ -3,13 +3,87 @@ title: Typesetting math
 category: Features
 mathjax: true
 ---
-Today, we'll "prove" that $$2 \times 2 \ne 5$$.
+First, let's determine that
 
 $$
-\begin{gather}
-2 \times 2 = 2 + 2 = 4 \\
-4 \ne 5 \implies 2 \times 2 \ne 5
-\end{gather}
+\newcommand\naturals{\mathbb{N}}
+\newcommand\rationals{\mathbb{Q}}
+\newcommand\reals{\mathbb{R}}
+
+\newcommand\xs{x_{1}x_{2}\dots x_n}
+\newcommand\nines{\overbrace{99\dots 9}^n}
+\newcommand\pq{\frac{p}{q}}
+\newcommand\qp{\frac{q}{p}}
+
+0.(9) = 1
+$$
+
+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
+
+$$
+\begin{align*}
+0.(\xs) \times \qp &= 0.(\xs) \times \frac{\nines}{\xs} \\
+&= \left(\frac{\xs}{10^n} + \frac{\xs}{10^{2n}} + \dots\right) \times \frac{\nines}{\xs} \\
+&= \left(\frac{1}{10^n} + \frac{1}{10^{2n}} + \dots\right) \times\,\nines \\
+&= 0.(9) \\
+&= 1
+\end{align*}
 $$
 
-Q.E.D.
+Finally, $$0.(\xs) \times \qp = 1 \implies \pq = 0.(\xs)$$.
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