From 6327ac306dbfce4d6a9500d61fb11dae4b192f4a Mon Sep 17 00:00:00 2001
From: Egor Tensin <Egor.Tensin@gmail.com>
Date: Mon, 30 Sep 2019 00:50:28 +0300
Subject: add "Recurring decimals"

---
 _posts/2019-09-30-recurring-decimals.md | 96 +++++++++++++++++++++++++++++++++
 1 file changed, 96 insertions(+)
 create mode 100644 _posts/2019-09-30-recurring-decimals.md

(limited to '_posts/2019-09-30-recurring-decimals.md')

diff --git a/_posts/2019-09-30-recurring-decimals.md b/_posts/2019-09-30-recurring-decimals.md
new file mode 100644
index 0000000..83d949c
--- /dev/null
+++ b/_posts/2019-09-30-recurring-decimals.md
@@ -0,0 +1,96 @@
+---
+title: Recurring decimals
+layout: post
+excerpt: >
+  Noob thoughts about recurring decimals.
+category: Math
+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 is 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
+
+$$
+\begin{align*}
+0.(\xs) \times \qp &= 0.(\xs) \times \frac{\nines}{\xs} \\
+&= \left(\frac{\xs}{10^n} + \frac{\xs}{10^{2n}} + \dots\right) \times \frac{\nines}{\xs} \\
+&= \left(\frac{1}{10^n} + \frac{1}{10^{2n}} + \dots\right) \times\,\nines \\
+&= 0.(9) \\
+&= 1
+\end{align*}
+$$
+
+Ergo,
+
+$$
+\pq = 0.(\xs)
+$$
-- 
cgit v1.2.3