From adb9b3fe169a92ea2da7b8c804c8b91f966c21a4 Mon Sep 17 00:00:00 2001
From: Egor Tensin <Egor.Tensin@gmail.com>
Date: Thu, 7 May 2020 01:48:11 +0000
Subject: remove everything except Jekyll theme stuff

I tagged the previous commit to explain what's going on.
---
 _posts/2019-09-30-recurring-decimals.md | 93 ---------------------------------
 1 file changed, 93 deletions(-)
 delete 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
deleted file mode 100644
index 741d083..0000000
--- a/_posts/2019-09-30-recurring-decimals.md
+++ /dev/null
@@ -1,93 +0,0 @@
----
-title: Recurring decimals
-excerpt: >
-  Eighth-grader thoughts on recurring decimals.
-category: Math
-mathjax: true
----
-{% include common/mathjax_workaround.md %}
-
-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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