a-conjecture-of-mine

An exercise on polyglossy: the same problem solved on multiple languages

git clone: git://git.pablopie.xyz/a-conjecture-of-mine
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Commit: 41564ac33218aa68e30cc0371aec383dd2ea6940
Parent: 7ad56dd1a0b9c51969d9740cc8d79468cb8c955f
Author: Pablo Emilio Escobar Gaviria <pablo-escobar@riseup.net>
Date: Sat, 15 Aug 2020 13:27:04 -0300

Updated the proof to make it more readable

Diffstat

3 files changed, 62 insertions, 11 deletions

Status	File Name	N° Changes	Insertions	Deletions
Modified	README.adoc	16	5	11
Modified	proof.pdf	0	0	0
Added	proof.tex	57	57	0

diff --git a/README.adoc b/README.adoc
@@ -2,20 +2,14 @@
 
 An exercise on _polyglossy_. The same problem solved on multiple languages.
 
-== The Problem - Mathematicians Version
+== The Problem - Mathematicians' Version
 
-Let latexmath:[$S : \mathbb{N} \rightarrow \mathbb{N}$]
-be the sum of the digits of a positive integer in a 
-https://en.wikipedia.org/wiki/Positional_notation[_base 10 positional number system_]. 
-We have:
-
-[latexmath]
-++++
-\forall a, b \in \mathbb{N}, S_{a + b} \equiv S_a + S_b \;(\textup{mod}\; 9 )
-++++
+Let latexmath:[$S : \mathbb{N} \rightarrow \mathbb{N}$] be the sum of the 
+digits of a natural number. Then 
+latexmath:[$S(n + m) \equiv S(n) + S(m) \; (\textup{mod} \; 9)$] for all
+natural numbers latexmath:[$n$] and latexmath:[$m$].
 
 This conjecture can be generalized for any _positional number system_. 
-Check out `proof.pdf` for more information.
 
 == The Problem - Programmers Verison

diff --git a/proof.pdf b/proof.pdf
Binary files differ.

diff --git a/proof.tex b/proof.tex
@@ -0,0 +1,57 @@
+\documentclass{article}
+\usepackage{amsmath, amssymb, amsthm}
+
+\newtheorem*{conjecture}{Conjecture}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\newcommand{\Mod}{\;(\textup{mod} \; 9)}
+
+\title{A Conjecture of Mine}
+\author{Pablo Emilio Escobar Gaviria}
+\date{August 15, 2020}
+
+\begin{document}
+\maketitle
+
+\begin{conjecture}
+  Let \(S : \mathbb{N} \rightarrow \mathbb{N}\) be the sum of the digits of a
+  natural number. Then \(S(n + m) \equiv S(n) + S(m) \Mod\) for all natural 
+  numbers \(n\) and \(m\).
+\end{conjecture}
+
+\begin{proof}
+  If \(n = n_0 \cdot 10^0 + \cdots + n_k \cdot 10^k \in \mathbb{N}\) then:
+
+  \[
+    \begin{split}
+      n
+      & = n_0 \cdot 10^0 + \cdots + n_k \cdot 10^k \\
+      & = n_0 \cdot (10^0 - 1 + 1) + \cdots + n_k \cdot (10^k - 1 + 1) \\
+      & = (n_0 \cdot (10^0 - 1) + \cdots + n_k \cdot (10^k - 1))
+        + (n_0 + \cdots + n_k) \\
+      & = S(n) + n_0 \cdot (10^0 - 1) + \cdots + n_k \cdot (10^k - 1)
+    \end{split}
+  \]
+
+  Therefore:
+
+  \[
+    \begin{split}
+      n 
+      & \equiv S(n) + n_0 \cdot (10^0 - 1) + \cdots + n_k \cdot (10^k - 1) \\
+      & \equiv S(n) + 0 + \cdots + 0 \\
+      & \equiv S(n) \Mod
+    \end{split}
+  \]
+
+  Similarly, if \(m \in \mathbb{N}\) then \(m \equiv S(m) \Mod\). Finally:
+
+  \[
+    \begin{split}
+      S(n + m)
+      & \equiv n + m \\
+      & \equiv S(n) + S(m) \Mod
+    \end{split}
+  \]
+\end{proof}
+\end{document}
+