From 6bf4b90c90f15f4ab60833bddf5b5756d1a6b1f6 Mon Sep 17 00:00:00 2001
From: Elizabeth Alexander Hunt <me@liz.coffee>
Date: Thu, 2 Jul 2026 11:55:17 -0700
Subject: Init

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 Homework/math4310/alg_structures_assn_3.tex | 183 ++++++++++++++++++++++++++++
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+% Created 2023-01-31 Tue 22:49
+% Intended LaTeX compiler: pdflatex
+\documentclass[11pt]{article}
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{graphicx}
+\usepackage{longtable}
+\usepackage{wrapfig}
+\usepackage{rotating}
+\usepackage[normalem]{ulem}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{capt-of}
+\usepackage{hyperref}
+\notindent \notag \usepackage{ dsfont } \usepackage{amsmath}
+\author{Logan Hunt}
+\date{\today}
+\title{Assignment Three}
+\hypersetup{
+ pdfauthor={Logan Hunt},
+ pdftitle={Assignment Three},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 2.2}
+\label{sec:orgfecc6e5}
+\subsection{Question One}
+\label{sec:orgd22e42b}
+\subsubsection{b}
+\label{sec:org757c583}
+\begin{center}
+\begin{tabular}{rrrrr}
+\(\oplus\) & 0 & 1 & 2 & 3\\[0pt]
+0 & 0 & 1 & 2 & 3\\[0pt]
+1 & 1 & 2 & 3 & 0\\[0pt]
+2 & 2 & 3 & 0 & 1\\[0pt]
+3 & 3 & 0 & 1 & 2\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{rrrrr}
+\(\odot\) & 0 & 1 & 2 & 3\\[0pt]
+0 & 0 & 0 & 0 & 0\\[0pt]
+1 & 0 & 1 & 2 & 3\\[0pt]
+2 & 0 & 2 & 0 & 2\\[0pt]
+3 & 0 & 3 & 2 & 1\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsubsection{c}
+\label{sec:org603554a}
+\begin{center}
+\begin{tabular}{rrrrrrrr}
+\(\oplus\) & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+0 & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+1 & 1 & 2 & 3 & 4 & 5 & 6 & 0\\[0pt]
+2 & 2 & 3 & 4 & 5 & 6 & 0 & 1\\[0pt]
+3 & 3 & 4 & 5 & 6 & 0 & 1 & 2\\[0pt]
+4 & 4 & 5 & 6 & 0 & 1 & 2 & 3\\[0pt]
+5 & 5 & 6 & 0 & 1 & 2 & 3 & 4\\[0pt]
+6 & 6 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{rrrrrrrr}
+\(\odot\) & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\[0pt]
+1 & 0 & 1 & 2 & 3 & 4 & 5 & 6\\[0pt]
+2 & 0 & 2 & 4 & 6 & 1 & 3 & 5\\[0pt]
+3 & 0 & 3 & 6 & 2 & 5 & 1 & 4\\[0pt]
+4 & 0 & 4 & 1 & 5 & 2 & 6 & 3\\[0pt]
+5 & 0 & 5 & 3 & 1 & 6 & 4 & 2\\[0pt]
+6 & 0 & 6 & 5 & 4 & 3 & 2 & 1\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsection{Question Three}
+\label{sec:org6ac99a4}
+\(x = [1], [3], [5], [7]\)
+
+\subsection{Question Five}
+\label{sec:org69e894a}
+\(x = [1], [2], [4], [5]\)
+
+\subsection{Question Eight}
+\label{sec:org26350fa}
+\(x = [1], [2], [6], [7]\)
+
+\subsection{Question Eleven}
+\label{sec:orgcd603cc}
+\subsubsection{b}
+\label{sec:org2896606}
+\(x = [0], [1], [2], [3]\)
+\subsection{Question Fifteen}
+\label{sec:org5bca8ba}
+\subsubsection{c}
+\label{sec:org830aeb3}
+From the Binomial Theorem,
+
+\begin{align*}
+(a + b)^5 &= a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5
+\end{align*}
+
+Then,
+\begin{equation*}
+(a + b)^5 &\equiv_5 (a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5ab^4 + b^5) \\
+&\equiv_5 (a^5 + b^5) \\
+\end{equation*}
+
+since each of the terms \(5a^4 b, 10a^3 b^2, 10a^2 b^3, 5ab^4\) are zero as \(5 \equiv_5 0\) and \(10 \equiv_5 0\).
+
+
+
+\section{Section 2.3}
+\label{sec:org7ebf66c}
+\subsection{Question One}
+\label{sec:orge906711}
+\subsubsection{b}
+\label{sec:orgddb4d23}
+\([1], [3], [5], [7]\) since \([7 * 7] = [49] = [1]\), \([5 * 5] = [25] = [1]\), \([3 * 3] = [9] = [1]\), and
+\([1 * 1] = [1]\).
+\subsection{Question Two}
+\label{sec:org4efa77f}
+\subsubsection{b}
+\label{sec:org41d490d}
+\([2], [4], [6]\) since \([2 * 4] = [8] = [0]\), \([4 * 4] = [16] = [0]\), and \([6 * 4] = [24] = [0]\).
+\subsection{Question Eight}
+\label{sec:org27e3ed4}
+\subsubsection{a}
+\label{sec:orgeac97d5}
+\begin{enumerate}
+\item \(2x = 1\)
+\item \(2x = 3\)
+\item \(2x = 5\)
+\end{enumerate}
+\subsubsection{b}
+\label{sec:orgb99f90c}
+Yes, each one is equivalent to 0 when \(x = 6\).
+\subsection{Question Nine}
+\label{sec:orge8d0de5}
+\subsubsection{a}
+\label{sec:org8045c18}
+By definition, there exists \(b\), the inverse of \(a\), such that \(ab = 1\).
+Assume that \(a\) is a zero divisor, then there exists \(c \neq 0\)
+such that \(ac = 0\). Then, \((ab)c = 0b \Rightarrow (1)(c) = 0\) implies that \(c = 0\), which is a contradiction.
+\subsubsection{b}
+\label{sec:orgc5282b2}
+By definition, there exists \(b\), with \(b \neq 0\) such that \(ab = 0\).
+Assume that \(a\) is a unit, then there exists \(c\) such that \(ac = 1\).
+Then, \((b)ac = 1b \Rightarrow 0c = b\) implies that \(b = 0\), which is a contradiction.
+
+\subsection{Question Eleven}
+\label{sec:org6981e66}
+By definition of \(a\) being a unit, there exists \(ay = 1\) with \(y\) being an inverse of \(a\).
+By multiplying our target \((y)ax = (y)b \Rightarrow x = yb\).
+
+To prove this is unique, assume that \(k\) and \(l\) are solutions of \(ax = b\). Then, \(ak = b\) and \(al = b\). Since \(a\) is a unit, by using our
+previous strategy, \((y)ak = (y)b\) and \((y)al = (y)b\), so \(k = yb\) and \(l = yb\) and thus \(k = l\).
+
+
+\section{Chapter 13}
+\label{sec:org62ae8b0}
+\subsection{A2}
+\label{sec:org0b1499e}
+As \(p | c \Rightarrow c = pk\), and \(p | c \Rightarrow c = ql\) then \(pk = ql\) and thus by Theorem 1.5 since \(p\) and \(q\) are prime \(p | l\)
+or \(p | q\) and \(q | p\) or \(q | k\), but since \(p\) and \(q\) are prime, then it must be that only \(p | l\) and \(q | k\).
+
+Since \(p | l\) then \(l = mp\) and \(c = ql \Rightarrow c = qmp\) and \(qp\) is a factor of \(c\).
+
+\subsection{A3 (a)}
+\label{sec:orge4f13f2}
+GO = 0715
+
+715\textsuperscript{3} (mod 2773) = 107
+\end{document}
\ No newline at end of file
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