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\author{Logan Hunt}
\date{\today}
\title{Assignment Three}
\hypersetup{
 pdfauthor={Logan Hunt},
 pdftitle={Assignment Three},
 pdfkeywords={},
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 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}