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+% Created 2023-01-18 Wed 12:13
+% 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}
+\noindent \notag \usepackage{ dsfont }
+\author{Logan Hunt}
+\date{\today}
+\title{Assignment One}
+\hypersetup{
+ pdfauthor={Logan Hunt},
+ pdftitle={Assignment One},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.5.5)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\tableofcontents
+
+
+\section{Section 1.1}
+\label{sec:org35f7eb6}
+\subsection{Question 5}
+\label{sec:orge503259}
+By reforming the given expression to show \(ca\) as a multiple of \(q\) and some remainder:
+
+\begin{align}
+a = bq + r \\
+ca = (cb)q + (c)r
+\end{align}
+
+and that \(0 \leq r < b \Rightarrow 0 \leq cr < cb\), Theorem 1.1 tells us that there is only one unique quotient:
+which is \(q\), and the remainder is \((c)r\).
+
+\subsection{Question 7}
+\label{sec:org6afde38}
+By Theorem 1.1, \(a = bq + r, 0 \leq r \lt b\) implies that when b = 3, \(a = 3q + r\) where \(0 \leq r \lt b\),
+thus \(a \in \mathds{Z} \Rightarrow a = 3q + r\) where \(r \in {0,1,2}\).
+
+By squaring \(a\), we have three options which simplify to the form \(3k\) or \(3k + 1\):
+\begin{enumerate}
+\item \(a^2 = (3q)^2 = 9q^2\) which is \(3k \ni k = 3q^2\)
+\item \(a^2 = (3q + 1)^2 = 9q^2 + 3q + 1 = 3(3q^2 + q) + 1\) which is \(3k + 1 \ni k = (3q^2 + q)\)
+\item \(a^2 = (3q + 2)^2 = 9q^2 + 6q + 4 = 3(3q^2 + 2q) + 4\) which is
+\(3l + 4 \ni l = (3q^2 + 2q)\) which is also \(3l + 1 + 3 = 3(l+1) + 1\),
+which again is \(3k + 1 \ni k = l+1\)
+\end{enumerate}
+
+\subsection{Question 8}
+\label{sec:orgfe8a512}
+By Theorem 1.1, \(a = bq + r, 0 \leq r \lt b\) implies that when b = 4, \(a = 4q + r\) where \(0 \leq r \lt b\),
+and thus \(a = 4q + r\) and \(r \in {0,1,2,3}\).
+
+Therefore we have four cases:
+
+\begin{enumerate}
+\item \(a = 4q\), and \(a\) must be even which is invalid
+\item \(a = 4q + 1\), and \(4q + 1 = 2k + 1 \ni k = 2q\) which fits the definition of an odd number
+\item \(a = 4q + 2\), thus \(a\) must be even as \(a = 2k \ni k = 2q + 1\)
+\item \(a = 4q + 3\), can be rewritten to \(4q + 1 + 2\), which is odd: \(2k + 1 \ni k = 2q + 1\)
+\end{enumerate}
+
+And thus any odd number can be rewritten as \(4q + 1\) or \(4q + 3\).
+
+\subsection{Question 10}
+\label{sec:org3c66372}
+The division of \(a\) and \(c\) by \(n\) can each be represented by
+
+\begin{align}
+a = q_{a}n + r_a \\
+c = q_c_{}n + r_c
+\end{align}
+
+where \(0 \le r_a < n\) and \(0 \le r_c < n\) by Theorem 1.1, and we can subtract each side of the
+second equation from the first:
+
+\begin{align}
+a - c = (q_{a}n + r_a) - (q_c_{}n + r_c)
+\end{align}
+
+With this work now in hand, we will prove by showing that the conjecture is true both ways:
+
+For the first, we will suppose that \(r_a = r_c\). Then, we find that
+\begin{align}
+a - c = n(q_a - q_c) \\
+a - c = nk
+\end{align}
+for some integer \(k\).
+
+For the second, we will prove that if \(a - c = nk\), when \(a\) and \(c\) are divided by \(n\),
+they leave the same remainder \(r\).
+
+From the work we did previously, and by substituting \(a-c = nk\), we find that
+\begin{align}
+a - c = (q_{a}n + r_a) - (q_c_{}n + r_c) \\
+nk = (q_{a}n + r_a) - (q_{c}n + r_c) \\
+nk = n(q_a - q_c) + (r_a - r_c) \\
+n(k - q_a + q_c) = r_a - r_c
+\end{align}
+and thus \(r_a - r_c\) is a multiple of \(n\).
+
+From the inequalities produced previously by Theorem 1.1, if \(r_a \geq r_c\) then \(0 \le r_a - r_c < n\)
+and \(-n < r_c - r_a \leq 0\). Else if \(r_c \geq r_a\) then \(0 \leq r_c - r_a < n\) and \(-n < r_a - r_c \leq 0\).
+
+By combining the two cases, it must be that \$ -n < r\textsubscript{a} - r\textsubscript{c} < n \$, and from the above fact that
+\(r_a - r_c\) is a multiple of n, the only multiple of n in \((-n, n)\) is 0. Thus \(r_a = r_c\).
+
+\section{Section 1.2}
+\label{sec:org4226083}
+\subsection{Question 1}
+\label{sec:org4ccc206}
+\subsubsection{c}
+\label{sec:org802af33}
+1,  57 and 112 are co-prime
+\subsection{Question 3}
+\label{sec:orgb341781}
+\begin{align}
+a | b \Rightarrow \exists n \in \mathds{Z} \ni an = b \\
+b | c \Rightarrow \exists m \in \mathds{Z} \ni bm = c \\
+(an)m = c \Rightarrow a | c
+\end{align}
+\subsection{Question 4}
+\label{sec:orge55e932}
+\subsubsection{a}
+\label{sec:org7540edc}
+\begin{align}
+a | b \Rightarrow \exists n \in \mathds{Z} \ni an = b \\
+a | c \Rightarrow \exists m \in \mathds{Z} \ni am = c \\
+a | (b + c) \Rightarrow \exists l \in \mathds{Z} \ni al = b + c \\
+al = an + am \\
+b + c = a(n + m)
+\Rightarrow a | b+c
+\end{align}
+\subsubsection{b}
+\label{sec:orge72f7c4}
+By continuing from "a":
+
+\begin{align}
+br + ct = (an)r + (am)t \\
+\Rightarrow br + ct = a(nr + mt) \\
+\Rightarrow a | (br + ct)
+\end{align}
+
+
+\subsection{Question 7}
+\label{sec:org7273a90}
+\(|a|\) since \(|a| \textbar a\) and \(|a| \textbar 0\), and there cannot be an integer larger than
+\(|a|\) that divides \(a\).
+
+\subsection{Question 9}
+\label{sec:org6b4815b}
+No, not every multiple of two factors of an integer divides that integer.
+
+For example, \(3 | 9\) and \(9 | 9\) but \(27 \nmid 9\).
+
+\subsection{Question 15}
+\label{sec:orgf96c6cd}
+\subsubsection{c}
+\label{sec:orgad14e2b}
+\begin{align}
+1003 = (2)(456) + 91 \\
+456 = (5)(91) + 1 \\
+91 = (91)(1) + 0 \\
+\Rightarrow (1003, 456) = 1
+\end{align}
+\subsubsection{d}
+\label{sec:orgf89fea4}
+\begin{align}
+322 = (2)(148) + 26 \\
+148 = (5)(26) + 18 \\
+26 = (1)(18) + 8 \\
+18 = (2)(8) + 2 \\
+8 = (4)(2) + 0 \\
+\Rightarrow (322, 148) = 2
+\end{align}
+
+\subsection{Question 17}
+\label{sec:org66d011a}
+
+\begin{align}
+a | c \Rightarrow \exists n \in \mathds{Z} \ni an = c \\
+b | c \Rightarrow b | an
+\end{align}
+
+And by Theorem 1.4, \((a, b) = 1 \Rightarrow b | n \Rightarrow ab | an \Rightarrow ab | c\):
+
+\subsection{Question 19}
+\label{sec:org905b244}
+Given some \(d\) such that \(d | a\) and \(d | b\), then \(a = nd\) and \(b = md\).
+\begin{align}
+a | (b + c) \Rightarrow \exists p \in \mathds{Z} \ni ap = b + c \\
+c = ap - md \Rightarrow c = (nd)p - md \Rightarrow c = d(np - md) \Rightarrow d | c
+\end{align}
+
+Since we're given that \((b, c) = 1 \Rightarrow (md, c) = 1\) and from the above fact that \(d | c\), we can conclude
+that \(md = 1\). Therefore \((a, b) = (a, md) = (a, 1) = 1\).
+
+Given some \(e\) such that \(e | a\) and \(e | c\), then \(a = ue\) and \(c = we\).
+\begin{align}
+a | (b + c) \Rightarrow \exists o \in \mathds{Z} \ni ao = b + c \\
+b = ao - we \Rightarrow b = (ue)o - we \Rightarrow b = e(uo - w) \Rightarrow e | b
+\end{align}
+
+And from similar reasoning we can conclude that \((a, c) = (a, we) = (a, 1) = 1\).
+
+\subsection{Question 31}
+\label{sec:orgc12b666}
+\subsubsection{a}
+\label{sec:org4312735}
+\([6, 10] = 60\)
+
+\([4, 5, 6, 10] = 60\)
+
+\([20, 42] = 840\)
+
+\([2, 3, 14, 36, 42] = 252\)
+
+\subsubsection{b}
+\label{sec:orgb964c2c}
+Let \(m = [a_1, a_2, \ldots, a_k]\), then by the Division Algorithm, \(t = mq + r\) for integers \(q\) and \(r\),
+with \(0 \leq r < m\).
+
+As given that all \(a_i | t\) \(\Rightarrow\) \(a_i | mq + r\), and as \(mq\) is a multiple of the LCM including
+a\textsubscript{i}, \(a_i\) must divide \(mq\), and thus \(a_i | r\).
+
+Because \(a_i | r\), \(r = na_i\) and is thus a multiple of all \(a_i\), but the above statement that
+\(0 \leq r < m\) shows that \(m\) - the LCM by definition - is greater than \(r\). Therefore, \(r = 0\) and
+\(t = mq \Rightarrow m | t\).
+\end{document}
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