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+% Created 2023-02-08 Wed 09:17
+% 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{Lizzy Hunt}
+\date{\today}
+\title{Assignment Four}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Four},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 3.1}
+\label{sec:org5ad4880}
+\subsection{Question One}
+\label{sec:org4dd5421}
+\subsubsection{a}
+\label{sec:orgfe8dc40}
+\(a \in R, b \in R \Rightarrow a + b \in R\)
+\subsubsection{b}
+\label{sec:orgccbb819}
+\(a \in R \Rightarow x \in R \ni a + x = 0_R\)
+\subsection{Question Three}
+\label{sec:orgc539ab7}
+\begin{enumerate}
+\item All operations are closed since only the elements in \({0, e, a, b}\) appear in the tables.
+\item From the second row and columns in the multiplication table we see that \(e\) is the multiplicative identity.
+\item From the first row and columns in the addition table we see that \(0\) is the zero element.
+\item In this field, each element is its own additive inverse.
+\item There is commutativity as the transpose of each table is identical to the original (symmetry along the diagonals).
+\end{enumerate}
+\subsection{Question Six}
+\label{sec:orge1c7f2c}
+\subsubsection{a}
+\label{sec:orgb356ff9}
+Since our addition and multiplication operators are the same in \(\mathds{Z}\), we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of 3 are also multiples of 3: \(3n + 3m = 3(n + m)\), so this set is closed under addition.
+
+Products of multiples of 3 are also multiples of 3: \((3n)(3m) = (3)(3nm)\), so this set is closed under multiplication.
+
+The additive inverse exists in the set for every element: \(3n + x = 0 \Rightarrow x = 3 \cdot (-n)\).
+
+\(0\) is the zero element and is a multiple of 3 since \(3 \cdot 0 = 0\).
+
+Therefore \({x : x = 3n \ni n \in \mathds{Z}}\) is a subring of \(\mathds{Z}\)
+
+\subsubsection{b}
+\label{sec:orga2b6add}
+Since our addition and multiplication operators are the same in \(\mathds{Z}\), we have
+associativity, commutativity, and distributivity.
+
+Sums of multiples of \(k\) are also multiples of \(k\): \(kn + km = k(n + m)\), so this set is closed under addition.
+
+Products of multiples of \(k\) are also multiples of \(k\): \((kn)(km) = (k)(knm)\), so this set is closed under multiplication.
+
+The additive inverse exists exists in the set for every element: \(kn + x = 0 \Rightarrow k = k \cdot (-n)\).
+
+\(0\) is the zero element and is a multiple of k since \(k \cdot 0 = 0\).
+
+Therefore \({x : x = kn \ni n \in \mathds{Z}}\) is a subring of \(\mathds{Z}\)
+
+\subsection{Question Nine}
+\label{sec:org4f6749e}
+\subsubsection{a}
+\label{sec:orgc523c2e}
+\({(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}\)
+\subsubsection{b}
+\label{sec:org970c88d}
+Since our addition and multiplication operators are the same as in \(R\), then we have
+associativity, commutativity, and distributivity.
+
+Sums of elements in \(R*\) are also in \(R*\) since any element \((r,r) \in R*, (j, j) \in R* = (r + j, r + j) \in R*\).
+
+Products of elements in \(R*\) are also in \(R*\) since any element \((r, r) \in R*, (j, j) \in R* = (rj, rj) \in R*\).
+
+The additive inverse exists in the set for every element: \((r, r) + x = 0 \Rightarrow x = (-r, -r) \in R*\).
+
+\((0, 0) \in R*\) is the zero element.
+
+Therefore \({(r, r) : (r, r) \in R*}\) is a subring of \(R*\).
+
+\subsection{Question Ten}
+\label{sec:orgfdaf968}
+Consider \(a \in S \ni a = (10, -10)\) and \(b \in S \ni b = (10, -10)\), then \(ab \notin S\) since
+\(ab = (100, 100)\) which does not follow the rule that \(100 + 100 = 0\).
+\subsection{Question Eleven}
+\label{sec:org3b794ec}
+\subsubsection{a}
+\label{sec:orgef5a91b}
+Addition is closed since
+\(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\) + \(\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+(a + c) & (a + c) \\
+(b + d) & (b + d)
+\end{smallmatrix}\) which of the form of the given rule.
+
+It is also closed under multiplication since
+\(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix}
+c & c \\
+d & d 
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+(ac + ad) & (ac + ad) \\
+(bc + bd) & (bc + bd)
+\end{smallmatrix}\) which is also of the form of the given rule.
+
+The zero element is the zero matrix \(\begin{smallmatrix}
+0 & 0 \\
+0 & 0
+\end{smallmatrix}\), trivially.
+
+Associativity and commutivity (of addition) come from these operations existing for 2x2 matrices in \(M(\mathds{R})\).
+
+\subsubsection{b}
+\label{sec:orgb4b3acf}
+\(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+(a(1) + a(0)) & (a(1) + a(0)) \\
+(b(1) + b(0)) & (b(1) + b(0))
+\end{smallmatrix}\)
+which is equivalent to \(\begin{smallmatrix}
+a & a \\
+b & b 
+\end{smallmatrix}\)
+
+\subsubsection{c}
+\label{sec:orga515fb6}
+\(\begin{smallmatrix}
+1 & 1 \\
+0 & 0 
+\end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix}
+1 & 1 \\
+2 & 2
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+((1)(1) + (1)(2)) & ((1)(1) + (1)(2)) \\
+((0)(1) + (0)(2)) & ((0)(1) + (0)(2))
+\end{smallmatrix}\) = \(\begin{smallmatrix}
+3 & 3 \\
+0 & 0
+\end{smallmatrix}\).
+\subsection{Question Twelve}
+\label{sec:orgb1926b2}
+\(\mathds{Z}[i]\) is closed under addition: \((a + bi) + (c + di) = (a + c) + (b + d)i\) and since
+\(\mathds{Z}\) is a ring itself, \((a + c) + (b + d)i\) is also in \(\mathds{Z}[i]\).
+
+\(\mathds{Z}[i]\) is closed under multiplication: \((a + bi) \cdot (c + di) = ac + adi + cbi + bdi^2 = (ac - bd) + (ad + bd)i\)
+by similar logic.
+
+The additive inverse always exists in the set: \((a + bi) + x = 0 \Rightarrow x = -a - bi\).
+
+Finally, the zero element is trivially \(0 + 0i\) since \((a + bi) + (0 + 0i) = a + bi\).
+
+\subsection{Question Fourteen}
+\label{sec:org31f60fc}
+The zero element is given as 2.
+
+\(S\) is closed under addition. For example given some \(f, g \in S\) then \((f + g)(x) = h\) and \(h\) will still satisfy the rule that
+\(h(2) = 0\) since \((f + g)(2) = f(2) + g(2) = 0 + 0\), and addition was already closed in the domain given that question 8 is a ring
+itself.
+
+Similarly, \(S\) is closed under multiplication. \((f \cdot g)(x) = h\) and h will still satisfy the rule as \(h(2) = f(2) \cdot g(2) = 0 \cdot 0\).
+
+The zero element is \(f \ni f(x) = 0\).
+
+Finally, the additive inverse \(g\) exists in the set for each \(f\) since \(f + g = 0\) implies that \(g\) is just \(-f\) (the rule still
+stands here).
+
+\subsection{Question Fifteen}
+\label{sec:org04d3ba1}
+\subsubsection{b}
+\label{sec:orgae94898}
+\begin{center}
+\begin{tabular}{lllllll}
+\(\odot\) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
+(0, 1) & (0, 0) & (0, 1) & (0, 2) & (0, 0) & (0, 1) & (0, 2)\\[0pt]
+(0, 2) & (0, 0) & (0, 2) & (0, 1) & (0, 0) & (0, 2) & (0, 1)\\[0pt]
+(1, 0) & (0, 0) & (0, 0) & (0, 0) & (1, 0) & (1, 0) & (1, 0)\\[0pt]
+(1, 1) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(1, 2) & (0, 0) & (0, 2) & (0, 1) & (1, 0) & (1, 2) & (1, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{center}
+\begin{tabular}{lllllll}
+\(\oplus\) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\[0pt]
+(0, 1) & (0, 1) & (0, 2) & (0, 0) & (1, 1) & (1, 2) & (1, 0)\\[0pt]
+(0, 2) & (0, 2) & (0, 0) & (0, 1) & (1, 2) & (1, 0) & (1, 1)\\[0pt]
+(1, 0) & (1, 0) & (1, 1) & (1, 2) & (2, 0) & (2, 1) & (2, 2)\\[0pt]
+(1, 1) & (1, 1) & (1, 2) & (1, 0) & (2, 1) & (2, 2) & (2, 0)\\[0pt]
+(1, 2) & (1, 2) & (1, 0) & (1, 1) & (2, 2) & (2, 0) & (2, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsection{Question Eighteen}
+\label{sec:orga487be8}
+No, by definition, as the distributive axiom is violated for this to be a ring.
+
+For example, \(1(1 + 1) = (1)(1) + (1)(1) = 2\) but \(1(1 + 1) = 1\).
+\end{document}
+\ No newline at end of file