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+% Created 2023-02-17 Fri 12:59
+% 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} \usepackage[a4paper,margin=1in,landscape]{geometry}
+\author{Lizzy Hunt}
+\date{\today}
+\title{Assignment Five}
+\hypersetup{
+ pdfauthor={Lizzy Hunt},
+ pdftitle={Assignment Five},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, 
+ pdflang={English}}
+\begin{document}
+
+\maketitle
+\setlength\parindent{0pt}
+
+\section{Section 3.2}
+\label{sec:org258154b}
+\subsection{Question One}
+\label{sec:orgf02bb85}
+\subsubsection{a}
+\label{sec:orgccb1c0c}
+\(a^2 - ab + ba - b^2\)
+\subsubsection{b}
+\label{sec:org9164b85}
+\(a^3 + ba^2 + 2a^2b + 2ab^2 + ab^2 + b^3\)
+\subsubsection{c}
+\label{sec:orgdaba485}
+(a) could be \(a^2 - b^2\)
+
+(b) could be \(a^3 + 3a^2b + 3ab^2 + b^3\)
+
+\subsection{Question Three}
+\label{sec:org4d02a16}
+\subsubsection{a}
+\label{sec:orgb743fca}
+\(\begin{smallmatrix}
+0 & 0 \\
+0 & 0 \\
+\end{smallmatrix}\), \(\begin{smallmatrix}
+1 & 0 \\
+0 & 1 \\
+\end{smallmatrix}\), \(\begin{smallmatrix}
+1 & 0 & 0 \\
+0 & 1 & 0 \\
+0 & 0 & 1 \\
+\end{smallmatrix}\), \(\begin{smallmatrix}
+0 & 0 & 0 \\
+0 & 0 & 0 \\
+0 & 0 & 0 \\
+\end{smallmatrix}\)
+\subsubsection{b}
+\label{sec:org8300d7e}
+\begin{verbatim}
+>>> set(filter(lambda y: all([y**2 % 12 == y for x in range(12)]), range(12)))
+{0, 1, 4, 9}
+\end{verbatim}
+
+\subsection{Question Seven}
+\label{sec:org9152891}
+S is closed under multiplication:
+\(i \in S, j \in S \Rightarrow i \cdot j = n1_R \cdot m1_R = (nm)1_R\).
+
+S is closed under subtraction:
+\(i \in S, j \in S \Rightarrow i - j = n1_R - m1_R = (n-m)1_R\)
+
+\subsection{Question Eight}
+\label{sec:org3ed31b0}
+Considering \(n,m \in T\) then \(n = xb, m = yb\) with \(x,y \in R\):
+
+\begin{enumerate}
+\item T is closed under multiplication: \(n \cdot m = xb \cdot yb = (x \cdot y)b\), which follows the rule.
+\item T is closed under subtraction: \(n - m = xb - yb = (x - y)b\), which also follows the rule.
+\end{enumerate}
+
+We also know \(T\) is not empty since it must have at least 0\textsubscript{R}.
+
+\subsection{Question Ten}
+\label{sec:orgc4efb36}
+\subsubsection{a}
+\label{sec:orgbc2ab9d}
+\(\bar{R} = {(0, 0), (1, 0), (2, 0)}\)
+
+\(\bar{S} = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4)}\)
+
+\subsubsection{b}
+\label{sec:org564675e}
+Considering \(n, m \in \bar{R}\), then \(n = (x, 0_S), n = (y, 0_S)\) with \(x,y \in R\).
+
+\begin{enumerate}
+\item \(\bar{R}\) is closed under multiplication: \(n \cdot m = (x, 0_S) \cdot (y, 0_S) = (x \cdot y, 0_S)\) and \(x \cdot y \in R\), so thus \(n \cdot m \in R \times S\).
+\item \(\bar{R}\) is closed under subtraction: \(n - m = (x, 0_S) - (y, 0_S) = (x - y, 0_S)\) and \(x - y \in R\), so thus \(n - m \in R \times S\).
+\end{enumerate}
+
+We also know that \(\bar{R}_0 = (0_R, 0_S) \in R \times S\) so \(\bar{R}\) is not empty.
+
+\subsubsection{c}
+\label{sec:org5905763}
+Considering \(n, m \in \bar{S}\), then \(n = (0_R, x), n = (0_R, y)\) with \(x,y \in S\).
+
+\begin{enumerate}
+\item \(\bar{S}\) is closed under multiplication: \(n \cdot m = (0_R, x) \cdot (0_R, y) = (x \cdot y, 0_S)\) and \(x \cdot y \in S\), so thus \(n \cdot m \in R \times S\).
+\item \(\bar{S}\) is closed under subtraction: \(n - m = (0_R, x) - (0_R, y) = (0_R, x - y)\) and \(x - y \in R\), so thus \(n - m \in R \times S\).
+\end{enumerate}
+
+We also know that \(\bar{S}_0 = (0_R, 0_S) \in R \times S\) so \(\bar{S}\) is not empty.
+
+\subsection{Question Thirteen}
+\label{sec:orgfc1192c}
+\subsubsection{a}
+\label{sec:orgdc9f5c9}
+Considering \(n, m \in S \cap T\), then \(n \in S\) and \(n \in T\), \(m \in S\) and \(m \in T\), therefore:
+
+\begin{enumerate}
+\item \(S \cap T\) is closed under multiplication as \(m \cdot n\) must also be in \(S \cap T\)
+\item \(S \cap T\) is closed under addition as \(m + n\) must also be in \(S \cap T\)
+\end{enumerate}
+
+Since \(S\) and \(T\) are both subrings of \(R\), \(0_R \in S \cap T\).
+
+\subsubsection{b}
+\label{sec:orgeb2e7d6}
+No, consider \(S\) being the integer multiples of 8 and \(T\) being the integer multiples of 3 being subrings of \(\mathds{Z}\) (proof of these being subrings is in
+Question Six of Section 3.1 in Assignment Four), then \(n \in S\) with \(n = 8\) and \(m \in T\) with \(m = 3\) then \(n + m = 11 \notin S \cup T\).
+
+\subsection{Question Fifteen - TODO}
+\label{sec:org675f8a8}
+
+\subsection{Question Twenty-One}
+\label{sec:org0c713ff}
+\subsubsection{a}
+\label{sec:org6158430}
+From \(ab = ac \Rightarrow ab - ac = 0_R \Rightarrow a(b - c) = 0_R\), we know \(b-c\) is equivalent to \(0_R\) since we're given that \(a\) is a non-zero element.
+
+\(b-c = 0_R \Rightarrow b = c\)
+
+\subsubsection{b}
+\label{sec:org4391edc}
+From \(ba = ca \Rightarrow ba - ca = 0_R \Rightarrow (b - c)(a) = 0_R\) we come to the same conclusion as (a)
+
+\(b-c = 0_R \Rightarrow b = c\)
+
+\section{Section 3.3}
+\label{sec:orgff5f7f5}
+\subsection{Question One}
+\label{sec:org08aff62}
+In the true nature of being a computer science student, automate something for 2 hours that you could've done in 20 minutes!
+\begin{verbatim}
+cartesian_prod = lambda zn, zm: list([(i, j) for i in range(zn) for j in range(zm)])
+mult_tup = lambda a, b, zn, zm: ((a[0] * b[0]) % zn, (a[1] * b[1]) % zm)
+add_tup = lambda a, b, zn, zm: ((a[0] + b[0]) % zn, (a[1] + b[1]) % zm)
+empty_table = lambda col, row, symbol: [[symbol] + [str(x) for x in col]] + [[str(x)] + [0 for i in range(len(col))] for x in row]
+
+def make_cartesian_congruence_table(zn, zm, op, mapping, symbol):
+  cp = cartesian_prod(zn, zm)
+  mapped_cp = [cp[mapping[i]] for i in range(len(cp))]
+  table = empty_table(mapped_cp, mapped_cp, symbol)
+  for i in range(len(cp)):
+    for j in range(len(cp)):
+      table[i+1][j+1] = str(op(mapped_cp[i], mapped_cp[j], zn, zm))
+  return table
+
+def make_normal_table(n, op, symbol):
+  table = empty_table(list(range(n)), list(range(n)), symbol)
+  for i in range(n):
+    for j in range(n):
+      table[i+1][j+1] = str(op(i, j) % n)
+  return table
+\end{verbatim}
+
+\subsubsection{Z\textsubscript{2} \texttimes{} Z\textsubscript{3} with bijection}
+\label{sec:orgcac27c3}
+\begin{verbatim}
+bijection = [0, 4, 2, 3, 1, 5]
+\end{verbatim}
+
+\begin{enumerate}
+\item Multiplication Tables
+\label{sec:org1f4aa9f}
+\begin{verbatim}
+make_cartesian_congruence_table(2, 3, mult_tup, bijection, '\odot')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllllll}
+\(\odot\) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
+(1, 1) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(0, 2) & (0, 0) & (0, 2) & (0, 1) & (0, 0) & (0, 2) & (0, 1)\\[0pt]
+(1, 0) & (0, 0) & (1, 0) & (0, 0) & (1, 0) & (0, 0) & (1, 0)\\[0pt]
+(0, 1) & (0, 0) & (0, 1) & (0, 2) & (0, 0) & (0, 1) & (0, 2)\\[0pt]
+(1, 2) & (0, 0) & (1, 2) & (0, 1) & (1, 0) & (0, 2) & (1, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\item Addition Tables
+\label{sec:org807c598}
+\begin{verbatim}
+make_cartesian_congruence_table(2, 3, add_tup, bijection, '\oplus')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllllll}
+\(\oplus\) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(0, 0) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2)\\[0pt]
+(1, 1) & (1, 1) & (0, 2) & (1, 0) & (0, 1) & (1, 2) & (0, 0)\\[0pt]
+(0, 2) & (0, 2) & (1, 0) & (0, 1) & (1, 2) & (0, 0) & (1, 1)\\[0pt]
+(1, 0) & (1, 0) & (0, 1) & (1, 2) & (0, 0) & (1, 1) & (0, 2)\\[0pt]
+(0, 1) & (0, 1) & (1, 2) & (0, 0) & (1, 1) & (0, 2) & (1, 0)\\[0pt]
+(1, 2) & (1, 2) & (0, 0) & (1, 1) & (0, 2) & (1, 0) & (0, 1)\\[0pt]
+\end{tabular}
+\end{center}
+\end{enumerate}
+
+\subsubsection{Z\textsubscript{6}}
+\label{sec:orgbe8adf1}
+\begin{enumerate}
+\item Multiplication Tables
+\label{sec:orga98b12b}
+\begin{verbatim}
+make_normal_table(6, lambda a, b: a * b, '\odot')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{rrrrrrr}
+\(\odot\) & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+0 & 0 & 0 & 0 & 0 & 0 & 0\\[0pt]
+1 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+2 & 0 & 2 & 4 & 0 & 2 & 4\\[0pt]
+3 & 0 & 3 & 0 & 3 & 0 & 3\\[0pt]
+4 & 0 & 4 & 2 & 0 & 4 & 2\\[0pt]
+5 & 0 & 5 & 4 & 3 & 2 & 1\\[0pt]
+\end{tabular}
+\end{center}
+
+\item Addition Tables
+\label{sec:orgfcf7bde}
+\begin{verbatim}
+make_normal_table(6, lambda a, b: a + b, '\oplus')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{rrrrrrr}
+\(\oplus\) & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+0 & 0 & 1 & 2 & 3 & 4 & 5\\[0pt]
+1 & 1 & 2 & 3 & 4 & 5 & 0\\[0pt]
+2 & 2 & 3 & 4 & 5 & 0 & 1\\[0pt]
+3 & 3 & 4 & 5 & 0 & 1 & 2\\[0pt]
+4 & 4 & 5 & 0 & 1 & 2 & 3\\[0pt]
+5 & 5 & 0 & 1 & 2 & 3 & 4\\[0pt]
+\end{tabular}
+\end{center}
+\end{enumerate}
+
+\subsection{Question Two}
+\label{sec:org9540aed}
+\begin{verbatim}
+bijection = [0, 3, 2, 1]
+make_cartesian_congruence_table(2, 2, mult_tup, bijection, '\odot')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllll}
+\(\odot\) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0)\\[0pt]
+(1, 1) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(1, 0) & (0, 0) & (1, 0) & (1, 0) & (0, 0)\\[0pt]
+(0, 1) & (0, 0) & (0, 1) & (0, 0) & (0, 1)\\[0pt]
+\end{tabular}
+\end{center}
+
+\begin{verbatim}
+bijection = [0, 3, 2, 1]
+make_cartesian_congruence_table(2, 2, add_tup, bijection, '\oplus')
+\end{verbatim}
+
+\begin{center}
+\begin{tabular}{lllll}
+\(\oplus\) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(0, 0) & (0, 0) & (1, 1) & (1, 0) & (0, 1)\\[0pt]
+(1, 1) & (1, 1) & (0, 0) & (0, 1) & (1, 0)\\[0pt]
+(1, 0) & (1, 0) & (0, 1) & (0, 0) & (1, 1)\\[0pt]
+(0, 1) & (0, 1) & (1, 0) & (1, 1) & (0, 0)\\[0pt]
+\end{tabular}
+\end{center}
+
+\subsection{Question Three}
+\label{sec:org421e770}
+\begin{enumerate}
+\item \(f\) is injective, since \(f(a) = f(b) \Rightarrow (a,a) = (b,b) \Rightarrow a = b\)
+\item \(f\) is surjective since every range element in \(R^*\), \((a,a)\) is mapped to \(a \in R\) by definition
+\item \(f(a) + f(b) = (a, a) + (b, b) = (a + b, a + b) = f(a + b)\) and \(f(a)f(b) = (a, a) \cdot (b, b) = (ab, ab) = f(ab)\)
+\end{enumerate}
+
+\subsection{Question Four}
+\label{sec:org1f8479a}
+\(f(1)f(3) = (2)(6) \equiv_{10} 2\) but \(f(3) = 6\) which doesn't hold the properties of homomorphism.
+
+\subsection{Question Five}
+\label{sec:orgeb656c3}
+Consider the given function:
+\begin{enumerate}
+\item \(f\) is injective, since \(f(a) = f(b) \Rightarrow\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) \(\Rightarrow a = b\)
+\item \(f\) is surjective, since every range element \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) has an element in \(\mathds{R}\), \(a\), such that \(f(a) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) by definition
+\item \(f(a) + f(b) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) + \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & (a + b) \end{smallmatrix}\) = \(f(a + b)\),
+and \(f(a) \cdot f(b) =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & 0 \\ 0 & b \end{smallmatrix}\) = \(\begin{smallmatrix} 0 & 0 \\ 0 & (a \cdot b) \end{smallmatrix}\) = \(f(a \cdot b)\),
+\end{enumerate}
+
+\subsection{Question Nine - TODO}
+\label{sec:org2f29e8e}
+
+\subsection{Question Eleven}
+\label{sec:org3e455ed}
+\subsubsection{b}
+\label{sec:org24734c1}
+\begin{verbatim}
+>>> f = lambda x: 3 * x
+>>> list(filter(lambda x: f(x * x) != (f(x) * f(x)), range(0, 20, 2)))
+[2, 4, 6, 8, 10, 12, 14, 16, 18]
+\end{verbatim}
+\subsubsection{d}
+\label{sec:orgc2ba394}
+\begin{verbatim}
+>>> k = lambda x: 0 if x == 0 else (x ** -1)
+>>> list(filter(lambda x: k(x * x) != (k(x) * k(x)), range(0, 20)))
+[5, 10, 13, 19]
+\end{verbatim}
+\subsection{Question Twelve}
+\label{sec:org95ce95e}
+\subsubsection{c}
+\label{sec:org77d2e96}
+Not a homomorphism. Just from reducing \(f(x + x)\) we find \(f(x + x) \neq f(x) + f(x)\):
+
+\begin{equation*}
+f(x + x) = \dfrac{1}{(x + x)^2 + 1} = \dfrac{1}{4x^2 + 1}
+\end{equation*}
+
+\begin{equation*}
+f(x) + f(x) = \dfrac{1}{x^2 + 1} + \dfrac{1}{x^2 + 1} = \dfrac{2}{x^2 + 1}
+\end{equation*}
+
+Thus \(f(x + x) \neq f(x) + f(x)\).
+\subsubsection{d}
+\label{sec:org1c2b201}
+\begin{verbatim}
+>>> import numpy as np
+>>> h = lambda x: [[-x, 0], [x, 0]]
+>>> list(filter(lambda x: not np.array_equiv(h(x * x), np.dot(h(x), h(x))), range(0, 20))
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
+\end{verbatim}
+\subsection{Question Thirteen}
+\label{sec:org57e2c18}
+\subsubsection{a}
+\label{sec:org7515001}
+If \(r \in R\) then \((r, 0_S) \in R \times S\), and \(f((r, 0_S)) = r\), so every range element has a domain element mapping to it.
+
+For some \(a = (r, m) \in R \times S\) and \(b = (t, v) \in R \times S\) then \(f((r, m)) + f((t, v)) = r + t = f((r + t, v + m))\), and
+\(f((r, m)) \cdot f((t, v)) = r \cdot t = f((r \cdot t, v \cdot m))\).
+
+\subsubsection{b}
+\label{sec:orgfb2a3ab}
+If \(s \in S\) then \((0_R, s) \in R \times S\), and \(f((0_R, s)) = s\), so every range element has a domain element mapping to it.
+
+For some \(a = (r, m) \in R \times S\) and \(b = (t, v) \in R \times S\) then \(f((r, m)) + f((t, v)) = m + v = f((r + t, m + v))\), and
+\(f((r, m)) \cdot f((t, v)) = m \cdot v = f((r \cdot t, m \cdot v))\).
+
+\subsection{Question Fifteen}
+\label{sec:org4046d12}
+Consider \(f: Z_4 \rightarrow Z_4 \ni f(x) = 0\), then 2 is a zero divisor, but 0 is not by definition.
+
+\subsection{Question Twenty One}
+\label{sec:org48d8cc7}
+\subsubsection{Lemma}
+\label{sec:org3598e64}
+We assume a multiplicative identity \(x\) exists in \(\mathds{Z}^{*}\):
+
+\(b \odot x = b \Rightarrow b + x - bx = b \Rightarrow x - bx = 0 \Rightarrow x(1-b) = 0\) and \(1-b \neq 0 \forall b \in \mathds{Z}^{*}\) so \(x = 0\).
+
+which is verified by:
+
+\(0 \odot b = 0 + b - 0 \cdot b = b\) and \(b \odot 0 = b + 0 - b \cdot 0 = b\).
+
+\subsubsection{Proof}
+\label{sec:orgee74248}
+
+Consider \(f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*}\) to be the isomorphism we so desire, then by Theorem 3.10,
+f(1 \(\in\) \mathds{Z}) = 0 \(\in\) \mathds{Z}\textsuperscript{*}.
+
+Therefore, f(2) = f(1 + 1) = f(1) \(\oplus\) f(1) = -1, f(3) = f(2 + 1) = -1 \(\oplus\) f(1) = -2.
+
+\(f : \mathds{Z}^{} \rightarrow \mathds{Z}^{*} \ni f(x) = 1 - x\) seems to fit the bill nicely.
+
+\begin{enumerate}
+\item \(f\) is a bijection since we have an inverse \(x = 1 - f(x)\)
+\item \(f(a \oplus b) = 1 - (a \oplus b) = 1 - (a + b - 1) = (1 - a) + (1 - b) = f(a) + f(b)\) and
+\(f(a \odot b) = 1 - (a \odot b) = 1 - (a + b - ab) = (1-a) \cdot (1-b) = f(a) \cdot f(b)\)
+\end{enumerate}
+
+\subsection{Question Twenty Four}
+\label{sec:org07a0b7a}
+\subsubsection{a}
+\label{sec:org897ff21}
+\begin{enumerate}
+\item By the usual coordinate addition we know that \(a, b \in R \Rightarrow a + b \in R\) and is associative, commutative, and the additive identity is \((0, 0)\).
+\item \(R\) is closed under multiplication: \(a, b \in R \Rightarrow a = (c, d) \wedge b = (e, f) \Rightarrow a \cdot b = (ce, de)\) and since \(c, d, e, f \in \mathds{R}\) then \((ce, de) \in \mathds{R} \times \mathds{R}\) by definition.
+\item Multiplication is associative: \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a \cdot b) \cdot c = (df, ef) \cdot (h, i) = (dfh, efh)\) and \(a \cdot (b \cdot c) = (d, e) \cdot (fh, gh) = (dfh, efh)\)
+\item Multiplication is distributive: \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow a(b + c) = (e, f) \cdot (f + h, g + i) = (ef + eh, ff + fh) = (ef, ff) + (eh, fh) = a \cdot b + a \cdot c\)
+and \(a, b, c \in R \Rightarrow a = (d, e) \wedge b = (f, g) \wedge c = (h, i) \Rightarrow (a + b) \cdot c = (d + f, e + g) \cdot (h, i) = (dh + fh, eh + gh) = (dh, eh) + (fh, gh) = a \cdot c + b \cdot c\)
+\end{enumerate}
+\subsubsection{b}
+\label{sec:org92e2744}
+Consider the function \(f: R \rightarrow M(\mathds{R})\) is an isomorphism, such that \(f((a, b)) =\) \(\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix}\), then:
+\begin{enumerate}
+\item \(f\) is surjective since every element in the range has a domain element \(\begin{smallmatrix} a & 0 \\ b & 0 \end{smallmatrix} = y \ni f((a, b)) = y\)
+\item \(f\) is injective since \(f((c, d) = x) = f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) \(\Rightarrow c = e \wedge d = f \Rightarrow x = y\)
+\item \(f((c, d) = x) + f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) + \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} c + e & 0 \\ d + f & 0 \end{smallmatrix}\)
+\(= f(x + y)\), and \(f((c, d) = x) \cdot f((e, f) = y) \Rightarrow\) \(\begin{smallmatrix} c & 0 \\ d & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} e & 0 \\ f & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} ce & 0 \\ de & 0 \end{smallmatrix}\)
+\(= f(x \cdot y)\)
+\end{enumerate}
+\end{document}
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