From 6bf4b90c90f15f4ab60833bddf5b5756d1a6b1f6 Mon Sep 17 00:00:00 2001 From: Elizabeth Alexander Hunt Date: Thu, 2 Jul 2026 11:55:17 -0700 Subject: Init --- Homework/math4310/alg_structures_assn_4.tex | 233 ++++++++++++++++++++++++++++ 1 file changed, 233 insertions(+) create mode 100644 Homework/math4310/alg_structures_assn_4.tex (limited to 'Homework/math4310/alg_structures_assn_4.tex') diff --git a/Homework/math4310/alg_structures_assn_4.tex b/Homework/math4310/alg_structures_assn_4.tex new file mode 100644 index 0000000..bf588dd --- /dev/null +++ b/Homework/math4310/alg_structures_assn_4.tex @@ -0,0 +1,233 @@ +% 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 -- cgit v1.3