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diff --git a/Homework/math4310/midterm_1.tex b/Homework/math4310/midterm_1.tex new file mode 100644 index 0000000..e1a4a53 --- /dev/null +++ b/Homework/math4310/midterm_1.tex @@ -0,0 +1,151 @@ +% Created 2023-02-20 Mon 22:30 +% 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{Midterm One} +\hypersetup{ + pdfauthor={Lizzy Hunt}, + pdftitle={Midterm One}, + pdfkeywords={}, + pdfsubject={}, + pdfcreator={Emacs 28.2 (Org mode 9.6.1)}, + pdflang={English}} +\begin{document} + +\maketitle +\setlength\parindent{0pt} + +\section{Question One} +\label{sec:org6315799} +\begin{verbatim} +In [1]: sols = lambda n: set(filter(lambda x: (x**2 + x) % n == 0, range(n))) + +In [2]: sols(5) +Out[2]: {0, 4} + +In [3]: sols(6) +Out[3]: {0, 2, 3, 5} +\end{verbatim} + +\section{Question Two - TODO} +\label{sec:org6c24b4c} + + +\section{Question Three} +\label{sec:org76bcd7b} +From the multiplication table we can see that \(e\) is the identity element \(1_R\), the zero +element \(0_R\) is \(z\). From the addition table we find that every element is its own +additive inverse. + +\subsection{a} +\label{sec:org983fdae} +The units in this ring are \({e, g}\), and the zero divisors are \({a, b, c, f}\). + +\subsection{b} +\label{sec:org9072029} +\(3bd = 3(bd) = 3z = b\) + +\(2ac = 2(ac) = 2b = b + b = z\) + +\(g^{-1}f^3 = (g)(fff) (gf)(ff) = (d)(f) = d\) + +Thus, \(3bd - 2ac + g^{-1}f^3 = b - z + d = b + d = f\). + +\section{Question Four} +\label{sec:orge571a7e} +To get a hunch, + +\begin{verbatim} +In [1]: import numpy as np + +In [2]: def find_counter(matrix_generator, in_ring, search_space=range(-100, 100)): + ...: for a1 in search_space: + ...: n = matrix_generator(a1) + ...: for a2 in search_space: + ...: m = matrix_generator(a2) + ...: dot = np.dot(n, m) + ...: add = np.add(n, m) + ...: + ...: if not in_ring(dot): + ...: return (n, m, dot, '*') + ...: if not in_ring(add): + ...: return (n, m, add, '+') + ...: + ...: return None + ...: + +In [3]: find_counter(lambda a: [[0, a], [0, -a]], lambda m: m[0][1] == -m[1][1] and m[0][0] == 0 and m[1][0] == 0) + +In [4]: In [31]: find_counter(lambda a: [[0, a], [a, 0]], lambda m: m[0][1] == m[1][0] and m[1][1] == 0 and m[0][0] == 0) +Out[4]: +([[0, -100], [-100, 0]], + [[0, -100], [-100, 0]], + array([[10000, 0], + [ 0, 10000]]), + '*') +\end{verbatim} +\subsection{a} +\label{sec:orge0ecef4} +Using Theorem 3.2, this is a ring: + +\begin{enumerate} +\item S\textsubscript{1} is closed under addition: \(x, y \in S_1 \Rightarrow x + y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x + y = \begin{smallmatrix} 0 & a + b \\ 0 & (-a) + (-b) \end{smallmatrix} = \begin{smallmatrix} 0 & a + b \\ 0 & -(a + b) \end{smallmatrix}\) which satisfies the rule +\item S\textsubscript{1} is closed under multiplication: \(x, y \in S_1 \Rightarrow x \cdot y =\) \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & b \\ 0 & -b \end{smallmatrix}\) \(\Rightarrow\) \(x \cdot y =\) \(\begin{smallmatrix} 0 + (a)(0) & 0(b) + a(-b) \\ 0 + (-a)(0) & 0b + (-a)(-b) \end{smallmatrix}\) \(=\) \(\begin{smallmatrix} 0 & -ab \\ 0 & ab \end{smallmatrix}\) which satisfies the rule +\item \(0_R_{} = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) +\item \(a \in S_1 \Rightarrow a + x = 0_R \wedge x \in S_1\) since \(\begin{smallmatrix} 0 & a \\ 0 & -a \end{smallmatrix}\) \(+\) \(x = 0_{S_1} =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) \(\Rightarrow x =\) \(\begin{smallmatrix} 0 & -a \\ 0 & a \end{smallmatrix}\) \(\in S_1\) +\end{enumerate} + +\subsection{b} +\label{sec:org1a8af9b} +\(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) \(\cdot\) \(\begin{smallmatrix} 0 & -100 \\ -100 & 0 \end{smallmatrix}\) = \(\begin{smallmatrix} 10000 & 0 \\ 0 & 10000 \end{smallmatrix}\) is not in the ring. + +\section{Question Five} +\label{sec:org85339f8} +No, it is not a homomorphism from the definition in 3.3. + +Consider \(n =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix}\) and \(m =\) \(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\). + +Then \(nm =\) \(\begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix}\) and thus \(Trace(n) \cdot Trace(m) = 1 \wedge Trace(nm) = 0 \Rightarrow f(n)f(m) \neq f(nm)\). + +\section{Question Six} +\label{sec:org405929b} +No, I don't believe \(f\) to be an isomorphism. +\subsection{Lemma} +\label{sec:org323f970} +We need to show that \(x \in Q \wedge x \sqrt{2} = y \in Q \Rightarrow x = 0\). + +If \(x \neq 0\) then \(y\) can be represented as \(\frac{a}{b}\) with \(a, b \neq 0 \wedge a,b \in \mathds{Z}\) and \((a, b) = 1\). + +So, \(x \sqrt{2} = \frac{a}{b} \Rightarrow x \sqrt{2} b = a \Rightarrow x^2(2b^2) = a^2\) and thus \(2 | aa\), and by applying Corollary 1.6, \(2 | a\). + +By substituting \(2c = a\) (a is divisible by 2), \(x^2(2b^2) = 4c^2 \Rightarrow x^2 b^2 = 2c^2\) which implies \(2 | (x^2 b^2) \Rightarrow 2 | x\) or \(2 | b\) (again by Corollary 1.6). + +But 2 cannot divide \(b\) as well as \(a\), since \((a, b) = 1\) and \(b \neq 0\). + +Thus \(2 | x\). So \(x = 2d \Rightarrow 4d^2(2b^2) = a^2 \Rightarrow 4 | a^2\), so \(4e = a\), and by similar logic for \(b\) above, \(4 | x\). +Then, subsituting \(x = 4f \Rightarrow 16f^2(2b^2) = a^2 \Rightarrow 16 | a^2\). Again, \(16 | x\). + +In fact, all range elements \(R\) of the function on the natural numbers \$f \(\ni\) \$ \(f(1) = 2, f(n) = f(n-1)^2 { n > 1 }\) must divide \(x\). + +Thus, x must be sup(\(R\)) or 0. Obviously \(R\) is unbounded on the right, so it follows \(x = 0\). + +\subsection{Proof} +\label{sec:orge723534} +\(a, b \in Q(\sqrt{2}) \Rightarrow f(a) = n + m \sqrt{2} \wedge b = x + y \sqrt{2} \Rightarrow f(a) = n - m \sqrt{2} \vee f(b) = x - y \sqrt{2}\). + +\(f\) is not injective; if we assume \(f(a) = f(b) \Rightarrow a = b\) then \(f(a) - f(b) = 0 \Rightarrow n - m \sqrt{2} - x - y \sqrt{2} = 0 \Rightarrow 0 = (n-x) - (m + y)\sqrt{2} \Rightarrow n-x = (m+y)\sqrt{2}\) and, as \(n-x \in Q\) since +Q is a ring and by definition, \((m + y)\sqrt{2} \in Q\) implies \((m + y) = 0\) by the lemma, thus \(m = -y\). But for \(a = b\), \(m = y\), a contradiction. +\end{document}
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