mirror of
https://github.com/Rushilwiz/math381.git
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265 lines
8.4 KiB
TeX
265 lines
8.4 KiB
TeX
\documentclass{article}
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\usepackage{fancyhdr}
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\usepackage{extramarks}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{amsthm}
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\usepackage{amsfonts}
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\usepackage{tikz}
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\usepackage[plain]{algorithm}
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\usepackage{algpseudocode}
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\usepackage[shortlabels]{enumitem}
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\usepackage{mathtools}
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\usetikzlibrary{automata,positioning}
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%
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% Basic Document Settings
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%
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\topmargin=-0.45in
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\evensidemargin=0in
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\oddsidemargin=0in
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\textwidth=6.5in
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\textheight=9.0in
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\headsep=0.25in
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\linespread{1.1}
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\pagestyle{fancy}
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\lhead{\hmwkAuthorName}
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\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
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\rhead{\firstxmark}
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\lfoot{\lastxmark}
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\cfoot{\thepage}
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\renewcommand\headrulewidth{0.4pt}
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\renewcommand\footrulewidth{0.4pt}
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\setlength\parindent{0pt}
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%
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% Create Problem Sections
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%
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\newcommand{\enterProblemHeader}[1]{
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\nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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}
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\newcommand{\exitProblemHeader}[1]{
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\nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
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\stepcounter{#1}
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\nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
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}
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\setcounter{secnumdepth}{0}
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\newcounter{partCounter}
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\newcounter{homeworkProblemCounter}
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\setcounter{homeworkProblemCounter}{1}
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\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
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%
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% Homework Problem Environment
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%
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% This environment takes an optional argument. When given, it will adjust the
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% problem counter. This is useful for when the problems given for your
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% assignment aren't sequential. See the last 3 problems of this template for an
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% example.
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%
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\newenvironment{homeworkProblem}[1][-1]{
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\ifnum#1>0
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\setcounter{homeworkProblemCounter}{#1}
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\fi
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\section{Problem \arabic{homeworkProblemCounter}}
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\setcounter{partCounter}{1}
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\enterProblemHeader{homeworkProblemCounter}
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}{
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\exitProblemHeader{homeworkProblemCounter}
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}
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\newcommand{\hmwkTitle}{Homework 1}
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\newcommand{\hmwkDueDate}{September 7, 2023}
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\newcommand{\hmwkClass}{Discrete Math}
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\newcommand{\hmwkClassTime}{Section 003}
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\newcommand{\hmwkClassInstructor}{Reese Lance}
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\newcommand{\hmwkAuthorName}{\textbf{Rushil Umaretiya}}
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%
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% Title Page
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%
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\title{
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\vspace{2in}
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\textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
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\normalsize\vspace{0.1in}\small{Tuesday/Thursday 11:00-12:15, Phillips 383}\\
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\vspace{0.1in}\large{\textit{\hmwkClassInstructor\ - \hmwkClassTime}}
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\vspace{3in}
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}
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\author{\hmwkAuthorName\\\small{rumareti@unc.edu}}
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\date{}
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\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
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%
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% Various Helper Commands
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%
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% Useful for algorithms
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\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
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% For derivatives
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\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
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% For partial derivatives
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\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
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% Integral dx
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\newcommand{\dx}{\mathrm{d}x}
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% Alias for the Solution section header
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\newcommand{\solution}{\textbf{\large Solution}}
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\newcommand{\unit}[1]{\section{Unit #1}}
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\newcommand{\problem}[1]{\textbf{\##1}}
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\newcommand{\prob}[1]{\problem{#1}}
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% Probability commands: Expectation, Variance, Covariance, Bias
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\newcommand{\E}{\mathrm{E}}
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\newcommand{\Var}{\mathrm{Var}}
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\newcommand{\Cov}{\mathrm{Cov}}
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\newcommand{\Bias}{\mathrm{Bias}}
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\begin{document}
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\maketitle
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\pagebreak
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\unit{1.1}
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\problem{2}
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\begin{enumerate}[a)]
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\item Not declarative, a command.
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\item Not declarative, a question.
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\item Is a proposition, not true: there are black flies in Maine.
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\item Not declarative, truth value can change based on \emph{x}.
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\item Is a proposition, not true; the moon is not made of cheese.
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\item Not declarative, truth value can change based on \emph{n}.
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\end{enumerate}
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\problem{4}
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\begin{enumerate}[a)]
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\item Janice does not have more Facebook friends than Juan.
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\item Quincy is not smarter than Venkat.
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\item Zelda does not drive more miles to school than Paola.
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\item Briana does not sleep longer than Gloria.
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\end{enumerate}
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\problem{10}
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\\Let \(p\) and \(q\) be the propositions:
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\begin{itemize}
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\item[\(p\):]I bought a lottery ticket this week.
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\item[\(q\):] I won the million dollar jackpot.
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\end{itemize}
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Express each of these propositions as an English sentence.
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\begin{enumerate}[a)]
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\item I did not buy a lottery ticket this week.
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\item I bought a lottery ticket this week or I won the million dollar jackpot.
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\item If I bought a lottery ticket this week, then I won the million dollar jackpot.
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\item I bought a lottery ticket this week and I won the million dollar jackpot.
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\item I bought a lottery ticket this week if and only if I won the million dollar jackpot.
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\item If I did not buy a lottery ticket this week, then I didn't win the million dollar jackpot.
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\item I did not buy a lottery ticket this week and I did not win the million dollar jackpot.
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\item I did not buy a lottery ticket this week, or I did buy a lottery ticket this week and won the million dollar jackpot.
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\end{enumerate}
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\problem{18}
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\begin{enumerate}[a)]
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\item Both equations are true, therefore True.
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\item One equation is false and one is true, therefore False.
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\item Both propositions are false, therefore True.
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\item One equation is false and one is true, therefore False.
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\end{enumerate}
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\pagebreak
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\problem{20}
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\begin{enumerate}[a)]
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\item Both conditions are false, therefore the statement is true.
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\item Both conditions are false, therefore the statement is true.
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\item Since \(p\) is true and \(q\) is false, the statement is false.
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\item Both conditions are true, therefore the statement is true.
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\end{enumerate}
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\problem{22}
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\begin{enumerate}[a)]
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\item Inclusive or, you need proficiency in either language or both.
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\item Exclusive or, you can have either soup or salad, but not both.
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\item Inclusive or, you need either form of identification or both.
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\item Exclusive or, luckily you cannot both publish and perish.
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\end{enumerate}
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\pagebreak
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\unit{1.3}
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\problem{4b}
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\begin{displaymath}
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\begin{array}{|c c c|c|c|}
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p & q & r & (p \wedge q)\wedge r & p \wedge(q \wedge r) \\
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\hline
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T & T & T & T & T \\
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T & T & F & F & F \\
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T & F & T & F & F \\
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T & F & F & F & F \\
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F & T & T & F & F \\
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F & T & F & F & F \\
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F & F & T & F & F \\
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F & F & F & F & F
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\end{array}
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\end{displaymath}
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\\
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Since the columns are identical the law is true.
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\problem{6}\\
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Use a truth table to verify the first De Morgan law
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\begin{displaymath}
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\neg(p \wedge q) \equiv \neg p \vee \neg q
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\end{displaymath}
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\begin{displaymath}
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\begin{array}{|c c|c|c|c|c|c|}
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p & q & \neg p & \neg q & p \wedge q & \neg(p \wedge q) & \neg p \vee \neg q \\
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\hline
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T & T & F & F & T & F & F \\
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T & F & F & T & F & T & T \\
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F & T & T & F & F & T & T \\
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F & F & T & T & F & T & T \\
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\end{array}
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\end{displaymath}
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\problem{8}\\
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Use De Morgan’s laws to find the negation of each of the
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following statements.\\
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\textbf{a)}
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Kwame will take a job in industry or go to graduate school. \\
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Kwame will not take a job in the industry and will not go to graduate school.\\
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\textbf{b)}
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Yoshiko knows Java and calculus.\\
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Yoshiko does not know Joava or does not know calculus.\\
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\prob{32}\\
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Show that \(p \leftrightarrow q\) and \(\neg p \leftrightarrow \neg q\) are logically equivalent.
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\begin{displaymath}
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\begin{array}{|c c|c|c|c|c|c|}
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p & q & \neg p & \neg q & p \leftrightarrow q & \neg p \leftrightarrow \neg q \\
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\hline
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T & T & F & F & T & T \\
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T & F & F & T & F & F \\
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F & T & T & F & F & F \\
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F & F & T & T & T & T \\
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\end{array}
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\end{displaymath}
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\begin{align*}
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\therefore p \leftrightarrow q \equiv \neg p \leftrightarrow \neg q
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\end{align*}
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\pagebreak
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\end{document} |