mirror of
https://github.com/Rushilwiz/math381.git
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299 lines
9.3 KiB
TeX
299 lines
9.3 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{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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\usepackage{amssymb}
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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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\textwidth=6.5in
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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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\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 4}
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\newcommand{\hmwkDueDate}{October 5, 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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% 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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\renewcommand{\And}{\wedge}
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\newcommand{\Or}{\vee}
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\newcommand{\Xor}{\oplus}
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\newcommand{\Not}{\neg}
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\newcommand{\Implies}{\rightarrow}
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\newcommand{\Iff}{\leftrightarrow}
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\newcommand{\AllIntegers}{\mathbb{Z}}
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\newcommand{\AllRationals}{\mathbb{Q}}
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\newcommand{\AllReals}{\mathbb{R}}
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\newcommand{\AllComplexes}{\mathbb{C}}
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\begin{document}
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\maketitle
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\pagebreak
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\unit{5.1}
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\prob{10}
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\begin{enumerate}[a)]
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\item Find a formula for
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\begin{align*}
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\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\ldots+\frac{1}{n(n+1)}
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\end{align*}
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by examining the values of this expression for small values of \(n\).\\
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Given the following equation,
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\begin{align*}
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P(n) &= \sum_{i=1}^{n}\frac{1}{n(n+1)}\\
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\end{align*}
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We can find the following values for \(P(x)\) and generalize a formula for \(P(n)\).
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\begin{align*}
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P(1) &&= \frac{1}{2}\\
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P(2) &= \frac{1}{2}+\frac{1}{6}&=\frac{2}{3}\\
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P(3) &= \frac{1}{2}+\frac{1}{6}+\frac{1}{12}&=\frac{3}{4}\\
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P(4) &= \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}&=\frac{4}{5}\\
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P(n) &&= \frac{n}{n+1}
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\end{align*}
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\item Prove the formula you conjectured in part (a).
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\begin{proof}
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We will prove the formula by induction.\\
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\textbf{Base Case:} \(n=1\)
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\begin{align*}
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P(1) = \frac{1}{1+1} = \frac{1}{2}
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\end{align*}
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\textbf{Inductive step:}
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Given P(n), we will prove P(n+1).
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\begin{align*}
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P(n+1) &= \sum_{i=1}^{n+1}\frac{1}{n(n+1)}\\
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&= \sum_{i=1}^{n}\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}\\
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&= \frac{n}{n+1}+\frac{1}{(n+1)(n+2)}\\
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&= \frac{n(n+2)+1}{(n+1)(n+2)}\\
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&= \frac{(n+1)^2}{(n+1)(n+2)}\\
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&= \frac{n+1}{n+2}
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\end{align*}
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\end{proof}
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\end{enumerate}
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\pagebreak
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\prob{34}
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Prove that 6 divides \(n^3 - n\) whenever \(n\) is a nonnegative integer.\\
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\begin{proof}
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Given \(P(n) = \exists k \in \AllIntegers, n^3-n = 6k\), we will show that \(\forall n \in \AllIntegers^+(P(n))\) by induction.\\
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\textbf{Base Case:} \(n=0\)
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\begin{align*}
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0^3 - 0 = 6(0)
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\end{align*}
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\textbf{Inductive step:}
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Given P(n), we will prove P(n+1).
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\begin{align*}(n+1)
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(n+1)^3-(n+1) &= n^3+3n^2+3n+1-(n+1)\\
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&= n^3+3n^2+2n\\
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&= n^3-n+(3n^2+3n)\\
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&= (n^3 - n) + 3(n)(n+1)
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\end{align*}
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Since we are assuming \(P(n)\), we can affirm that \(n^3-n\) is true. Now we can show that 6 also divides the second term,
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\begin{align*}
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\exists k \in \AllIntegers, 3(n)(n+1) &= 6k\\
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n(n+1) &= 2k
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\end{align*}
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Since any odd and even integer multiplied together is even, we can affirm that 6 divides \(3(n)(n+1)\). Since we know that,
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\begin{align*}
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\forall a,b,n \in \AllIntegers,\\n|a, n|b \implies n|(a+b).
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\end{align*}
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6 must divide \((n^3 - n) + 3(n)(n+1)\). Thus \(P(n+1)\) is true.
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\end{proof}
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\pagebreak
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\prob{64}
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Use mathematical induction to prove that if \(p\) is a prime and \(p | a_1a_2 \cdots a_n\), where \(a_i\) is an integer for \(i = 1, 2, 3, \dots , n\), then \(p | a_i\) for some integer \(i\).
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\begin{align*}
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P(n) = \forall a_1, a_2, \dots, a_n \in \AllIntegers, p | a_1a_2 \cdots a_n \implies p | a_i \text{ for some } i \in \AllIntegers .\\
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\end{align*}
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\textbf{Base Case:} \(n=1\)
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\begin{align*}
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p | a_1 \implies p | a_1
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\end{align*}
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\textbf{Inductive Step:} Given P(n), we will prove P(n+1).
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\begin{align*}
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P(n+1) &= p|a_1a_2\cdots a_na_{n+1} \implies p|a_i \text{ for some } i \in \AllIntegers .\\
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\end{align*}
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Since we know that \(p\) is prime,
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\begin{align*}
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p|a_1a_2\cdots a_na_{n+1} &\implies p|a_1a_2\cdots a_n \text{ or } p|a_{n+1}\\
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&\implies P(n) \text{ or } p|a_{n+1}\\
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&\implies p|a_i \text{ for some } i \in \AllIntegers \text{ or } p|a_{n+1}\\
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&\implies P(n+1)
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\end{align*}
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We have shown that \(P(n) \implies P(n+1)\), thus \(P(n+1)\) is true.
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\pagebreak
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\unit{2.1}
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\prob{2} Use set builder notation to give a description of each of these sets.
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\begin{enumerate}[a)]
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\item \{{0, 3, 6, 9, 12}\}\\
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\begin{align*}
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A = \{x \in \AllIntegers^+ | x = 3n, 0 \leq n \leq 4\}
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\end{align*}
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\item \{{-3,-2,-1, 0, 1, 2, 3}\}
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\begin{align*}
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B = \{x \in \AllIntegers | -3 \leq x \leq 3\}
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\end{align*}
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\item \(\{m,n,o,p\}\)
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\begin{align*}
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C = \{x | x \text{ is a lowercase letter in the alphabet from m to p}\}
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\end{align*}
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\end{enumerate}
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\pagebreak
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\prob{6} For each of these pairs of sets, determine whether the first is a subset of the second, the second is a subset of the first, or neither is a subset of the other.
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\begin{enumerate}[a)]
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\item the set of people who speak English, the set of people who speak English with an Australian accent
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\item the set of fruits, the set of citrus fruits
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\item the set of students studying discrete mathematics, the set of students studying data structures
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\end{enumerate}
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\pagebreak
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\prob{12} Determine whether these statements are true or false.
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\begin{enumerate}[a)]
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\item \(\emptyset \in \{\emptyset\}\)
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\item \(\emptyset \in \{\emptyset, \{\emptyset\}\}\)
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\item \(\{\emptyset\} \in \{\emptyset\}\)
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\item \(\{\emptyset\} \in \{\{\emptyset\}\}\)
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\item \(\{\emptyset\} \subset \{\emptyset, \{\emptyset\}\}\)
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\item \(\{\{\emptyset\}\} \subset \{\emptyset, \{\emptyset\}\}\)
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\item \(\{\{\emptyset\}\} \subset \{\{\emptyset\}, \{\emptyset\}\}\)
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\end{enumerate}
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\pagebreak
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\prob{18}
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Use a Venn diagram to illustrate the relationships \(A \subset B\) and \(A \subset C\).
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\pagebreak
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\prob{22}
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What is the cardinality of each of these sets?
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\begin{enumerate}[a)]
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\item \(\emptyset\)
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\item \(\{\emptyset\}\)
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\item \(\{\emptyset, \{\emptyset\}\}\)
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\item \(\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}\)
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\end{enumerate}
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\pagebreak
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\prob{32} Suppose that \(A \times B = \emptyset\), where \(A \text{ and } B\) are sets. What can you conclude?
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\pagebreak
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\prob{44} Prove or disprove that if \(A, B, \text{and } C\) are nonempty sets and \(A \times B = A \times C\text{, then } B = C\).
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\end{document} |