mirror of
https://github.com/Rushilwiz/math381.git
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297 lines
9.3 KiB
TeX
297 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{amssymb}
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\usetikzlibrary{automata,positioning}
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% Basic Document Settings
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\pagestyle{fancy}
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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\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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%
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% Homework Problem Environment
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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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\enterProblemHeader{homeworkProblemCounter}
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}{
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\exitProblemHeader{homeworkProblemCounter}
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}
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\newcommand{\hmwkTitle}{Notes}
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\newcommand{\hmwkDueDate}{Fall, 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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\newcommand{\AllNaturals}{\mathbb{N}}
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\newtheorem{proposition}{Proposition}
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\newtheorem{theorem}{Theorem}
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\begin{document}
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\maketitle
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\pagebreak
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\begin{proposition}
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The sum of the first n, positive odd integers is the equation \(n^2\).
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\end{proposition}
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\begin{proof}
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By induction
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\begin{enumerate}
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\item base case
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\begin{align*}
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P(1) &= 1^2 = 1\\
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\end{align*}
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\item inductive step
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\begin{align*}
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\text{Assume } P(n) = "1+3+5+...+2n-1 = n^2"\\
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\text{WTS: } P(n+1) = "1+3+5+...+2n-1+2(n+1)-1 = (n+1)^2"\\
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1+3+5+...+2n-1+2(n+1)-1 &= (n+1)^2\\
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\end{align*}
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\end{enumerate}
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\end{proof}
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\begin{proposition} All horses are the same color.\\
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\emph{Note:} It suffices to show: P(n) = "All sets of n horses have the same color"\\
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\begin{proof}
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By induction
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\begin{enumerate}
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\item base case:\\
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P(1) = "All sets of 1 horse have the same color"\\
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\item inductive step: Assume P(n), i.e. every step of n horses have the same color\\
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WTS: P(n+1), i.e. every set of n+1 horses have the same color\\
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H = \(\{H_1, H_2, ..., H_n,H_{n+1}\}\) is a set of n+1 horses\\
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\(H_1 = \{H_1, H_2, ..., H_n\}\) is a set of n horses\\
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\(H_2 = \{H_2, H_3, ..., H_n,H_{n+1}\}\) is a set of n horses\\
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\end{enumerate}
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\end{proof}
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\end{proposition}
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\begin{theorem}
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Given two sets, when \(n \neq 1\), when they both overlap and are disjoint, the union of the two sets is equal to the sum of the two sets.
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\end{theorem}
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\pagebreak
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\section{Strong Induction}
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This is what we were doing before:\\
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\textbf{Weak Induction:}
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\begin{enumerate}
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\item base case
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\item inductive step (\(P(n) \implies P(n+1)\))
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\end{enumerate}
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Trying to prove \(P(n) \forall n \in \AllNaturals\).\\\\
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If \(P(n) \implies P(n+1)\) is too hard to show, instead try strong induction:
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\begin{enumerate}
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\item base case (Assume all steps before \(n+1\))
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\item Assume \(P(k)\forall k \in \{1,2,...,n\}\) then try to show P(n+1)
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\end{enumerate}
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\begin{proposition}
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A chocolate bar with n \(\geq\) 1 pieces can be broken into individual pieces by making n-1 breaks.
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\end{proposition}
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\begin{proof}
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Using weak induction,
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\begin{enumerate}
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\item base case: n = 1\\
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1 piece can be broken into individual pieces by making 0 breaks.
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\item inductive step\\
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Assume P(n)="a bar with n pieces can be broken into individual pieces by making n-1 breaks"\\
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WTS: P(n+1)="a bar with n+1 pieces can be broken into individual pieces by making n breaks"\\
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The issue is that we need to know that everything from P(n) to P(1) works.\\
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\textbf{Since we cannot prove this for an arbitrary n, we must use strong induction.}\\
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\end{enumerate}
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\end{proof}
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\begin{proof}
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Using strong induction,
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\begin{enumerate}
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\item base case: n = 1\\
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1 piece can be broken into individual pieces by making 0 breaks.
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\item inductive step\\
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Assume P(k) \(\forall k \in \{1,2,...,n\}\)\\
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WTS: P(n+1)="a bar with n+1 pieces can be broken into individual pieces by making n breaks"\\
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\begin{enumerate}
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\item Consider an arbitrary bar of n+1 size. Break the bar into two pieces
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\begin{enumerate}
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\item One piece has k pieces
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\item The other piece has \((n+1)-k\) pieces
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\end{enumerate}
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\item Assuming \(P(k)\), the first piece can be broken into individual pieces by making \(k-1\) breaks.
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\item \(P(n+1-k)\) will require \(n+1-k-1=n-k\) breaks.
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\end{enumerate}
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\(\therefore\) The total number of breaks is \(1+(k-1)+(n-k)=n\).
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\end{enumerate}
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\end{proof}
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\pagebreak
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\begin{theorem}
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It is true that strong induction \(\rightarrow\) weak induction
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\end{theorem}
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\begin{theorem}
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Fundamental Theorem of Arithmetic:
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\begin{align*}
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\forall n \in \AllNaturals - {0,1}, \text{n is either prime or can be written as a product of primes.}\\
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\end{align*}
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\end{theorem}
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\begin{proof}
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Using strong induction,
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\begin{enumerate}
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\item base case: n = 2\\
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2 is prime.
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\item inductive step: Assume P(k) \(\forall k \in \{2,3,...,n\}\)\\
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WTS: P(n+1)\\
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We can prove this by cases:
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\begin{enumerate}
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\item n+1 is prime:
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It can be expressed as \(1 \times (n+1)\)
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\item n+1 is not prime\\
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\begin{align*}
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\exists l,w \in \AllIntegers, n+1 = lw\\
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1 < l, w < n+1\\
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\end{align*}
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So P(l) = T, P(w) = T, therefore:\\
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\(l = p_1^{k_1}\times p_2^{k_2} \times ... \times p_n^{k_n}\), where \(p_1, p_2, ..., p_n\) are primes and \(k\in \AllNaturals\).\\
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\(w = q_1^{j_1}\times q_2^{j_2} \times ... \times q_m^{j_m}\), where \(q_1, q_2, ..., q_m\) are primes and \(j\in \AllNaturals\).\\\\
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Given \(l\) and \(w\), we can find \(n+1\) by multiplying them together.\\
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\begin{align*}
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n+1 &= lw\\
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n+1 &= (p_1^{k_1}\times p_2^{k_2} \times ... \times p_n^{k_n}) (q_1^{j_1}\times q_2^{j_2} \times ... \times q_m^{j_m})\\
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\end{align*}
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\end{enumerate}
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\end{enumerate}
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\(n+1\) is a product of primes.\\
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\end{proof}
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\pagebreak
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\begin{proposition}
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Consider the sequence \(a_1=0, a_2=1, a_n=2a_{n-1}-a_{n-2}\). Prove \(a_n = n-1\).
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\end{proposition}
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\begin{proof}
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Using strong induction,
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\begin{enumerate}
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\item bsae case: n = 1, n=2\\
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\(n=1: a_1 = 0 = 1-1\)\\
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\end{enumerate}
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\end{proof}
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\end{document} |