CSHW1.tex

\documentclass{article}
% Change "article" to "report" to get rid of page number on title page
\usepackage{amsmath,amsfonts,amsthm,amssymb}
\usepackage{setspace}
\usepackage{fancyhdr}
\usepackage{lastpage}
\usepackage{extramarks}
\usepackage{chngpage}
\usepackage{soul}
\usepackage[usenames,dvipsnames]{color}
\usepackage{graphicx,float,wrapfig}
\usepackage{ifthen}
\usepackage{listings}
\usepackage{courier}

% Homework Specific Information
\newcommand{\hmwkTitle}{Homework 1}
\newcommand{\hmwkSubTitle}{}
\newcommand{\hmwkDueDate}{\today}
\newcommand{\hmwkClass}{CSCI 262}
\newcommand{\hmwkClassTime}{2:00}
\newcommand{\hmwkClassInstructor}{Kazemian}
\newcommand{\hmwkAuthorName}{Harlan Haskins}

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\pagestyle{fancy}                                                       %
\lhead{\hmwkAuthorName}                                                 %
\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}  %
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% Some tools
\newcommand{\enterProblemHeader}[1]{\nobreak\extramarks{#1}{#1 continued on next page\ldots}\nobreak%
                                    \nobreak\extramarks{#1 (continued)}{#1 continued on next page\ldots}\nobreak}%
\newcommand{\exitProblemHeader}[1]{\nobreak\extramarks{#1 (continued)}{#1 continued on next page\ldots}\nobreak%
                                   \nobreak\extramarks{#1}{}\nobreak}%

\newlength{\labelLength}
\newcommand{\labelAnswer}[2]
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   \addtolength{\labelLength}{0.25in}%
   \changetext{}{-\labelLength}{}{}{}%
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   \marginpar{\fbox{#1}}%

   % We put the blank space above in order to make sure this
   % \marginpar gets correctly placed.
   \changetext{}{+\labelLength}{}{}{}}%

\setcounter{secnumdepth}{0}
\newcommand{\homeworkProblemName}{}%
\newcounter{homeworkProblemCounter}%
\newenvironment{homeworkProblem}[1][Problem \arabic{homeworkProblemCounter}]%
  {\stepcounter{homeworkProblemCounter}%
   \renewcommand{\homeworkProblemName}{#1}%
   \section{\homeworkProblemName}%
   \enterProblemHeader{\homeworkProblemName}}%
  {\exitProblemHeader{\homeworkProblemName}}%

\newcommand{\problemAnswer}[1]
  {\noindent\fbox{\begin{minipage}[c]{\columnwidth}#1\end{minipage}}}%

\newcommand{\problemLAnswer}[1]
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  {% We put this space here to make sure we're not connected to the above.
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   \subsection{\homeworkSectionName}%
   \enterProblemHeader{\homeworkProblemName\ [\homeworkSectionName]}}%
  {\enterProblemHeader{\homeworkProblemName}%

   % We put the blank space above in order to make sure this margin
   % change doesn't happen too soon (otherwise \sectionAnswer's can
   % get ugly about their \marginpar placement.
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\newcommand{\sectionAnswer}[1]
  {% We put this space here to make sure we're disconnected from the previous
   % passage

   \noindent\fbox{\begin{minipage}[c]{\columnwidth}#1\end{minipage}}%
   \enterProblemHeader{\homeworkProblemName}\exitProblemHeader{\homeworkProblemName}%
   \marginpar{\fbox{\homeworkSectionName}}%

   % We put the blank space above in order to make sure this
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%%% I think \captionwidth (commented out below) can go away
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%}

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    \label{#1}
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% Make title
\title{\vspace{2in}\textmd{\textbf{\hmwkClass:\ \hmwkTitle\ifthenelse{\equal{\hmwkSubTitle}{}}{}{\\\hmwkSubTitle}}}\\\normalsize\vspace{0.1in}\small{Due\ on\ \hmwkDueDate}\\\vspace{0.1in}\large{\textit{\hmwkClassInstructor\ \hmwkClassTime}}\vspace{3in}}
\date{}
\author{\textbf{\hmwkAuthorName}}
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\begin{document}
\begin{spacing}{1.1}
\maketitle
% Uncomment the \tableofcontents and \newpage lines to get a Contents page
% Uncomment the \setcounter line as well if you do NOT want subsections
%       listed in Contents
%\setcounter{tocdepth}{1}
\newpage

% When problems are long, it may be desirable to put a \newpage or a
% \clearpage before each homeworkProblem environment

\clearpage
\begin{homeworkProblem}[Problem 2]

\begin{proof}
Assume $\exists S1 \land \exists S2$ \\
Given $|S1 \cup S2| = |S1| + |S2| - |S1 \cap S2|$, due to the definition of union. \\
Given $|S1 \cap S2| \ge 0$, due to the definition of intersection. \\
Given $x - y \le x$ when $y \ge 0$, due to the definition of subtraction. \\
$
n = |S1| \\
m = |S2| \\
z = |S1 \cap S2| \\
|S1 \cup S2| = n + m - z \\
(n + m) - z \le n + m \\
\therefore
|S1 \cup S2| \le n + m
$
\end{proof}
\end{homeworkProblem}

\begin{homeworkProblem}[Problem 3]
\begin{proof}
Assume $ \exists S1 \land \exists S2$ \\
Given $ |S1 \times S2| = \left\{ {(x,y)|x \in S1, y \in S2}\right\}
$
\end{proof}
\end{homeworkProblem}

\newcommand{\eqseven}{}

\begin{homeworkProblem}[Problem 7]
\begin{proof}
\end{proof}
\end{homeworkProblem}

\begin{homeworkProblem}[Problem 25]
\begin{proof}
Assume $n=1$ \\
\begin{align*}
\sum\limits_{i=1}^n 1^2 &= \frac{1(1+1)(2(1)+1)}{6} \\
1 &= \frac{1(6)}{6} \\
1 &= \frac{6}{6} \\
\end{align*}
$\therefore$ the proposition holds for $1$\\
\\
Assume $P(k)$ is true.
\begin{align*}
P(k) = \sum\limits_{i=1}^k i^2 &= \frac{k(k+1)(2k+1)}{6}
\end{align*}

Prove $P(k+1)$ is true.
\begin{align*}
P(k+1) = \sum\limits_{i=1}^k+1 i^2 &= \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} \\
P(k+1) = \sum\limits_{i=1}^k+1 i^2 &= \frac{(k+1)(k+2)((2k+2)+1)}{6} \\
P(k+1) = \sum\limits_{i=1}^k+1 i^2 &= \frac{(k+1)(2k^2 + 7k + 6)}{6} \\
P(k+1) = \sum\limits_{i=1}^k+1 i^2 &= \frac{(k+1)(2k(k+1) + 6(k+1))}{6} \\
P(k+1) = \sum\limits_{i=1}^k+1 i^2 &= \frac{k(k+1)(2k+1) + 6(k+1)^2)}{6} \\
P(k+1) = \sum\limits_{i=1}^k+1 i^2 &= \frac{k(k+1)(2k+1))}{6}+(k+1)^2 \\
P(k+1) = \sum\limits_{i=1}^k i^2+(k+1)^2 &= \frac{k(k+1)(2k+1))}{6}+(k+1)^2 \\
P(k+1) = \sum\limits_{i=1}^k i^2 &= \frac{k(k+1)(2k+1))}{6} \\
\end{align*}
$\therefore$ the proposition holds for all $n \ge 1$
\end{proof}
\end{homeworkProblem}

\begin{homeworkProblem}[Problem 27]
\begin{proof}
Assume $n=4$ \\
$2^n=16$ \\
$n!=24$ \\
$\therefore$ the proposition holds for $n$ \\
$2^{n+1} = 2^n * 2$ \\
$(n+1)! = n! * (n+1)$ \\
because $n \ge 4$, $n \ge 2$ \\
$\therefore$ the proposition holds for $n+1$ \\
$\therefore$ the proposition holds for all $n \ge 4$
\end{proof}
\end{homeworkProblem}

\begin{homeworkProblem}[Bob and Alice]
\begin{proof}
Assume Bob is telling the truth and Alice is lying. \\
If Alice is lying, then Bob must be lying. \\
$\therefore$ This case is false. \\
\\
Assume Bob is lying and Alice is telling the truth. \\
If Alice is telling the truth, then Bob must be telling the truth. \\
$\therefore$ This case is false.\\
\\
Assume both Bob and Alice are telling the truth. \\
If both are telling the truth, then Bob must be lying. \\
$\therefore$ This case is false. \\
\\
Assume both Bob and Alice are lying. \\
If Bob is lying, then both Alice and Bob must be lying. \\
If Alice is lying, then Bob must be lying. \\
$\therefore$ This case is true.
\end{proof}
\end{homeworkProblem}

\end{spacing}
\end{document}

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