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\usepackage{amsmath,amssymb,amsfonts,amsthm}

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% Choose the correct class.  Use % to comment out the one you don't
% need, or just delete the line.

%\markright{MAT230 Discrete Mathematics Solutions}
\markright{MAT231 Transition to Higher Mathematics Solutions}

% Add other new commands here as needed, but initially these can be
% left alone

\newcommand{\NN}{\mathbb{N}}              % Natural numbers
\newcommand{\QQ}{\mathbb{Q}}              % Rationals
\newcommand{\RR}{\mathbb{R}}              % Reals
\newcommand{\ZZ}{\mathbb{Z}}              % Integers
\newcommand{\powerset}{\mathcal{P}}       % power set
\newcommand{\setcomp}[1]{\overline{#1}}   % set complement
\newcommand{\negate}{{\sim}}              % negation (logic)

\newcommand{\modulo}{\text{ mod }}
\newcommand{\congruent}[3]{{#1} \equiv {#2} \,\, (\text{mod } {#3})}

\newtheorem*{proposition}{Proposition}

\begin{document}

% Place your name and assignment due date here

Jackie Q. Student\\
September 5, 2014\\

% Now start your solutions

%===========================================================================
\bigskip\rule{\textwidth}{0.005in}
\section*{Section 2.1}
%===========================================================================

\begin{enumerate}

\item[2.] This is the first assigned problem. You would replace this
  text with your solution.  Notice in the \LaTeX\ source how the
  \emph{enumerate} environment is used.  In particular, since the
  problem numbers are not consecutive, they are given as arguments to
  the \texttt{{\textbackslash}item} tags.

\item[4.] This is the second assigned problem.  If you had to use the
  names of common sets they would look like $\NN$, $\ZZ$, $\QQ$, and
  $\RR$.  These are not natively part of \LaTeX\ but are defined in
  the preamble of this file.

\item[6.] Here are some more set symbols: $A \cup B$, $A \cap B$, $A
  \subseteq B$, and $\bigcup_{\alpha \in I} A_{\alpha}$.  Also, don't
  forget $x \in A$ and $x \notin B$.  The power set of $S$ is denoted
  $\powerset (S)$ and Cartesian products look like $A \times B$.


\item[8.] Logic symbols that you might need include $P \wedge Q$, $P
  \vee Q$, $\negate P$.  Notice that \texttt{{\textbackslash}negate}
  is not part of \LaTeX\ but is defined in the preamble of this file.
  Another commonly used symbol for the negation of $P$ is $\neg P$; you may
  use it if you prefer.
\end{enumerate}

\newpage
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\bigskip\rule{\textwidth}{0.005in}
\section*{Chapter 4}
%===========================================================================

% **************************************************************************
% **************************************************************************
% ***   Note: to be able to use the proof environment, you must have     ***
% ***   "amsthm" in the "\usepackage{amsmath,...} line near the top of   ***
% ***   this file.                                                       ***
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\begin{enumerate}
%
% Here the proposition is listed right after the \item.  ALWAYS
% state the proposition/theorem you are proving as part of your
% proof.  No proof will be accepted unless you fully state the
% proposition first!
%
\item[1.] If $x$ is an even integer, then $x^2$ is even.
  \begin{proof}
    Suppose $x$ is an even integer.  There is an integer $a$ for which
    $x = 2a$ by the definition of even numbers.  Then
    \begin{displaymath}
      x^2 = (2a)^2 = 4a^2 = 2(2a^2) = 2b
      \end{displaymath}
    where $b = 2a^2$ is an integer by the closure property for
    integers.  We have shown that $x^2 = 2b$ where $b$ is an integer
    so $x^2$ must be even.
  \end{proof}
%
% This answer uses the proposition environment.  To be able to use
% this, you need the line
%     \newtheorem*{proposition}{Proposition}
% near the top of this file before \begin{document}.  Note: the * in
% the \newtheorem* command will mean the propositions will be
% unnumbered, if * is not there then the propositions used in this
% fill will be numbered starting with 1.
%
% As you'll see when viewing the output, using the proposition
% environment in numbered homework problems is not very pretty.  In a
% paper, however, using numbered propositions and/or theorems followed
% by their proof is the norm.
%
\item[x.]
  \begin{proposition}
    The sum of an even integer and an odd integer is odd.
  \end{proposition}
  \begin{proof}
    Suppose $a$ is an even integer and $b$ is an odd integer.  Then,
    by our definitions of even and odd numbers, we know that integers
    $m$ and $n$ exist so that $a = 2m$ and $b=2n+1$.  Then
    \begin{displaymath}
      a + b = (2m) + (2n + 1) = 2(m+n)+1 = 2c+1
    \end{displaymath}
    where $c = m+n$ is an integer by the closure property of
    addition.  We have shown that $a+b = 2c+1$ for an integer $c$ so
    $a+b$ must be odd.
  \end{proof}
%
% This proof uses more mathematical notation than the previous two,
% and provides and demonstrates the symbol for "divides".
%
\item[y.] (This is the second proof presented in the lecture slides.)
  \begin{proposition}
    Suppose $a,b,c \in \ZZ$.  If $a|b$ and $b|c$, then $a|c$.
  \end{proposition}
  \begin{proof}
    Let $a,b,c \in \ZZ$.  If $a|b$ and $b|c$ then by the definition
    of divisibility there are integers $m$ and $n$ for which $b = am$
    and $c = bn$.  In this case
    \begin{displaymath}
      c = bn = (am)n = a(mn) = ak
    \end{displaymath}
    where $k = mn$.  Since $c$ is a multiple of $a$, by the definition of
    divisibility we see that $a|c$.
  \end{proof}
\end{enumerate}

\bigskip
Here are some ways that modulo operator and congruence statements can
be handled.  Note that some ``newcommand'' definitions are needed to
use these; you can find them at the top of the
source of this file.

\bigskip
Suppose $a,b \in  \ZZ$ and $n \in \NN$.  The statements
\begin{displaymath}
  \congruent{a}{b}{n}
\end{displaymath}
and
\begin{displaymath}
  a \modulo n = b \modulo n
\end{displaymath}
are logically equivalent.

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\end{document}
