\documentclass[12pt,a4paper]{amsart}
%%%%%%%%%%% Here is where you tell LaTeX to use packages. Each
%%%%%%%%%%% package gives you access to different commands and
%%%%%%%%%%% functionality. Most of them are unnecessary for the
%%%%%%%%%%% mathematics in this document.

%%%%%%%%%%% If you hadn't figured out, the % symbol automatically
%%%%%%%%%%% nulls out all subsequent characters on that line. So
%%%%%%%%%%% it's used for commenting code.
\usepackage{hyperref}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{framed}
\usepackage{multicol}
\usepackage[OT2,T1]{fontenc}
\usepackage{graphicx}
\usepackage[all]{xy}
\usepackage{color}
\usepackage{mathrsfs}
\usepackage{mathtools}

%%%%%%%%%%% Determining margins and spacing.
\setlength\topmargin{-0.3in}
\addtolength{\oddsidemargin}{-.625in}
	\addtolength{\evensidemargin}{-.625in}
	\addtolength{\textwidth}{1.25in}
\setlength\footskip{1in}

%%%%%%%%%%% I think this command enables `sections'
\numberwithin{equation}{section}

%%%%%%%%%%% This command allows you to label things as `theorem',
%%%%%%%%%%% `conjecture', etc.
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{question}{Question}[theorem]
\newtheorem{example}{Example}[theorem]


\usepackage[utf8]{inputenc}
\title{Math 220.201 Class 3 Questions - Solutions}
\begin{document}
\maketitle
\begin{enumerate}
\item For each of the following sentences, say whether it is a \emph{statement} or an \emph{open sentence}. Can you state its negation?
\begin{enumerate}
\item 5 is even or 3 is prime.

{\bf Solution:} It is a statement, and its negation is `5 is not even and 3 is not prime.'
\item At least one of my two friends misplaced his/her homework assignment.

{\bf Solution:} It is a statement, and its negation is `Neither of my two friends misplaced their homework assignment.'
\item For any polyhedron, the number of vertices plus the number of faces equals the number of edges plus 2.

{\bf Solution:} It is a statement, and its negation is `There is a polyhedron such that the number of vertices plus the number of faces is not equal to the number of edges plus 2.'
\item If $x^4=1$, then $x=1$ or $x=-1$.

{\bf Solution:} It is an open sentence, because there is an unquantified variable.
\end{enumerate}
\item Construct a truth table in $P, Q$ for the compound statement $(P\vee Q)\wedge \sim(P\wedge Q)$.\footnote{This is sometimes called `exclusive or', or `xor'.}
\vskip 0.1in
{\bf Solution:} The values in the truth table are $F, T, T, F$.
\vskip 0.1in
\item Construct a truth table in $P, Q$ for the statement $P\implies Q$. Can you construct a compound statement with the same truth table using only $\sim, \vee,$ and $\wedge$?
\vskip 0.1in
{\bf Solution:} The truth table is the same as that of $\sim P \vee Q$.
\vskip 0.1in
\item Let $A=\{3, 6, 8, 9, 11\}$ and $B=\{6, 9, 10\}$. Find all sets $S$ of integers such that the following statement holds true for all integers $x$.
\vskip 0.1in
$$(x\in S) \implies (x\in A)\wedge (x\in B)$$
Is there a set $S$ such that $(x\in S) \iff (x\in A)\wedge (x\in B)$?
\vskip 0.1in
{\bf Solution:} $S$ satisfies the given open sentence for all integers $x$ iff $S \in \mathcal{P}(A\cap B)$, i.e. $S \subseteq A \cap B=\{6,9\}$. That is,
$$S=\emptyset, \{6\},\{9\},\{6,9\}$$
satisfy the property $\forall x\in\mathbb{Z},(x\in S) \implies (x\in A)\wedge (x\in B)$. When $S=\{6,9\}$, we have
$$\forall x\in\mathbb{Z},(x\in S) \iff (x\in A)\wedge (x\in B)$$
\end{enumerate}
\end{document}