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\lhead{MATH 311-02}
\chead{Problem Set \#5}
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\rfoot{\bf due March 25}
\begin{document}
\begin{enumerate}
\item \textbf{(9 points)} \emph{The following questions will explore
this slightly obscure relation property.}
\begin{definition}
A relation $R$ on a set $S$ is \emph{antireflexive} if and only
if, for all $a\in S$, $(a,a)\notin R$; in other words, $aRa$ is
false for all $a\in S$.
\end{definition}
\begin{enumerate}
\item \textbf{(2 points)} \emph{Demonstrate, either by example or
explanation, that there exist relations which are neither
reflexive nor antireflexive.}
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\item \textbf{(7 points)} Prove that for a set $S$, if $R\subseteq
S\times S$ is an antireflexive, symmetric, and transitive
relation, then $R=\emptyset$.
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\end{enumerate}
\item \textbf{(25 points)} \emph{The following proofs concern unions
and intersections of relations; if $R_1$ and $R_2$ are considered
as subsets of $S$, we may take $R_1\cup R_2$ and $R_1\cap R_2$ to
represent the ordinary operations on these sets.}
\begin{enumerate}
\item \textbf{(5 points)} \emph{Prove or disprove that, for
relations $R_1$ and $R_2$ on $S$, if either $R_1$ or $R_2$ is
reflexive, then the relation $R_1\cup R_2$ is reflexive.}
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\item \textbf{(5 points)} \emph{Prove or disprove that, for
relations $R5_1$ and $R_2$ on $S$, if both $R_1$ and $R_2$ are
symmetric, then the relation $R_1\cup R_2$ is symmetric.}
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\item \textbf{(5 points)} \emph{Prove or disprove that, for
relations $R_1$ and $R_2$ on $S$, if both $R_1$ and $R_2$ are
transitive, then the relation $R_1\cup R_2$ is transitive.}
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\item \textbf{(5 points)} \emph{Prove that for equivalence relations
$R_1$ and $R_2$ on $S$, $R_1\cap R_2$ is an equivalence relation
\emph{(note: this is the intersection, whereas the previous
questions discussed the union)}.}
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\item \textbf{(5 points)} \emph{If $x\in S$ and $R_1$ and $R_2$ are
equivalence relations of $S$, what is the relationship between
the equivalence classes of $x$ with respect to $R_1$, $R_2$, and
$R_1\cap R_2$?}
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\end{enumerate}
\item \textbf{(6 points)} \emph{Prove or disprove and salvage if
possible: for $[a],[b]\in\mathbb Z_n$ for a positive integer $n$,
if $[a]\cdot [b]=[0]$, then either $[a]=[0]$ or $[b]=[0]$.}
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%\item \textbf{(4 point bonus)} \emph{Prove that for a positive integer
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\end{enumerate}
\end{document}