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\usepackage[utf8]{inputenc}
\title{Math 220.201 Class 13 Questions}
\begin{document}
\maketitle

\begin{enumerate}
\item $S=\mathbb{Z}$; $a\sim b$ if $a\mid b$. This is not an equivalence relation. It is reflexive and transitive, but it is not symmetric. For example, $2\sim 6$ but $6\not\sim 2$.
\vskip 0.1in
\item $S=\mathbb{Z}$; $a\sim b$ if either $a\nmid b$ or $b\nmid a$. This is not an equivalence relation, because it is not reflexive. $n\not\sim n$ for every $n$. (In fact, this relation is exactly $a\sim b$ if $a\neq b$.)
\vskip 0.1in
\item $S=\mathbb{Z}$; $a\sim b$ if $a\equiv b\pmod{6}$. This is an equivalence relation.
\vskip 0.1in
\item $S=\mathbb{N}$; $a\sim b$ if $a$ and $b$ have a common prime factor. This is not an equivalence relation. It is reflexive and symmetric, but it is not transitive. $4\sim 6$ and $6\sim 9$ but $4\not\sim 9$.
\vskip 0.1in
\item $S=\mathbb{R}$; $a\sim b$ if $b-a \in \mathbb{Z}$. This is an equivalence relation.
\vskip 0.1in
\item $S=\mathbb{R}$; $a\sim b$ if $|b-a|<1$. This is not an equivalence relation. It is reflexive and symmetric, but it is not transitive. For example, $0\sim \frac{2}{3}$ and $\frac{2}{3}\sim \frac{4}{3}$ but $0\not\sim \frac{4}{3}$.
\vskip 0.1in
\item $S=\mathbb{R}$; $a\sim b$ if $\sqrt{e^a+2}=\sqrt{e^b+2}$. This is an equivalence relation. In fact, for any function $f(x)$, the relation $a\sim b \iff f(a)=f(b)$ is an equivalence relation.
\vskip 0.1in
\item $S=\mathbb{Z}$; $a\sim b$ if $3a+5b$ is even. This is an equivalence relation. It is equivalent to saying $a\equiv b\pmod{2}$.
\vskip 0.1in
\item $S=\mathbb{Z}$; $a\sim b$ if $a+b$ is odd. This is not an equivalence relation, because it is not reflexive or transitive. It is equivalent to saying that $a$ and $b$ have opposite parity.
\vskip 0.1in
\item $S$ is the set of lines in the plane; $\ell_1\sim \ell_2$ if either $\ell_1=\ell_2$ or $\ell_1\mid\mid\ell_2$. This is an equivalence relation.
\vskip 0.1in
\item $S$ is the set of lines in the plane; $\ell_1\sim \ell_2$ if either $\ell_1=\ell_2$ or $\ell_1\perp\ell_2$. This is not an equivalence relation. It is reflexive and symmetric, but not transitive. For example, the line $x=0$ is perpendicular to the line $y=0$, which is equivalent to the line $x=1$, but the lines $x=0$ and $x=1$ are not equivalent.
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\item $S$ is the set of lines in the plane; $\ell_1\sim \ell_2$ if $\ell_1=\ell_2$ or $\ell_1\perp\ell_2$ or $\ell_1\mid\mid\ell_2$. This is an equivalence relation.
\vskip 0.1in
\item $S$ is the set of lines in $\mathbb{R}^3$ containing $(0,0,0)$; $\ell_1\sim \ell_2$ if $\ell_1=\ell_2$ or $\ell_1 \perp \ell_2$. This is not an equivalence relation. It is reflexive and symmetric, but not transitive. The lines going through the points $(0, 1, 0)$ and $(0, 1, 1)$ are both orthogonal to the line going through $(1, 0, 0)$ but these are not orthogonal to each other.

\end{enumerate}
\end{document}