handouts/Advanced/Quotient Groups/parts/0 mod.tex

\section{Modular Arithmetic}

I'm sure you're all familiar with modular arithmetic.
In this section, our goal is to meet \textit{equivalence relations},
\textit{equivalence classes}, and use them to formally define arithmetic in mod $n$.


\problem{}
Compute the following:

\begin{itemize}
	\item $5 + 3 \pmod{4}$
	\item $7 \times 4 \pmod{9}$
	\item $-4 \pmod{5}$
	\item $3^{-1} \pmod{7}$
\end{itemize}

\vfill


\definition{}
An \textit{equivalence relation} on a set $A$
is a symbol that makes a statement about two elements of $A$.
For example, $=$ is an equivalence relation on the set of integers.

\vspace{2mm}

An equivalence relation must satisfy the following properties:
\begin{itemize}
	\item Reflexivity: $x \sim x$ for all $x \in A$
	\item Symmetry: if $x \sim y$, $y \sim x$ for any $x, y \in A$
	\item Transitivity: if $x \sim y$ and $y \sim z$, then $x \sim z$
\end{itemize}


\problem{}<abseq>
Which of the following are equivalence relations on $\mathbb{Z}$?
\begin{itemize}
	\item $>$
	\item $\leq$
	\item $\Bumpeq$, where $a \Bumpeq b$ if $|a| = |b|$
	\item $\neq$
\end{itemize}


\vfill
\pagebreak








\problem{}
Consider the relation $\equiv_n$ on $\mathbb{Z}$, where $a \equiv_n b$ holds iff $a \equiv b \pmod{n}$. \par
Show that $\equiv_n$ is an equivalence relation.

\vfill


\definition{}
Say we have an equivalence relation $\sim$ on a set $A$. \par
The \textit{equivalence class} of $x$ is the set of all elements that are $\sim$ to $x$. \par
Here are a few examples: \par
\begin{itemize}[itemsep=2mm]
	\item
	The equivalence class of $2$ in $\mathbb{Z}$ under the relation $=$ is $\{2\}$, \par
	since the only $x$ that satisfies $x = 2$ is $2$.

	\item
	The equivalence class of $9$ in $\mathbb{Z}$ under the relation $\Bumpeq$
	from \ref{abseq} is $\{-9, 9\}$.
\end{itemize}


\problem{}
What is the equivalence class of $3$ in $\mathbb{Z}$ under $\equiv_5$? \par
\hint{Remember that $\mathbb{Z}$ contains both positive and negative numbers.}

\begin{solution}
	$\{..., -7, -2, 3, 8, 12, ... \}$
\end{solution}


\vfill

\problem{}
Let $A$ be a set and $\sim$ an equivalence relation. \par
Show that every element of $A$ is in \textit{exactly one} equivalence class\footnotemark{}\hspace{-1ex}. \par
\hint{What properties does an equivalence relation satisfy?}

\footnotetext{
	We could also say \say{$A$ is partitioned by $[A ~/ \sim]$}
	or \say{$A$ is the disjoint union of $[A ~/ \sim]$,} \par
	where $[A ~/ \sim]$ is the set of equivalence classes of $\sim$.
}

\vfill



We now have a proper definition of \say{mod $n$:} \par
it is the equivalence relation $a \equiv_n b$, which is usually written as $a \equiv b \pmod{n}$. \par
We will use this definition thoughout this handout.

\note[Note]{
	This is different than the \say{mod} operator $a ~\%~ b $,
	which is defined as the remainder of $a \div b$.
}


\pagebreak


\definition{}
Given any $x \in \mathbb{Z}$, $[x]_n$ is the equivalence class of $x$ under $\equiv_n$.


\problem{}
Compute the following:
\begin{itemize}[itemsep = 1mm]
	\item $[5]_3 + [4]_3$
	\item $[-2]_7 + [9]_7$
\end{itemize}

\vfill

\problem{}
Does $[4]_3 + [7]_5$ make sense?

\vfill

\problem{}
Find all $n$ that satisfy
$[5]_n \times [17]_n = [3]_n + [2]_n$ \par
\hint{$[a]_n = [b]_n$ iff $n$ divides $a - b$, by definition of mod.}

\begin{solution}
	$[85] = [12] ~\implies~ n ~|~ 85 - 12 ~\implies~ n ~|~ 73 ~\implies~ n \in \{1, 73\}$
\end{solution}


\vfill

\definition{}
$\znz{n}$ (pronounced \say{$\mathbb{Z}$ mod $n \mathbb{Z}$}) is the set of equivalence classes of $\equiv_n$ on $\mathbb{Z}$. \par
For example, $\znz{5} = \{~ [0]_5,~ [1]_5,~ [2]_5,~ [3]_5,~ [4]_5 ~\}$. \par

\vspace{2mm}

This notation may seem a bit odd, but don't let it confuse you. \par
One of our goals today is to understand what exactly $\znz{n}$ means.

\problem{}
What is $\znz{6}$?

\vfill


\pagebreak