162 lines
3.8 KiB
TeX
162 lines
3.8 KiB
TeX
\section{Modular Arithmetic}
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I'm sure you're all familiar with modular arithmetic.
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In this section, our goal is to meet \textit{equivalence relations},
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\textit{equivalence classes}, and use them to formally define arithmetic in mod $n$.
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\problem{}
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Compute the following:
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\begin{itemize}
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\item $5 + 3 \pmod{4}$
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\item $7 \times 4 \pmod{9}$
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\item $-4 \pmod{5}$
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\item $3^{-1} \pmod{7}$
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\end{itemize}
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\vfill
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\definition{}
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An \textit{equivalence relation} on a set $A$
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is a symbol that makes a statement about two elements of $A$.
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For example, $=$ is an equivalence relation on the set of integers.
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\vspace{2mm}
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An equivalence relation must satisfy the following properties:
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\begin{itemize}
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\item Reflexivity: $x \sim x$ for all $x \in A$
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\item Symmetry: if $x \sim y$, $y \sim x$ for any $x, y \in A$
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\item Transitivity: if $x \sim y$ and $y \sim z$, then $x \sim z$
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\end{itemize}
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\problem{}<abseq>
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Which of the following are equivalence relations on $\mathbb{Z}$?
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\begin{itemize}
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\item $>$
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\item $\leq$
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\item $\Bumpeq$, where $a \Bumpeq b$ if $|a| = |b|$
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\item $\neq$
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\end{itemize}
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\vfill
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\pagebreak
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\problem{}
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Consider the relation $\equiv_n$ on $\mathbb{Z}$, where $a \equiv_n b$ holds iff $a \equiv b \pmod{n}$. \par
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Show that $\equiv_n$ is an equivalence relation.
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\vfill
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\definition{}
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Say we have an equivalence relation $\sim$ on a set $A$. \par
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The \textit{equivalence class} of $x$ is the set of all elements that are $\sim$ to $x$. \par
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Here are a few examples: \par
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\begin{itemize}[itemsep=2mm]
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\item
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The equivalence class of $2$ in $\mathbb{Z}$ under the relation $=$ is $\{2\}$, \par
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since the only $x$ that satisfies $x = 2$ is $2$.
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\item
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The equivalence class of $9$ in $\mathbb{Z}$ under the relation $\Bumpeq$
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from \ref{abseq} is $\{-9, 9\}$.
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\end{itemize}
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\problem{}
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What is the equivalence class of $3$ in $\mathbb{Z}$ under $\equiv_5$? \par
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\hint{Remember that $\mathbb{Z}$ contains both positive and negative numbers.}
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\begin{solution}
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$\{..., -7, -2, 3, 8, 12, ... \}$
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\end{solution}
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\vfill
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\problem{}
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Let $A$ be a set and $\sim$ an equivalence relation. \par
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Show that every element of $A$ is in \textit{exactly one} equivalence class\footnotemark{}\hspace{-1ex}. \par
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\hint{What properties does an equivalence relation satisfy?}
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\footnotetext{
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We could also say \say{$A$ is partitioned by $[A ~/ \sim]$}
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or \say{$A$ is the disjoint union of $[A ~/ \sim]$,} \par
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where $[A ~/ \sim]$ is the set of equivalence classes of $\sim$.
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}
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\vfill
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We now have a proper definition of \say{mod $n$:} \par
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it is the equivalence relation $a \equiv_n b$, which is usually written as $a \equiv b \pmod{n}$. \par
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We will use this definition throughout this handout.
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\note[Note]{
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This is different than the \say{mod} operator $a ~\%~ b $,
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which is defined as the remainder of $a \div b$.
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}
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\pagebreak
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\definition{}
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Given any $x \in \mathbb{Z}$, $[x]_n$ is the equivalence class of $x$ under $\equiv_n$.
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\problem{}
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Compute the following:
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\begin{itemize}[itemsep = 1mm]
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\item $[5]_3 + [4]_3$
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\item $[-2]_7 + [9]_7$
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\end{itemize}
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\vfill
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\problem{}
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Does $[4]_3 + [7]_5$ make sense?
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\vfill
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\problem{}
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Find all $n$ that satisfy
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$[5]_n \times [17]_n = [3]_n + [2]_n$ \par
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\hint{$[a]_n = [b]_n$ iff $n$ divides $a - b$, by definition of mod.}
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\begin{solution}
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$[85] = [12] ~\implies~ n ~|~ 85 - 12 ~\implies~ n ~|~ 73 ~\implies~ n \in \{1, 73\}$
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\end{solution}
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\vfill
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\definition{}
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$\znz{n}$ (pronounced \say{$\mathbb{Z}$ mod $n \mathbb{Z}$}) is the set of equivalence classes of $\equiv_n$ on $\mathbb{Z}$. \par
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For example, $\znz{5} = \{~ [0]_5,~ [1]_5,~ [2]_5,~ [3]_5,~ [4]_5 ~\}$. \par
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\vspace{2mm}
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This notation may seem a bit odd, but don't let it confuse you. \par
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One of our goals today is to understand what exactly $\znz{n}$ means.
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\problem{}
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What is $\znz{6}$?
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\vfill
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\pagebreak |