92 lines
2.5 KiB
TeX
Executable File
92 lines
2.5 KiB
TeX
Executable File
\section{Modular Arithmetic}
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\definition{}
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$\mathbb{Z}_n$ is the set of integers mod $n$. For example, $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$. \par
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\vspace{2mm}
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Multiplication in $\mathbb{Z}_n$ works much like multiplication in $\mathbb{Z}$: \par
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If $a, b$ are elements of $\mathbb{Z}_n$, $a \times b$ is the remainder of $a \times b$ when divided by $n$. \par
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\note{For example, $2 \times 2 = 4$ and $3 \times 4 = 12 = 2$ in $\mathbb{Z}_5$}
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\problem{}
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Create a multiplication table for $\mathbb{Z}_4$:
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\begin{center}
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\begin{tabular}{c | c c c c}
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$\times$ & 0 & 1 & 2 & 3 \\
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\hline
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0 & ? & ? & ? & ? \\
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1 & ? & ? & ? & ? \\
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2 & ? & ? & ? & ? \\
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3 & ? & ? & ? & ? \\
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\end{tabular}
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\end{center}
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\definition{}
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Let $a, b$ be elements of %\mathbb{Z}_n$. \par
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If $a \times b = 1$, we say that $b$ is the \textit{inverse} of $a$ in $\mathbb{Z}_n$.
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\vspace{2mm}
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We usually write \say{$a$ inverse} as $a^{-1}$. \par
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Inverses are \textbf{not} guaranteed to exist.
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\theorem{}<mod_has_inverse>
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$a$ has an inverse in $\mathbb{Z}_n$ if and only if $\gcd(a, n) = 1$ \par
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\problem{}
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Find the inverse of $3$ in $\mathbb{Z}_4$, if one exists. \par
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Find the inverse of $20$ in $\mathbb{Z}_{14}$, if one exists. \par
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Find the inverse of $4$ in $\mathbb{Z}_7$, if one exists.
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\begin{solution}
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\begin{itemize}
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\item $3^{-1}$ in $\mathbb{Z}_{4}$ is $3$
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\item $20^{-1}$ in $\mathbb{Z}_{14}$ doesn't exist.
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\item $4^{-1}$ in $\mathbb{Z}_{7}$ is $2$
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\end{itemize}
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\end{solution}
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\vfill
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\problem{}
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Show that if $n$ is prime, every element of $\mathbb{Z}_n$ (except 0) has an inverse.
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\vfill
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\problem{}
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Show that if $n$ is not prime, $\mathbb{Z}_n$ has at least one element with no inverse.
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\vfill
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\pagebreak
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\problem{}<general_inverse>
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In general, how can we find the inverse of $a$ in $\mathbb{Z}_n$? Assume $a$ and $n$ are coprime.\par
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\hint{You can find that $34^{-1}$ is $-175$ in $\mathbb{Z}_{541}$ by looking at a previous problem.}
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\begin{solution}
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We need an $a^{-1}$ so that $a \times a^{-1} = 1$. \par
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This means that $aa^{-1} - mk = 1$. \par
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Since $a$ and $m$ are coprime, $\gcd(a, m) = 1$ and $aa^{-1} - mk = \gcd(a, m)$ \par
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Now use the extended Euclidean algorithm from \ref{extendedeuclid} to find $a^\star$.
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\end{solution}
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\vfill
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\definition{}
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Elements in $\mathbb{Z}_n$ that have an inverse are called \textit{units}. \par
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The set of units in $\mathbb{Z}_n$ is called $\mathbb{Z}_n^\times$, which is read \say{$\mathbb{Z}$ mod $n$ cross}.
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\problem{}
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What is $\mathbb{Z}_5^\times$? \par
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What is $\mathbb{Z}_{12}^\times$? \par
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\vfill
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\pagebreak
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