\documentclass[../main.tex]{subfiles}

\begin{document}
	\section{Modular Arithmetic}

	\definition{}
	We say that $a, b$ are equivalent mod $m$ if $m$ divides $a - b$. \\
	If $a$ is equivalent to $b$ mod $m$, we write $a \equiv b \pmod{m}$. \\
	You can think of $b$ as the remainder of $a \div m$:
	\begin{itemize}
		\item[] $32 \equiv 2 \pmod{6}$
		\item[] $4 \equiv 4 \pmod{6}$
		\item[] $-2 \equiv 4 \pmod{6}$
	\end{itemize}

	\problem{}
	Complete the following:
	\begin{itemize}
		\item[] $87 \equiv ? \pmod{12}$ \hspace{3em} \textcolor{gray}{(Your answer should be between $0$ and $12$)}
		\item[] $13 \equiv 2 \pmod{?}$
		\item[] $? \equiv 1 \pmod{9}$
	\end{itemize}

	\begin{solution}
		\begin{itemize}
			\item[] $87 \equiv 3 \pmod{12}$
			\item[] $13 \equiv 2 \pmod{11}$
			\item[] $(9k + 1) \equiv 1 \pmod{9}\ \forall k \in \mathbb{Z}$
		\end{itemize}
	\end{solution}

	\vfill

	\definition{}
	The inverse of $a$ mod $m$ is an integer $a^\star$ so that \\
	$a \times a^\star \equiv 1 \pmod{m}$. \\
	Note that not every $a$ has an inverse mod $m$.

	\theorem{}<mod_has_inverse>
	$a$ has an inverse mod $m$ iff $\gcd(a, m) = 1$ \\
	\textcolor{gray}{\textit{The proof of this theorem is left as a challenge problem.}}

	\problem{D\'ej\`a vu?}<find_inverse>
	Find the inverse of $20 \pmod{14}$, if one exists. \\
	Find the inverse of $34 \pmod{541}$, if one exists.

	\begin{solution}
		$20^\star$ does not exist mod $14$, by \ref{mod_has_inverse}. \\
		$34^\star \equiv -175 \equiv 366 \pmod{541}$.
		See \ref{general_inverse} for an explanation.
	\end{solution}

	\vfill

	\problem{}<general_inverse>
	In general, how can we find the inverse of $a \pmod{p}$?\\
	(Assume $p$ is prime.)

	\begin{solution}
		We need an $a^\star$ so that $a \times a^\star \equiv 1 \pmod{m}$. \\
		This means that $aa^\star - mk = 1$. \\
		Since $p$ is prime, $\gcd(a, m) = 1$, and $aa^\star - mk = \gcd(a, m)$ \\
		Now use the Extended Euclidean Algorithm from \ref{extend_e_algorithm} to find $a^\star$.
	\end{solution}

	\vfill
	\pagebreak
\end{document}
