147 lines
3.8 KiB
TeX
Executable File
147 lines
3.8 KiB
TeX
Executable File
\section{The Euclidean Algorithm}
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\definition{}
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The \textit{greatest common divisor} of $a$ and $b$ is the greatest integer that divides both $a$ and $b$. \par
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We denote this number with $\gcd(a, b)$. For example, $\gcd(45, 60) = 15$.
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\problem{}
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Find $\gcd(20, 14)$ by hand.
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\begin{solution}
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$\gcd(20, 14) = 2$
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\end{solution}
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\vfill
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\theorem{The Division Algorithm}<divalgo>
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Given two integers $a, b$, we can find two integers $q, r$, where $0 \leq r < b$ and $a = qb + r$. \par
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In other words, we can divide $a$ by $b$ to get $q$ remainder $r$.
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\begin{instructornote}
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\ref{divalgo} looks scary on paper, but it's quite simple. \par
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Doing a small example on the board (like $14 \div 3$) may be a good idea. \par
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\vspace{2mm}
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For those that are new to modular arithmetic, you may want to explain how remainders,
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clock-face counting, division algorithm, and modular arithmetic are all the same.
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\end{instructornote}
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\theorem{}<gcd_abc>
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For any integers $a, b, c$, \par
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$\gcd(ac + b, a) = \gcd(a, b)$
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\problem{The Euclidean Algorithm}<euclid>
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Using the two theorems above, detail an algorithm for finding $\gcd(a, b)$. \par
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Then, compute $\gcd(1610, 207)$ by hand. \par
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\begin{solution}
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Using \ref{gcd_abc} and the division algorithm,
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% Minipage prevents column breaks inside body
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\begin{multicols}{2}
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\begin{minipage}{\columnwidth}
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$\gcd(1610, 207)$ \par
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$= \gcd(207, 161)$ \par
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$= \gcd(161, 46)$ \par
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$= \gcd(46, 23)$ \par
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$= \gcd(23, 0) = 23$ \par
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\end{minipage}
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\columnbreak
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\begin{minipage}{\columnwidth}
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$1610 = 207 \times 7 + 161$ \par
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$207 = 161 \times 1 + 46$ \par
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$161 = 46 \times 3 + 23$ \par
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$46 = 23 \times 2 + 0$ \par
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\end{minipage}
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\end{multicols}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}<extendedeuclid>
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Using the output of the Euclidean algorithm,
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\begin{itemize}
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\item[-] find a pair $(u, v)$ that satisfies $20u + 14v = \gcd(20, 14)$
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\item[-] find a pair $(u, v)$ that satisfies $541u + 34v = \gcd(541, 34)$
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% gcd = 1
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% u = 11; v = -175
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\end{itemize}
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This is called the \textit{extended Euclidean algorithm}. \par
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\hint{
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You don't need to fully solve the last part of this question. \\
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Understand how you \textit{would} do it, then move on.
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Don't spend too much time on arithmetic.
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}
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%For which numbers $c$ can we find a $(u, v)$ so that $541u + 34v = c$? \\
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%For every such $c$, what are $u$ and $v$?
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\vspace{2mm}
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\textbf{Hint:}
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After running the Euclidean algorithm, you have a table similar to the one shown below. \par
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You can use a bit of algebra to rearrange these statements to get what you need. \par
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\vspace{5mm}
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\newdimen\mywidth
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\setbox0=\hbox{Using the Euclidean Algorithm to find that $\gcd(20, 14) = 2$:}
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\mywidth=\wd0
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\begin{minipage}{\mywidth}
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\begin{center}
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Using the Euclidean Algorithm to find that $\gcd(20, 14) = 2$: \par
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$20 = 14 \times 1 + 6$ \par
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$14 = 6 \times 2 + 2$ \par
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$6 = 2 \times 3 + 0$ \par
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\end{center}
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\end{minipage}\par
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\vspace{2mm}
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We now want to write the 2 in the last equation in terms of 20 and 14.
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\begin{solution}
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Using the output of the Euclidean Algorithm, we can use substitution and a bit of algebra to solve such problems. Consider the following example:
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\begin{multicols}{2}
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\begin{minipage}{\columnwidth}
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\textit{Euclidean Algorithm:} \par
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$20 = 14 \times 1 + 6$ \par
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$14 = 6 \times 2 + 2$ \par
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$6 = 2 \times 3 + 0$ \par
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\end{minipage}
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\columnbreak
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\begin{minipage}{\columnwidth}
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\textit{Rearranged:} \par
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$6 = 20 - 14 \times 1$ \par
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$2 = 14 - 6 \times 2 = \gcd(20, 14)$ \par
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\end{minipage}
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\end{multicols}
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Using the right table, we can replace $6$ in $2 = 14 - 6 \times 2$ to get
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$2 = 14 - (20 - 14) \times 2$, \par
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which gives us $2 = \gcd(20, 14) = (3)14 + (-2)20$. \par
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\linehack{}
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$\gcd(20, 14) = 20(-2) + 14(3)$ \par
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$\gcd(541, 34) = 541(11) + 34(-175)$
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\end{solution}
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\vfill
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\pagebreak
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