\section{Multiplication}

We'll use the C and D scales of your slide rule to multiply. \par
Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index}
of the C scale over the smaller number, $2$:

\def\sliderulewidth{10}
\begin{center}
\begin{tikzpicture}[scale=1]
	\cdscale{\cdscalefn(2)}{1}{C}
	\cdscale{0}{0}{D}
\end{tikzpicture}
\end{center}

Then we'll find the second number, $3$ on the C scale, and read the D scale under it:

\begin{center}
\begin{tikzpicture}[scale=1]
	\cdscale{\cdscalefn(2)}{1}{C}
	\cdscale{0}{0}{D}

	\slideruleind
		{\cdscalefn(6)}
		{1}
		{6}

\end{tikzpicture}
\end{center}

Of course, our answer is 6.

\problem{}
What is $1.15 \times 2.1$? \par
Use your slide rule.

\begin{solution}
	\begin{center}
		\begin{tikzpicture}[scale=1]
			\cdscale{\cdscalefn(1.15)}{1}{C}
			\cdscale{0}{0}{D}

			\slideruleind
				{\cdscalefn(1.15)}
				{1}
				{1.15}

			\slideruleind
				{\cdscalefn(1.15) + \cdscalefn(2.1)}
				{1}
				{2.415}

		\end{tikzpicture}
		\end{center}
\end{solution}

\vfill

Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within
two decimal places is close enough for most practical applications.

\pagebreak




Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
What should we do if we want to calculate $32 \times 210$? \par

\problem{}
Using your slide rule, calculate $32 \times 210$.


\begin{solution}
	\begin{center}
	\begin{tikzpicture}[scale=1]
		\cdscale{\cdscalefn(2.1)}{1}{C}
		\cdscale{0}{0}{D}

		\slideruleind
			{\cdscalefn(2.1)}
			{1}
			{2.1}

		\slideruleind
			{\cdscalefn(2.1) + \cdscalefn(3.2)}
			{1}
			{6.72}

	\end{tikzpicture}
	\end{center}

		Placing the decimal point correctly is your job. \par
		$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
\end{solution}

\vfill

\problem{}
Compute the following:
\begin{enumerate}
	\item $1.44 \times 52$
	\item $0.38 \times 1.24$
	\item $\pi \times 2.35$
\end{enumerate}

\begin{solution}
	\begin{enumerate}
		\item $1.44 \times 52 = 74.88$
		\item $0.38 \times 1.24 = 0.4712$
		\item $\pi \times 2.35 = 7.382$
	\end{enumerate}
\end{solution}

\vfill

\problem{}<provemult>
Note that the numbers on your C and D scales are logarithmically spaced.

\def\sliderulewidth{13}
\begin{center}
\begin{tikzpicture}[scale=1]
	\cdscale{0}{1}{C}
	\cdscale{0}{0}{D}
\end{tikzpicture}
\end{center}

Why does our multiplication procedure work?

\vfill
\pagebreak

Now we want to compute $7.2 \times 5.5$:

\def\sliderulewidth{10}
\begin{center}
\begin{tikzpicture}[scale=0.8]
	\cdscale{\cdscalefn(5.5)}{1}{C}
	\cdscale{0}{0}{D}

	\slideruleind
		{\cdscalefn(5.5)}
		{1}
		{5.5}

	\slideruleind
		{\cdscalefn(5.5) + \cdscalefn(7.2)}
		{1}
		{???}

\end{tikzpicture}
\end{center}

No matter what order we go in, the answer ends up off the scale. There must be another way.

\vspace{2mm}

Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}.
Move it over the \textit{larger} number, $7.2$:

\begin{center}
\begin{tikzpicture}[scale=1]
	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
	\cdscale{0}{0}{D}

	\slideruleind
		{\cdscalefn(7.2)}
		{1}
		{7.2}

\end{tikzpicture}
\end{center}

Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:

\begin{center}
\begin{tikzpicture}[scale=1]
	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
	\cdscale{0}{0}{D}


	\slideruleind
		{\cdscalefn(7.2)}
		{1}
		{7.2}

	\slideruleind
		{\cdscalefn(3.96)}
		{1}
		{3.96}

\end{tikzpicture}
\end{center}

Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$.
We can do this by hand to verify our answer.

\vspace{2mm}

\problem{}
Why does this work?

\begin{solution}
	Consider the following picture, where we have two D scales next to each other:

	\begin{center}
	\begin{tikzpicture}[scale=0.7]
		\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
		\cdscale{0}{0}{}
		\cdscale{-10}{0}{}

		\draw[
			draw=black,
		]
			(0, 0)
			--
			(0, -0.3)
			node [below] {D};

		\draw[
			draw=black,
		]
			(-10, 0)
			--
			(-10, -0.3)
			node [below] {D};

		\slideruleind
			{-10 + \cdscalefn(7.2)}
			{1}
			{7.2}

		\slideruleind
			{\cdscalefn(7.2)}
			{1}
			{7.2}

		\slideruleind
			{\cdscalefn(3.96)}
			{1}
			{3.96}

	\end{tikzpicture}
	\end{center}

	\vspace{2mm}

	The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$
	smaller than it should be.

	\vspace{2mm}

	\vspace{2mm}
	In other words, the answer we get from reverse multiplication is $\log{a} + \log{b} - \log{10}$. \par
	This reduces to $\log{(\frac{a \times b}{10})}$, and explains the misplaced decimal point in $7.2 \times 5.5$.
\end{solution}


\vfill
\pagebreak

\problem{}
Compute the following using your slide rule:
\begin{enumerate}
	\item $9 \times 8$
	\item $15 \times 35$
	\item $42.1 \times 7.65$
\end{enumerate}

\begin{solution}
	\begin{enumerate}
		\item $9 \times 8 = 72$
		\item $15 \times 35 = 525$
		\item $42.1 \times 7.65 = 322.065$
	\end{enumerate}
\end{solution}

\vfill
\pagebreak