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# How to multiply numbers - Part 3 Napier's Logarithms
Holding on to the idea of "sums are easier than products", the ancients look for another way that simplifies the prosthaphaeresis process. And Napier came up with a method now we understand as logarithms, roughly as follows.
## Napier's logarithm (ca. 1614).
First we take a number really close to $1$, say $0.99$. This will serve as our **base**. Now we construct a table that are powers of our base $0.99$, [[1 teaching/summer program 2023/week 3/powers-of-0.99|like this table here : powers of 0.99]]
Now for two numbers $x$ and $y$ in the interval $[0.1,1)$, find the corresponding numbers $N_{x}$ and $N_{y}$ (interpolate if you have to) such that $x \approx 0.99^{N_{x}}$ and $y \approx 0.99^{N_{y}}$. Now, add $N_{x}+N_{y}$, then find the corresponding value to this new $N$ in the table. The result is the product! See if you can work this out.
A worked example.
Say we want to multiply $x=0.42$ with $y=0.72$.
(1) Using the table, we see that $N_{x}\approx 86$, and $N_{y}\approx 32.5$ (linear interpolate if we have to).
(2) This gives the sum $N_{x}+N_{y}\approx 118.5$.
(3) Using the table we interpolate between $118$ and $119$ to get a value between $0.305$ and $0.302$, which is about $0.3035$ (we can be more precise if we want to).
So we conclude that $xy=(0.42)(0.72)\approx 0.3035$.
![[---images/---assets/---icons/question-icon.svg]] Compute the actual product $(0.42)(0.72)$, and comment on the accuracy. How can we improve our method?
![[---images/---assets/---icons/question-icon.svg]] Can you try to interpolate $N_{x}$ above more accurately to get a better result?
![[---images/---assets/---icons/question-icon.svg]] Try estimating the product of $0.82$ and $0.77$ using this method.
![[---images/---assets/---icons/question-icon.svg]] Can you explain in modern notation, why does this work? Recall the rules of exponentiation $b^N b^M = b^{N+M}$ under suitable assumptions. Can you also describe it using logarithms that you know?
![[---images/---assets/---icons/question-icon.svg]] Can you adopt this to perform the product $4313$ and $59312$? Notice these numbers are bigger than 1, but...
![[---images/---assets/---icons/question-icon.svg]] What happens when you try to do $(0.13)(0.11)$ with this method?
In this situation, for $x=0.13$ we have $N_{x}\approx 203$, and for $y=0.11$ we have $N_{y}\approx 220$. The sum $N_{x}+N_{y}\approx 423$, but this is not in our table. How can we get around this? Can you find a "fix" without needing more values of the power of $0.99$?
Hint: For $N=229$, we have $0.99^N \approx 0.1$.
![[---images/---assets/---icons/question-icon.svg]] Compare this Napier's method with prosthaphaeresis from last time.
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## Bonus activity : The slide rule.
This idea of changing adding to multiplication allows a physical device to do these directly. Since we can simplify "adding" by physically matching up two lengths together. You can print out a [[1 teaching/summer program 2023/week 3/slide-rule|bigger copy here of a slide rule here]] (cut along the long horizontal lines, so you have two strips, one labeled C and the other labeled D.)
![[1 teaching/summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 13.28.00.excalidraw.svg]]
%%[[1 teaching/summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 13.28.00.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 13.28.00.excalidraw.dark.svg|dark exported image]]%%
The scales and the markings $x$ ranges from 1 to 10 in increments of 0.1, but the horizontal distance are $\log_{10}(x)$. That is to say, if we take the graph of $y=\log_{10}(x)$ over the domain $[1,10]$, the spacing that we get are the $y$ values:
![[1 teaching/summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 14.43.42.excalidraw.svg]]
%%[[1 teaching/summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 14.43.42.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 14.43.42.excalidraw.dark.svg|dark exported image]]%%
Can you figure out how it works to multiply two numbers together?
Let us say you have two numbers $x,y$ that you want to multiply together. Take your two stripes labeled C and D:
(1) Match the position of 1 on C to $x$ on D
(2) Find the position of $y$ on C
(3) Find the number on D that corresponds to $y$ on C. This value is the product $xy$.
(4) If the position of $y$ on C will exceed the length of D, then instead do the following: Match the position of 10 on C to $x$ on D.
(5) Find the position of $y$ on C
(6) Read the value on D that corresponds to $y$ on C. This value is the product $xy$ up to a factor of 10.
Try it! Try multiplying $4\times 2$ using this method, and $3\times 3$. Also try $5\times 7$, with the modification.
Here is what $4\times 2$ would look like:
![[1 teaching/summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 18.07.11.excalidraw.svg]]
%%[[1 teaching/summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 18.07.11.excalidraw|🖋 Edit in Excalidraw]], and the [[summer program 2023/week 3/---files/How-to-multiply-numbers-3 2023-05-25 18.07.11.excalidraw.dark.svg|dark exported image]]%%
![[---images/---assets/---icons/question-icon.svg]] How would you use this to perform products of bigger numbers, such as $24\times 39$?
(By the way, you can always guess the last digit correctly easily...)
![[---images/---assets/---icons/question-icon.svg]] Can you try to explain why this works?
On an actual slide rule, people add various different scales in addition to the fundamental scales C and D to do other kinds of calculations, such as squaring, cubing, trigonometric values. People use it before the days of electronic calculators, enough to build airplanes and send people to the moon! Slide rules are essentially antiques now, and they are kind of hard to find these days, however...