Assignment 6b (Due Thursday/Friday) 1

QUADRIVIUM Assignment 6b: Algorism II

Due: -Annandale- Tuesday, March 7th -Fishkill- Wednesday, March 15th

Read the following: folios Fibonacci [Leonardo Pisano, Leonardo of Pisa (ca. 1170-ca. 1250)]. Fibonacci's Liber Abaci: A

Translation into Modern English of Leonardo Pisano's Book of Calculation. (First appeared in 1202.) Translation and Notes by Laurence Sigler. Springer. 2002. Excerpts.

-pp. 23-30: Read chapter 2 on multiplication, etc. -p. 189: "Imperial Denari for Genoese Denari." [Handout]

2

Suzuki, Jeff. Mathematics in Historical Context. Mathematical Association of America. 2009. -pp. 134-135: Chuquet, short biography

½

Chuquet, Nicolas, Graham Flegg, Cynthia Hay, and Barbara Moss. Nicolas Chuquet, Renaissance Mathematician. Excerpts.

-From p. 204, do problems 19, 20, and 21. -Extra Credit: Problem 22 on p. 204.

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Homework 6b: The Chuquet problems from above and the Tiling Problem below.

Tiling Problem

<---See images at left---[ Covering a full chessboard with dominos such that they don't overlap and all stay within the grid is easy. 64 squares are tiled with 32 dominos.

]---See images at right---> Now remove the two white corner squares, leaving 62 squares on the

chessboard. Try to cover this board of 62 squares with dominos. [Don't

cover the removed squares.] Can you do it?

Can you make an argument for why you can or cannot do it?

Assignment 6b (Due Thursday/Friday) 2

Here are some more boards to experiment on.

Assuming that you fail multiple times, like most people do, is there a consistency to what colors the final

uncoverable squares will be? Can you use this consistency to come up with an explanation? Because a domino has to cover 1 white and 1 gray square, the total number of white squares must be equal to the total number

of gray squares, and of course the number of squares must be even.

Now selectively remove two squares from the chessboard. Remove any two squares, anywhere, but with the added condition that the two squares must be different colors. Meaning, remove a white square and

a gray square... from anywhere. See below for an example. Can you tile this without overlap or over-extension? Formulate a theory as to why or why not.

E.g. These three boards have squares of different colors removed.

If you draw a continuous looped path onto the chess board that covers every square once, you can follow this path even after

you cross out two different-colored squares. [Caveat: You may have to shift your starting square one place to get it all to line up.]

Now you choose. Eliminate any 2 squares, your choice, one of each color. [I suggest scribbling out your chosen squares to pitch black.]

Assignment 6b (Due Thursday/Friday) 3

My paraphrase and comments on Fibonacci's explanation:

Lines 1-5: Fibonacci gives us the initial information.

I abbreviated the currencies for convenience.

11!!𝐼 = 31!!𝑃 13!!𝐺 = 23!!𝑃

?𝐺 = 8!!𝐼

Lines 4-6: Here he shows how this problem is set up. In modern algebraic terms he is simply stating that 1×1×1 = 1. See next...

8!! ∙ 31!! ∙ 13

!!

11!! ∙ 23!!

=?𝐺

Assignment 6b (Due Thursday/Friday) 4

11!!𝐼 = 31!!𝑃 → !"!!!

!!!!!= 1

13!!𝐺 = 23!!𝑃 → !"!!!

!"!!!= 1

?𝐺 = 8!!𝐼 → !!!!

?!= 1

8!!𝐼 ∙ 31!!𝑃 ∙ 13!!𝐺 ?𝐺 ∙ 11!!𝐼 ∙ 23!!𝑃

= 1

8!!𝐼 ∙ 31!!𝑃 ∙ 13!!𝐺11!!𝐼 ∙ 23!!𝑃

=?𝐺

Modern analysis: These can all be rewritten to equal 1

using simple division.

And these equations can be multiplied together to equal one.

Multiply both sides by ?G and you have Fibonacci's equation.

Lines 6-9: Here Fibonacci makes improper fractions out of these mixed fractions.

Note: Mixed fractions (also called mixed numbers) are in the form 8!!, eight and one sixth. The improper fraction would be !"! . I personally don't like the name "improper fraction." It

implies that there is something wrong with it. If anything, the mixed fraction is improper, because it is hard to do anything

with it until you turn it into an "improper fraction" (as Fibonacci is showing us).

11!!=11 ∙ 2 + 1

2=232

31!!=34 ∙ 4 + 3

4=1274

13!!=13 ∙ 3 + 1

3=403

23!!=23 ∙ 5 + 3

5=1185

8!!=8 ∙ 6 + 1

6=496

Lines 9-14: And then into the first equation described in lines 4-6 he substitutes in all the improper fractions.

Then he multiplies the numerators in the numerator: 49 ∙ 127 ∙ 40. To that product he then multiplies the denominators of the

denominator: 2 ∙ 5. To this product he then divides by 23, 118, 6, 4, and 3.

Note: He uses a trick for dividing by 118, noting that !!!"

= !!∙!"

. This trick is notated in his weird fractional notation where

! !!∙!"

= !!∙!"

+ !!"= !

!!".

8!! ∙ 31!! ∙ 13

!!

11!! ∙ 23!!=

496 ∙ 1274 ∙ 403232 ∙ 1185

=

=49 ∙ 127 ∙ 40 ∙ 2 ∙ 523 ∙ 118 ∙ 6 ∙ 4 ∙ 3

=

=49 ∙ 127 ∙ 40 ∙ 2 ∙ 523 ∙ 2 ∙ 59 ∙ 6 ∙ 4 ∙ 3

=

Line 15: Here Fibonacci slightly digresses to explain his weird fractional notation. He focuses on just the denominator of the fraction and rearranges it by factoring it into 2 ∙ 8 ∙ 9 ∙ 23 ∙ 59. On the right I've shown this process.

I rearranged the denominator in ascending order.

Then I prime factorized it.

Then I recombined the pieces to match Fibonacci's denominator.

=1

23 ∙ 2 ∙ 59 ∙ 6 ∙ 4 ∙ 3=

=1

2 ∙ 3 ∙ 4 ∙ 6 ∙ 23 ∙ 59=

=1

2 ∙ 3 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 23 ∙ 59=

=1

2 ∙ 8 ∙ 9 ∙ 23 ∙ 59

Lines 16-23: Fibonacci suggests that you cancel whatever you can in order to simplify the final calculation. [I reordered the numerator for clarity.] He cancels out the 2s and reduces !"

! to 5.

He then walks the reader through his rather strange method of finishing off this calculation. He first fully multiplies out the numerator. I won't bore myself or you with a thorough explanation of the denominator. Suffice it to say that this can be done manually (or by calculator). Note: See how the denominator now looks like

the denominator in his final solution in the box.

=2 ∙ 5 ∙ 40 ∙ 49 ∙ 1272 ∙ 8 ∙ 9 ∙ 23 ∙ 59

=

=5 ∙ 5 ∙ 49 ∙ 1279 ∙ 23 ∙ 59

=

=1555759 ∙ 23 ∙ 59

Line 24: The final result is .... 12.738475395 Genoese denari equal 8!! Imperial denari. Or as Fibonacci notates it: ! !" !"

! !" !"12. You gotta admit that his notation is somewhat elegant.