European Mathematics in the Middle Ages

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European Mathematics in the Middle Ages
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The period known as the Dark Ages, the Middle
Ages, or the medieval period in Europe, saw very
little mathematical activity. During this time
mathematical advances were we being made by
Arabs, Persians, and Indians. There were,
however, a few Europeans who made meaningful
contributions to mathematics despite the general
lack of interest in intellectual pursuits.
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Perhaps the best known medieval European
mathematician is Leonardo Pisano Bigollo.
Centuries after his death, mathematicians began
referring to him as Fibonacci (short for son of
Bonaccio despite the fact that his father was
not called Bonaccio).
Fibonaccis father was a traveling salesman, and
the two of them followed business up and down the
Mediterranean coast. Since half of this region
was Arab-controlled, Fibonacci was exposed to the
Islamic mathematics we learned in Chapter 9. In
particular he learned the Hindu-Arabic number
system, and he wrote the Book of the Abacus with
the goal of introducing this new system to Europe.
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In the Book of the Abacus Leonardo introduced his
own system of fractions. It was particularly
useful for dealing with quantities measured by
multiple units such as yards, feet, and inches.
Recall that 1 yard 3 feet 12 3 inches, so
that 1 foot yard, and 1 inch foot
yard.
A quantity such as 5 yards, 2 feet, and 8 inches
is therefore a 5 yards.
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Fibonacci would have written the quantity a
5 as 8 2 12 3 This system
made division easy. Note that a
5 where 212 17 12 8, and 17 5 3 2.
5.
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Fibonacci would have written the quantity a
5 as 8 2 12 3 This system
made division easy. Note that a
5 where 212 17 12 8, and 17 5 3 2.
5.
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Fibonacci would have written the quantity a
5 as 8 2 12 3 This system
made division easy. Note that a
5 where 212 17 12 8, and 17 5 3 2.
5.
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To divide a number such as 212 by 36, Fibonacci
first found a rule for 36. Since 36 arose as 12
3, we factor 36 as 12 3 and use the rule 1
0 12 3 for division by 36. For division by
other numbers, it was typical to first find their
prime factors. A list of prime factors formed
the bottom half of a rule. Our book gives the
rule for 75 on page 280.
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
1 0 12 3
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 12 17 remainder 8
1 0 12 3
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 12 17 remainder 8
8 0 12 3
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 12 17 remainder 8
Next divide 17 by 3, and write the remainder
above the 3.
8 0 12 3
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 12 17 remainder 8
Next divide 17 by 3, and write the remainder
above the 3.
17 3 5 remainder 2
8 0 12 3
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 12 17 remainder 8
Next divide 17 by 3, and write the remainder
above the 3.
17 3 5 remainder 2
8 2 12 3
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 12 17 remainder 8
Next divide 17 by 3, and write the remainder
above the 3.
17 3 5 remainder 2
8 2 12 3
Write the 5 to the right of the above.
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To use the rule below to divide 212 by 36, first
divide 212 by 12 and write the remainder above
the 12.
212 inches 17 feet, 8 inches
212 12 17 remainder 8
Next divide 17 by 3, and write the remainder
above the 3.
17 3 5 remainder 2
17 feet 5 yards, 2 feet
8 2 12 3
5 yards, 2 feet, 8 inches 5 yards, 2/3 yard,
8/36 yard
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Write the 5 to the right of the above.
The process makes sense when you think of writing
212 inches in terms of inches, feet, and yards.
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One of the homework problems ask you to find
rules and perform division for other numbers. On
pages 280 281 you find examples of division by
75 and 24. Lets do 355 75.
The most famous problem in the Book of the Abacus
led to the sequence we now call the Fibonacci
numbers. The problem is
How many pairs of rabbits can be bred from one
pair in a year, if the pairs produce a pair every
month and begin to breed in the second month
after birth?
How does the sequence arise?
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Why does the sequence a01, a11, a22, a33,
of Fibonacci numbers satisfy the relationship an
an-1 an-2?
The number of pairs at the start of month n is
the number of pairs that were there at the start
of month n 1 (assuming no rabbits die) together
with the number of new pairs of babies. So an
an-1 number of new pairs of babies at the
start of month n.
Why does the number of new pairs of babies at
the start of month n an-2, the total number of
pairs alive an that start of month n-2?
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Fibonacci also wrote The Book of Squares in which
he dealt with problems such as those found on
pages 284 285.
On page 286 the author of our textbook indicates
that Fibonacci proved Proposition 10.3 by
induction, but he doesnt introduce the concept
of proof by mathematical induction until page
291. One of the homework problems asks what it
means to prove something by induction. The
discussion on pages 291 292 will help you with
this.
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Lets consider Proposition 10.5. It introduces
congruous numbers and congruent squares.
Rewritten in modern language, Proposition 10.5
asks us to find a number c such that for some
numbers x, y, and z, we have x2 c y2 and z2
c y2 in other words x2 c y2 and y2 c z2.
Square numbers are sums of odd numbers, so c must
be a sum of odd numbers in two different ways.
y2
x2
z2
y2
c
c
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The number c is a congruous number and x2, y2,
and z2 are its congruent squares. We know 24
is a congruous number and 12, 52, and 72 are its
congruent squares since 12 24 52 and 72 24
52.
In Proposition 10.6 Fibonacci shows how to form
new congruous numbers and congruent squares from
old ones. How does he suggest we do this? One
of the homework problems asks you to illustrate
this proposition with an example.
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Fibonacci investigated congruous numbers and
congruent squares because he wanted to solve
Problem 10.6. This problem was suggested by a
fellow mathematician in the court of Frederick I.
The congruous number in Problem 10.6 is c 5.
The investigation and solution of this problem
make up The Book of Squares.
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When interest in mathematics returned to Europe
at the close the Dark Ages, Fibonaccis books
were closely studied. His Book of the Abacus
helped introduced the number system, while The
Book of Squares contained results of theoretical
interest.
Theoretical results produced in the Middle Ages
can also be found in the work of Jordan de
Nemore. His Regarding the Given Numbers consists
of a list of different types of equations
together with instructions for reducing and
solving them. Examples appear as Propositions
10.7 10.10.
One of the homework problems asks you to
translate Proposition 10.7 and its proof into a
modern form. Lets start it now.
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Lets also translate Proposition 10.9. The
homework asks you to translate Proposition 10.10.
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The Art of the Calculator by Ben Gershon was
another piece of theoretical mathematics to
emerge from Europe during the Dark Ages. It
contained the result which our book labels
Proposition 10.12 (page 292). Our author calls
it a combination of the associative and the
commutative laws of multiplication. How would
you translate the theorem? How do the
associative and commutative laws appear?
Gershons Astronomy contains rules for finding
approximate solutions to equations using double
false position. The rules produce exact results
for linear equations and appear in multiple,
similar forms to avoid the use of negative
numbers.
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Rule 10.1 on page 293 can be translated as
follows Suppose we are given the equation
f(x) c for some constant c. Suppose we also
know that f(q1) c q2 for a given q1, and that
f(q3) c q4 for a given q3. Then we should
find q5 with q3 lt q5 lt q1 such that (q1
q5)/(q1 q3) q2/(q2 q4). If f(x) c is
linear, then f(q5) c, otherwise q5 is an
approximate solution to the equation. Example
10.2 shows how to apply the rule for f(x) 24
when we are given that f(10) 30 and f(6) 15.
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Example 10.3 also illustrates Rule 10.1 and
contains two typos the difference is 1 and the
fifth quantity is 4. The homework asks you to
explain Rule 10.2 in a similar way. It also asks
you to rephrase the examples. Exercise 7 on page
295 asks you to apply the Rules 10.1 10.3.
Which rule should we use for part (a)? For part
(d)?
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The last section of Chapter 10 shows how the work
of Nicholas Oresme gave the first steps towards
the concepts of graphs and integration of
functions.
In his Treatise of the Configuration of Qualities
and Motions, Oresme discusses how qualities such
as velocity can be measured and represented
geometrically.
To represent the velocity of a moving object
geometrically, he would first draw a line
representing time. At each point on the time
line he would draw a perpendicular line
representing the velocity at that point.
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This representation depicts an object moving with
initial velocity v at time 0. At time 1 the
object has the same velocity it started with. At
time 2 the velocity has doubled, and by time 3 it
has tripled. The velocity at time 4 is also 3v,
and at time 5 it has dropped back to v.
v
0
1
2
3
4
5
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The height of a line drawn at point t represents
the intensity, or longitude, of the quality (in
this case velocity) at that point (in this case
time). He referred to the horizonal line as the
extension, or latitude, of the quality. The
quantity of quality was the resulting place
figure.
v
0
1
2
3
4
5
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The height of a line drawn at point t represents
the intensity, or longitude, of the quality (in
this case velocity) at that point (in this case
time). He referred to the horizonal line as the
extension, or latitude, of the quality. The
quantity of quality was the resulting place
figure.
v
0
1
2
3
4
5
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The height of a line drawn at point t represents
the intensity, or longitude, of the quality (in
this case velocity) at that point (in this case
time). He referred to the horizonal line as the
extension, or latitude, of the quality. The
quantity of quality was the resulting place
figure.
the quantity of velocity
v
0
1
2
3
4
5
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What does the area of the quantity of velocity
below tell you?
velocity time distance
The area gives you the total distance traveled by
the object.
the quantity of velocity
v
0
1
2
3
4
5
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What does the area of the quantity of velocity
below tell you?
velocity time distance
The area gives you the total distance traveled by
the object.
v
0
1
2
3
4
5
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What does the area of the quantity of velocity
below tell you?
velocity time distance
The area gives you the total distance traveled by
the object.
2v
2v
v
v
5v
0
1
2
3
4
5
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The author of our textbook writes on page 297,
In a sense, Oremse skipped Cartesian graphs and
went straight to the definite integral. Why?
2v
2v
v
v
5v
0
1
2
3
4
5
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Oresme dealt with quantities of qualities that
were combinations of figures with straight sides.
He called these uniform if the qualities were
constant, or uniformly difform if they increased
or decreased linearly. Other quantities were
difformly difform.
uniform
difformly difform
uniformly difform
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Exercise 1 on page 300 asks you to sketch the
configuration of velocities for various objects.
Lets do parts (a) and (c) now.
Exercise 2 asks you to find the uniform velocity
another body would have to travel during the same
time interval, so that the quantity of quality of
the uniformly moving body is equal to the
quantity of quality of the difformly moving body
for the bodies considered in the previous
problem. What does this mean? Lets try do it
for (a) and (c).
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In the Treatise, Oresme dealt with two
counterintuitive situations (see page 298).
With Example 10.8 he shows how an object can
accelerate to infinite velocity yet cover a
finite distance.
Oresme also claims that an object can travel for
an infinite amount of time, yet cover a finite
distance. How is this possible? This concept
appears as exercise 3 on page 300.
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