1 π It is quite curious that π is related to probability. The probability that two natural numbers...
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Transcript of 1 π It is quite curious that π is related to probability. The probability that two natural numbers...
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πIt is quite curious that π is related to probability. The probability that two natural numbers selected at random will be relatively prime is
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This is quite astonishing since π is derived from a geometric setting.
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Pertinent Dates• ca. 546 BC Thales
• ca. 585 – 500 B.C. Pythagoras
•485 – 410 B.C. Proclus
• 469 - 399 B.C. Socrates
• 455 – 385 B.C. Aristophanes
• 427 – 347 B.C. Plato
• 384 – 322 B.C. Aristotle
• ca. 300 B.C. Euclid
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3 – Pythagorean Mathematics
The student will learn about
Greek mathematics before the time of Alexander the Great.
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Cultural ConnectionThe Philosophers of the Agora
Hellenic Greece – ca. 800 – 336 B.C.
Student led discussion.
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§3-1 Birth of Demonstrative Mathematics
Student Discussion.
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§3-1 Birth of Demonstrative Mathematics
Statement Reason
Theorem: Vertical angles formed by two intersecting lines are equal.
Proof: < 1 = < 3
< 1 + < 2 = 180 Straight line
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< 2 + < 3 = 180 Straight line
< 1 + < 2 = < 2 + < 3 Substitution
< 1 = < 3 Subtraction
QED w 5
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§3-2 Pythagoras and the Pythagoreans
Student Discussion.
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§3-3 Pythagorean Arithmetic 1
Student Discussion.
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§3-3 Pythagorean Arithmetic 2
Perfect numbers equal the sum of their proper divisors. 6
Abundant numbers exceed the sum of their proper divisors. 12
Deficient numbers are less than the sum of their proper divisors. 8
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§3-3 Pythagorean Arithmetic 3Euclid proved that if 2n – 1 is prime then
2n – 1(2n – 1) is perfect.
n 2n 2n – 1 2n – 1 2n – 1(2n – 1)
1 2 1 1 1
2 4 3 2 6
3 8 7 4 28
4 16 15 8 240
5 32 31 16 496
6 64 63 32 2016
7 128 127 64 8128
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§3-3 Pythagorean Arithmetic 3Figurative numbers
Triangular numbers
Tn = =
n
1ii n
n + 1
Square numbers
1 3 6
1 4 9
Sn = n2 = n (n + 1) / 2 + n (n - 1) / 2 = Tn + Tn - 1
Tn = = n (n + 1) / 2
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§3-4 Pythagorean Theorem 1
Student Discussion.
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§3-4 Pythagorean Theorem 2Pythagorean dissection proof.
a
a
a
a
b
b
bb
cc
c
c
c
=
a
a
a
a
b
b
b
b
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§3-4 Pythagorean Theorem 3Bhaskara’s dissection proof.
a
a
a
bb
b
b
a
c
c
c
c
c2 = 4 · ½ · a · b + (b – a)2
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§3-4 Pythagorean Theorem 4Garfield’s dissection proof.
a
b
b
ac
c
½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c2
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§3-5 Irrational Magnitudes 1
Student Discussion.
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§3-5 Irrational Magnitudes 2Geometric interpretation of 2/3.
0 1
2/3
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§3-5 Irrational Magnitudes 3
as the diagonal of a unit square.
Proof that is irrational. (Aristotle 384 – 322 B.C.)
Assume is rational. I.e. = a/b & a and b are relatively prime.
2
2
2 2
Then 2 = a2/b2 and a2 = 2 b2 and hence a is even.
Let a = 2k since it is even and then
4k2 = 2 b2 and hence b is even.
a contradiction hence is not rational.2
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§3-6 Algebraic Identities 1
Student Discussion.
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§3-6 Algebraic Identities 2(a + b)2 = a2 + 2ab + b2.
a
a
a
a
b
b
b b
a + b
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a 2
§3-6 Algebraic Identities 3(a - b)2 = a2 - 2ab + b2.
a - b
a
b
b
b b
a - b
a - b a - b1 2
3 4a 2 – aba 2 – ab – aba 2 – ab – ab + b 2
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§3-7 Geometric Solutions of Equations 1
Student Discussion.
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§3-7 Geometric Solutions of Equations 2
Linear equations a x = b c
ab
c
x
bcaxorx
c
b
a
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§3-7 Geometric Solutions of Equations 3
Quadratic equations x2 = a b.
a b
x
abxorb
x
x
a 2
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§3-8 Transformation of Areas 1
Student Discussion.
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§3-8 Transformation of Areas 1
Construct a square equal in area to a given polygon.
Given ABCDE
Construct BR AC with R on DC
Area ABC = area ARC
Hence area of ABCDE = ARDE
A B
C
D
E
R
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§3-8 Transformation of Areas 2
Construct a square equal in area to a given polygon.
Given ARDE = ABCDE
Construct RS AD with S on ED
Area ARD = area ASD
Hence area of ABCDE = ARDE = ASE A
S
C
D
E
R
Make it a square!
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§3-9 The Regular Solids 1
Student Discussion.
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§3-9 The Regular Solids 2
• Tetrahedron
• Hexahedron
• Octahedron
• Dodecahedron
• Icosahedron
Show models
Show imbedded model.
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§3-10 Postulational Thinking
Student Discussion.
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Assignment
Read Chapter 4.
Outline of Paper 1 due on Monday!