1 π It is quite curious that π is related to probability. The probability that two natural numbers selected at random will be relatively prime is This.

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Presentation transcript:

1 π It is quite curious that π is related to probability. The probability that two natural numbers selected at random will be relatively prime is This is quite astonishing since π is derived from a geometric setting.

2 Pertinent Dates ca. 546 BCThales ca. 585 – 500 B.C. Pythagoras 485 – 410 B.C. Proclus B.C. Socrates 455 – 385 B.C. Aristophanes 427 – 347 B.C. Plato 384 – 322 B.C. Aristotle ca. 300 B.C.Euclid

3 3 – Pythagorean Mathematics The student will learn about Greek mathematics before the time of Alexander the Great.

4 Cultural Connection The Philosophers of the Agora Hellenic Greece – ca. 800 – 336 B.C. Student led discussion.

5 §3-1 Birth of Demonstrative Mathematics Student Discussion.

6 §3-1 Birth of Demonstrative Mathematics StatementReason Theorem: Vertical angles formed by two intersecting lines are equal. Proof: < 1 = < 3 < 1 + < 2 = 180 Straight line < 2 + < 3 = 180 Straight line < 1 + < 2 = < 2 + < 3 Substitution < 1 = < 3 Subtraction QEDw 5

7 §3-2 Pythagoras and the Pythagoreans Student Discussion.

8 §3-3 Pythagorean Arithmetic 1 Student Discussion.

9 §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

10 §3-3 Pythagorean Arithmetic 3 Euclid proved that if 2 n – 1 is prime then 2 n – 1 (2 n – 1) is perfect. n2n2n 2 n – 1 2 n – 1 (2 n – 1)

11 §3-3 Pythagorean Arithmetic 3 Figurative numbers Triangular numbers T n = = n n + 1 Square numbers S n = n 2 = n (n + 1) / 2 + n (n - 1) / 2 = T n + T n - 1 T n = = n (n + 1) / 2

12 §3-4 Pythagorean Theorem 1 Student Discussion.

13

14 §3-4 Pythagorean Theorem 2 Pythagorean dissection proof. a a a a b b b b c c c c c = a a a a b b b b

15 §3-4 Pythagorean Theorem 3 Bhaskara’s dissection proof. a a a b b b b a c c c c c 2 = 4 · ½ · a · b + (b – a) 2

16 §3-4 Pythagorean Theorem 4 Garfield’s dissection proof. a b b a c c ½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c 2

17

18 §3-5 Irrational Magnitudes 1 Student Discussion.

19 §3-5 Irrational Magnitudes 2 Geometric interpretation of 2/ /3

20 §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. Then 2 = a 2 /b 2 and a 2 = 2 b 2 and hence a is even. Let a = 2k since it is even and then 4k 2 = 2 b 2 and hence b is even.   a contradiction hence is not rational.

21 §3-6 Algebraic Identities 1 Student Discussion.

22 §3-6 Algebraic Identities 2 (a + b) 2 = a 2 + 2ab + b 2. a a a a b b bb a + b

23 a 2 §3-6 Algebraic Identities 3 (a - b) 2 = a 2 - 2ab + b 2. a - b a b b bb a 2 – aba 2 – ab – aba 2 – ab – ab + b 2

24 §3-7 Geometric Solutions of Equations 1 Student Discussion.

25 §3-7 Geometric Solutions of Equations 2 Linear equations a x = b c a b c x

26 §3-7 Geometric Solutions of Equations 3 Quadratic equations x 2 = a b. ab x

27 §3-8 Transformation of Areas 1 Student Discussion.

28 §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 AB C D E R

29 §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!

30 §3-9 The Regular Solids 1 Student Discussion.

31 §3-9 The Regular Solids 2 Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron Show models Show imbedded model.

32 §3-10 Postulational Thinking Student Discussion.

33 Assignment Read Chapter 4. Outline of Paper 1 due on Monday!