1

Yes, I read a large magnitude of papers.

Great and happy scribing. Perfectly written.

Wonderful for me – Jim Troutman

3.141592653589793238

2

Cultural ConnectionThe Asian Empires

China before 1260 A.D.

Student led discussion.

India before 1206 A.D.

Rise of Islam 622 A.D. - 1250 A.D.

3

7 – Chinese, Hindu, and Arabian Mathematics

The student will learn about

Early mathematics from the east and middle east regions.

4

§7-1 China – sources and periods

Student Discussion.

5

§7-2 China – Shang to the Tang

Student Discussion.

6

§7-3 China – Tang to the Ming

Student Discussion.

7

§7-3 China –Tang to Ming

1 11 21

2 12 22

3 13 28

4 14 56

5 15

6 16 70

7 17

8 18 See page 218

9 19

10 20

Chinese Rod System

8

§7-4 China – Concluding Remarks

Student Discussion.

9

§7-5 India - General Survey

Student Discussion.

10

§7-5 India - Ramanujan

Srinivasa Ramanujan (1887 – 1920)

Taxi Number 1 3 +12 3 = 1729 = 9 3 + 10 3

is very nearly an integer.

It is about 2.62537 10 17 which is an integer followed by twelve zeros. That is, it is eighteen digits a decimal point followed by twelve zeros before the next non zero digit!!!!

163e

*

*

11

§7-5 India - Ramanujan 2

Srinivasa Ramanujan (1887 – 1920)

8dx

x1

)x(log 3

02

2

n4

0n4 )994(

n263901103

)!n(

!)n4(

9801

221

*

*

*

21415926526.322

199

4

12

2

12

§7-6 Number Computing

Student Discussion.

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§7-6 Multiplication237 x 45 = 237 x 5 + 237 x 40

2 3 7 x 51 0

158

52 3 7 x 4 089

24

8 0

1 1 8 5+ 9 4 8 0

1 0 56

6 5

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§7-6 Multiplication

237 x 45 By lattice

2 3 7

4

53 5

2 8

1 5

1 2

1 0

0 8

5661 0

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§7-7 Arithmetic and Algebra

Student Discussion.

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§7–8 Geometry and Trigonometry

Student Discussion.

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§7–8 Geometry and Trigonometry

Area of a cyclic quadrilateral.

K 2 = (s – a)(s – b)(s – c)(s – d)

Diagonals of a cyclic quadrilateral.

bcad

)bdac)(cdab(m2

bcad

)bcad)(bdac(n2

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§7–9 Contrast Greek & Hindu

Student Discussion.

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§7–10 Arabia – Rise of Moslem Culture

Student Discussion.

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§7–11 Arabia – Arithmetic & Algebra

Student Discussion.

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§7–11 Arabia – Arithmetic & Algebra

Casting out nines.

Method of false positioning.

x + x/4 = 30

Try ____ ?

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§7–12 Arabia – Geometry & Trigonometry

Student Discussion.

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§7– 13 Arabia Etymology

Student Discussion.

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§7–14 Arabia – Contribution

Student Discussion.

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Assignment

Read Chapter 8

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Quadratic Solutions 1

The Greeks and Hindu’s used a method similar to our completing the square.

x 2 + p x = q

The unshaded portion to the right is x 2 + p x or q.

If we add the shaded region (p/2) 2 we get a square with side of x + p/2.

(x + p/2) 2 = q + (p/2) 2

2

2

pq

2

px

OR

2

p

2

pqx

2

x

x

p/2

p/2

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Quadratic Solutions 1b

x 2 + p x = q

P = 10, q = 39, p/2 = 5 and (p/2) 2 = 25

(x + 5) 2 = 8 2

x

x

10/2

10/2x 2 + 10 x = 39

x 2 + 10 x = 39

x 2 + 10 x + 25 = 39 + 25 = 64

x + 5 = 8

x = 8 – 5 = 3

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Quadratic Solutions 2

Al-Khowârizmî’s Solution Method

x 2 + p x = q

The unshaded portion to the right is x 2 + p x or q.

If we add the shaded region 4 (p/4) 2 we get a square with side of x + p/2.

(x + 2(p/4)) 2 = q + 4 · (p/4) 2

2

2

pq

2

px

OR

2

p

2

pqx

2

x p/4p/4

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Quadratic Solutions 2b

x 2 + p x = q

P = 10, q = 39, p/4 = 5/2 and (p/4) 2 = 25/4

(x + 5) 2 = 8 2

x 2 + 10 x = 39

x 2 + 10 x = 39

x 2 + 10 x + 4 · (25/4) = 39 + 4 · 25/4 = 64

x + 5 = 8

x = 8 – 5 = 3

x p/4p/4

30

Chinese Remainder Theorem 2

Let n 1, n 2, . . . , n r be positive integers so that the g.c.d. (n i, n j) = 1 for i j, then the system of linear congruencies - x a 1 (mod n 1), x a 2 (mod n

2), . . . , x a r (mod n r). Has a simultaneous solution, which is unique modulo (n 1 · n 2 · . . . · nr), namely,

= a 1N1x 1+ a 2N2x 2 + . . . + a r N rx r where

N k = n 1 · n 2 · · · n r | n k and

x k = solution to N k x k 1 (mod n k)

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Chinese Remainder Theorem 2b

= a 1N1x 1+ a 2N2x 2 + a 3 N 3x 3

n 1 = 3n 2 = 5n 3 = 8

thenx 2 (mod 3)x 1 (mod 5)x 4 (mod 8)

has a unique solution.

a 1 = 2a 2 = 1a 3 = 4

n = n 1· n 2 · n3 = 3 · 5 · 8 = 120

N 1 = 120 | 3 = 40N 2 = 120 | 5 = 24N 3 = 120 | 8 = 15

and

40 · x 1 1 (mod 3) yields x 1 = 124 · x 2 1 (mod 5) yields x 2 = 415 · x 3 1 (mod 8) yields x 3 = 7

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Chinese Remainder Theorem 2b

= a 1N1x 1+ a 2N2x 2 + a 3 N 3x 3

then

is a unique solution to

a 1 = 2a 2 = 1a 3 = 4

N 1 = 40N 2 = 24N 3 = 15

Substitute intox 1 = 1x 2 = 4x 3 = 7

= 2 · 40 · 1 + 1 · 24 · 4 + 4 · 15 · 7 =

x 2 (mod 3)x 1 (mod 5)x 4 (mod 8)

Note: Unique modulo 120 (3 · 5 · 8) So 596, 476, 356, 236, 116 are all solutions

Confirm 116 is a solution.

596