1 Yes, I read a large magnitude of papers. Great and happy scribing. Perfectly written. Wonderful for me – Jim Troutman

2 Cultural Connection The Asian Empires China before 1260 A.D. Student led discussion. India before 1206 A.D. Rise of Islam 622 A.D 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 See page 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 = 1729 = is very nearly an integer. It is about  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!!!! * *

11 §7-5 India - Ramanujan 2 Srinivasa Ramanujan (1887 – 1920) * * *

12 §7-6 Number Computing Student Discussion.

13 §7-6 Multiplication 237 x 45 = 237 x x x x

14 §7-6 Multiplication 237 x 45By lattice

15 §7-7 Arithmetic and Algebra Student Discussion.

16 §7–8 Geometry and Trigonometry Student Discussion.

17 §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.

18 §7–9 Contrast Greek & Hindu Student Discussion.

19 §7–10 Arabia – Rise of Moslem Culture Student Discussion.

20 §7–11 Arabia – Arithmetic & Algebra Student Discussion.

21 §7–11 Arabia – Arithmetic & Algebra Casting out nines. Method of false positioning. x + x/4 = 30 Try ____ ?

22 §7–12 Arabia – Geometry & Trigonometry Student Discussion.

23 §7– 13 Arabia Etymology Student Discussion.

24 §7–14 Arabia – Contribution Student Discussion.

25 Assignment Read Chapter 8

26 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 OR x x p/2

27 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 x x = 39 x x + 25 = = 64 x + 5 = 8 x = 8 – 5 = 3

28 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 OR xp/4

29 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 x = 39 x x + 4 · (25/4) = · 25/4 = 64 x + 5 = 8 x = 8 – 5 = 3 xp/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 ·... · n r ), namely,  = a 1 N 1 x 1 + a 2 N 2 x a r N r x 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 )

31 Chinese Remainder Theorem 2b  = a 1 N 1 x 1 + a 2 N 2 x 2 + a 3 N 3 x 3 n 1 = 3 n 2 = 5 n 3 = 8 then x  2 (mod 3) x  1 (mod 5) x  4 (mod 8) has a unique solution. a 1 = 2 a 2 = 1 a 3 = 4 n = n 1 · n 2 · n 3 = 3 · 5 · 8 = 120 N 1 = 120 | 3 = 40 N 2 = 120 | 5 = 24 N 3 = 120 | 8 = 15 and 40 · x 1  1 (mod 3) yields x 1 = 1 24 · x 2  1 (mod 5) yields x 2 = 4 15 · x 3  1 (mod 8) yields x 3 = 7

32 Chinese Remainder Theorem 2b  = a 1 N 1 x 1 + a 2 N 2 x 2 + a 3 N 3 x 3 then is a unique solution to a 1 = 2 a 2 = 1 a 3 = 4 N 1 = 40 N 2 = 24 N 3 = 15 Substitute into x 1 = 1 x 2 = 4 x 3 = 7  = 2 · 40 · · 24 · · 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