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3.141592653589793238

Yes, I read a large magnitude of papers. Great and happy scribing. Perfectly written. Wonderful for me – Jim Troutman. 3.141592653589793238. Cultural Connection. The Asian Empires. China before 1260 A.D. India before 1206 A.D. Rise of Islam 622 A.D. - 1250 A.D. Student led discussion.

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3.141592653589793238

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  1. Yes, I read a large magnitude of papers. Great and happy scribing. Perfectly written. Wonderful for me – Jim Troutman 3.141592653589793238

  2. Cultural Connection The Asian Empires China before 1260 A.D. India before 1206 A.D. Rise of Islam 622 A.D. - 1250 A.D. Student led discussion.

  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 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!!!!

  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 5 + 237 x 40 Left to Right 1 8 9 4 1 0 5 5 8 2 8 0 2 3 7 x 5 2 3 7 x 4 0 6 1 0 5 6 5 1 1 8 5 + 9 4 8 0

  14. §7-6 Multiplication 237 x 45 By lattice 2 3 7 0 8 1 2 2 8 4 1 0 1 5 3 5 5 1 0 6 6 5

  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 p/2 x x p/2 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

  27. Quadratic Solutions 1b 10/2 x x 10/2 x 2 + p x = q x 2 + 10 x = 39 P = 10, q = 39, p/2 = 5 and (p/2) 2 = 25 x 2 + 10 x = 39 x 2 + 10 x + 25 = 39 + 25 = 64 (x + 5) 2 = 8 2 x + 5 = 8 x = 8 – 5 = 3

  28. Quadratic Solutions 2 p/4 x p/4 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

  29. Quadratic Solutions 2b p/4 x p/4 x 2 + p x = q x 2 + 10 x = 39 P = 10, q = 39, p/4 = 5/2 and (p/4) 2 = 25/4 x 2 + 10 x = 39 x 2 + 10 x + 4 · (25/4) = 39 + 4 · 25/4 = 64 (x + 5) 2 = 8 2 x + 5 = 8 x = 8 – 5 = 3

  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)

  31. Chinese Remainder Theorem 2b N 1 = 120|3 = 40 a 1 = 2 n 1 = 3 40 · x 1 1 (mod 3) yields x 1 = 1 x  2 (mod 3) n 2 = 5 N 2 = 120|5 = 24 a 2 = 1 x  1 (mod 5) 24 · x 2 1 (mod 5) yields x 2 = 4 a 3 = 4 N 3 = 120|8 = 15 n 3 = 8 x  4 (mod 8) 15 · x 3 1 (mod 8) yields x 3 = 7 then has a unique solution.  = a 1N1x 1+ a 2N2x 2 + a 3 N 3x 3 n = n 1· n 2· n3 = 3 · 5 · 8 = 120 and

  32. Chinese Remainder Theorem 2b a 1 = 2 N 1 = 40 x  2 (mod 3) x 1 = 1 a 2 = 1 N 2 = 24 x  1 (mod 5) x 2 = 4 N 3 = 15 a 3 = 4 x  4 (mod 8) x 3 = 7 then Substitute into  = a 1N1x 1+ a 2N2x 2 + a 3 N 3x 3  = 2· 40 · 1 + 1· 24 · 4 + 4· 15 · 7 = 596 is a unique solution to Note: Unique modulo 120 (3 · 5 · 8) So 596, 476, 356, 236, 116 are all solutions Confirm 116 is a solution.

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