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Greek Mathematics II Fall 2021 - R Greek Mathematics II Fall 2021 - R. L. Herman Greek Number Theory • Pythagorean Theorem 2 2 2 x + y = z ; x; y; z integers. n = 3 • Diophantine Equations Solve 3x + 5y = 1; x; y integers. n = 6 • Euclid • Proved # primes infinite, Book IX, Prop 20 n = 10 • Perfect Numbers, Book VII, Def 22, Book IX, Prop 36 Euclid proves: If 2n − 1 is prime, then n = 15 (2n − 1) 2n−1 is perfect. n Mersenne prime: 2 − 1: Figure 1: Polygonal Numbers • Polygonal Numbers History of Math R. L. Herman Fall 2021 1/20 Euclidean Algorithm - Book VII, Prop 1 • gcd(a; b): Greatest common divisor of a and b. Continuing the computation: • Algorithm ai bi 210 45 a1 = max(a; b) − min(a; b) 165 45 b1 = min(a; b) 120 45 repeat 75 45 30 45 Terminates when ai+1 = bi+1: 15 30 Example: Find gcd(210; 45): 15 15 We find gcd(210; 45) = 15: a1 = 210 − 45 = 165 b1 = 45 History of Math R. L. Herman Fall 2021 2/20 Euclidean Algorithm - Another Approach Find the greatest common divisor of positive integers, a and b: • If a < b; exchange a and b: • Divide a by b and get the remainder, r: Thus, a = qb + r: • If r 6= 0, replace a by b and b by r: Repeat the division. • If r = 0, report gcd(a; b) = b: Example: Find gcd(210; 45): 210 = 4 · 45 + 30 45 = 1 · 30 + 15 30 = 2 · 15 Thus, gcd(210; 45) = 15: History of Math R. L. Herman Fall 2021 3/20 Pell's Equation (1611-1685) • x 2 − Ny 2 = 1; N is a nonsquare integer, and x; y are integer solutions. • Example of a Diophantine equation. p • Related to 2 : x 2 − 2y 2 = 0; p y = 1 ) x = 2: • x 2 − 2y 2 = 1; p x If x; y large, then y ≈ 2: • Known to Pythagoreans, Diophantus, and • Archimedes' Cattle Problem can be reduced to solving Pell's Equation. • Brahmagupta (598-570) first to solve. History of Math R. L. Herman Fall 2021 4/20 Let p z = x1 + y1 n; p w = x2 + y2 n; p zw = x3 + y3 n; Then • Example: x 2 − 3y 2 = 1 x = x x + ny y ; • Guess (2; 1): 3 1 2 1 2 p So, z = 2 + 3 = w: y3 = x1y2 + x2y1: p 2 zw = 2 + 3 Since Norm(zw) = 1; p (x3; y3) is a solution. = 7 + 4 3: Then, (7; 4) is a solution. Pell's Equation General Solution • x 2 − ny 2 = 1; n is a nonsquare integer and x; y are integer solutions. p • Let z = x + y n; x; y 2 p Z andz ¯ = x − y n: • Norm(z) = zz¯ = x 2 − ny 2 = 1: • Norm(zw) = Norm(z)Norm(w): History of Math R. L. Herman Fall 2021 5/20 • Example: x 2 − 3y 2 = 1 • Guess (2; 1): p So, z = 2 + 3 = w: p 2 zw = 2 + 3 p = 7 + 4 3: Then, (7; 4) is a solution. Pell's Equation General Solution • x 2 − ny 2 = 1; n is a nonsquare integer and Let x; y are integer solutions. p p z = x + y n; • Let z = x + y n; x; y 2 1 1 p Z p andz ¯ = x − y n: w = x2 + y2 n; p • Norm(z) = zz¯ = x 2 − ny 2 = 1: zw = x3 + y3 n; • Norm(zw) = Norm(z)Norm(w): Then x3 = x1x2 + ny1y2; y3 = x1y2 + x2y1: Since Norm(zw) = 1; (x3; y3) is a solution. History of Math R. L. Herman Fall 2021 5/20 Then, (7; 4) is a solution. Pell's Equation General Solution • x 2 − ny 2 = 1; n is a nonsquare integer and Let x; y are integer solutions. p p z = x + y n; • Let z = x + y n; x; y 2 1 1 p Z p andz ¯ = x − y n: w = x2 + y2 n; p • Norm(z) = zz¯ = x 2 − ny 2 = 1: zw = x3 + y3 n; • Norm(zw) = Norm(z)Norm(w): Then • Example: x 2 − 3y 2 = 1 x = x x + ny y ; • Guess (2; 1): 3 1 2 1 2 p So, z = 2 + 3 = w: y3 = x1y2 + x2y1: p 2 zw = 2 + 3 Since Norm(zw) = 1; p (x3; y3) is a solution. = 7 + 4 3: History of Math R. L. Herman Fall 2021 5/20 Pell's Equation General Solution • x 2 − ny 2 = 1; n is a nonsquare integer and Let x; y are integer solutions. p p z = x + y n; • Let z = x + y n; x; y 2 1 1 p Z p andz ¯ = x − y n: w = x2 + y2 n; p • Norm(z) = zz¯ = x 2 − ny 2 = 1: zw = x3 + y3 n; • Norm(zw) = Norm(z)Norm(w): Then • Example: x 2 − 3y 2 = 1 x = x x + ny y ; • Guess (2; 1): 3 1 2 1 2 p So, z = 2 + 3 = w: y3 = x1y2 + x2y1: p 2 zw = 2 + 3 Since Norm(zw) = 1; p (x3; y3) is a solution. = 7 + 4 3: Then, (7; 4) is a solution. History of Math R. L. Herman Fall 2021 5/20 Eudoxus of Cnidus (c.390 { c. 337 BCE) • Studied under Plato • Taught Aristotle • Astronomer, Mathematician N = 8 • Theory of Proportions Circles: A / r 2, Spheres: V / r 3, Volume of a pyramid N = 18 Volume of a cone • Studied reals, continuous quantities • Method of Exhaustion Due to Antiphon (480{411 BCE) N = 38 Area from a sequence of inscribed polygons Figure 2: Method of Exhaustion History of Math R. L. Herman Fall 2021 6/20 Archimedes of Syracuse (287-212 BCE) • Went to Alexandria, Egypt then, back to Syracuse, Sicily • Greatest Mathematician of Antiquity • Mathematician, Engineer, Inventor • Archimedean screw, lever, pulley • King Heiro II's crown - Eureka Archimedes Principle of Bouyancy • According to Plutarch (46-120) • Marcellus - Syracuse 212 BCE • Claw of Archimedes • Heat Ray • Prone to intense concentration • Death of Archimedes History of Math R. L. Herman Fall 2021 7/20 • Regular Polygons (n-gon) 1 Ap = 2 hQ; Q = Perimeter • Inscribed Polygons 1 Ap = 2 anh < area of circle • Circumscribed Polygon q 180 2 a2 a = 2R sin n ; h = R − 4 : • Approximation of π; Ap Ap R2 < π < h2 Archimedes' Mathematics • Mastered Euclid and Eudoxus' (c. 390-337 BCE) Method of Exhaustion C • Measurement of a Circle C A D = const.; D2 = const. D A = area History of Math R. L. Herman Fall 2021 8/20 • Inscribed Polygons 1 Ap = 2 anh < area of circle • Circumscribed Polygon q 180 2 a2 a = 2R sin n ; h = R − 4 : • Approximation of π; Ap Ap R2 < π < h2 Archimedes' Mathematics • Mastered Euclid and Eudoxus' (c. 390-337 BCE) Method of Exhaustion • Measurement of a Circle C A D = const.; D2 = const. • Regular Polygons (n-gon) h 1 Ap = 2 hQ; Q = Perimeter R History of Math R. L. Herman Fall 2021 8/20 • Circumscribed Polygon q 180 2 a2 a = 2R sin n ; h = R − 4 : • Approximation of π; Ap Ap R2 < π < h2 Archimedes' Mathematics • Mastered Euclid and Eudoxus' (c. 390-337 BCE) Method of Exhaustion • Measurement of a Circle C A D = const.; D2 = const. • Regular Polygons (n-gon) h 1 Ap = 2 hQ; Q = Perimeter R • Inscribed Polygons 1 Ap = 2 anh < area of circle History of Math R. L. Herman Fall 2021 8/20 • Approximation of π; Ap Ap R2 < π < h2 Archimedes' Mathematics • Mastered Euclid and Eudoxus' (c. 390-337 BCE) Method of Exhaustion • Measurement of a Circle C A a D = const.; D2 = const. • Regular Polygons (n-gon) h 1 Ap = 2 hQ; Q = Perimeter R • Inscribed Polygons 1 Ap = 2 anh < area of circle • Circumscribed Polygon q 180 2 a2 a = 2R sin n ; h = R − 4 : History of Math R. L. Herman Fall 2021 8/20 Archimedes' Mathematics • Mastered Euclid and Eudoxus' (c. 390-337 BCE) Method of Exhaustion • Measurement of a Circle C A D = const.; D2 = const. • Regular Polygons (n-gon) h 1 Ap = 2 hQ; Q = Perimeter R • Inscribed Polygons 1 Ap = 2 anh < area of circle • Circumscribed Polygon q 180 2 a2 a = 2R sin n ; h = R − 4 : • Approximation of π; Ap Ap R2 < π < h2 History of Math R. L. Herman Fall 2021 8/20 Estimating π • Approximation of π; Ap Ap 2 < π < 2 ; R h Circumscribed • Recall 4 Ap a = 2R sin 180 ; 2 Inscribed q n h = R2 − a2 = R cos 180 ; 0 4 n 0 2 4 6 8 10 12 14 16 18 1 180 Ap = 2 anh = nhR sin n : n • Therefore, n 360 180 sin < π < n tan ; 2 n n h • Hexagon (n = 6); R 2:598 < π < 3:464: • Archimedes - up to 96-gon 3:1394 < π < 3:1427: History of Math R. L. Herman Fall 2021 9/20 Archimedes' Inscribed and Circumscribed n-gons Consider a fixed circle of radius R: 180 • Inscribed n-gon: h = R cos n ; 180 n = 5 n = 6 Ai = nhR sin n : • Circumscribed n-gon: R 180 r = 180 ; Ac = nHr sin n : cos n • Thus, 180 A = nR2 tan ; n = 7 n = 8 i n n 360 A = R2 sin : c 2 n • This gives n 360 180 n = 9 20 sin < π < n tan ; 2 n n History of Math R. L. Herman Fall 2021 10/20 Early Approximations of π • Bible, π ≈ 3: 4 4 256 • Egyptians, 3 = 81 ≈ 3:1604938 • Ptolemy (150), 360-gon, 3.1416 • Chinese (430-501) 355 ≈ 3:14159292 • Leibniz-Madhava 113 π 1 1 = 1 − + − · · · 3927 4 3 5 • Hindu (1100) 1250 ≈ 3:1416 • Euler • Viete', 393,216-gon, π to 9 places.
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