14 – The Later 19 th Century – Arithmetization of Analysis

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14 – The Later 19 th Century – Arithmetization of Analysis. The student will learn about. the contributions to mathematics and mathematicians of the late 19 th century. §14-1 Sequel to Euclid. Student Discussion. §14-1 Sequel to Euclid. - PowerPoint PPT Presentation

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14 – The Later 19th Century – Arithmetization of Analysis

The student will learn about

the contributions to mathematics and mathematicians of the late 19th century.

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§14-1 Sequel to Euclid

Student Discussion.

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§14-1 Sequel to Euclid

“. . . A course in this material is very desirable for every perspective teacher of high-school geometry. The material is definitely elementary, but not easy, and is extremely fascinating.”

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§14-2 Three Famous Problems

Student Discussion.

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§14-2 Construction Limits1. Can construct only algebraic numbers.

i.e. solutions to polynomial equations with rational coefficients. Example : x2 – 2 = 0

Note: non-algebraic numbers are transcendental numbers.

2. Can not construct roots of cubic equations with rational coefficients but with no rational roots.

Descartes’ rational root test. Example 8x3 – 6x –1 = 0

Possible rational roots are ± 1, ± ½, ± ¼, and ± 1/8.

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§14-2 Quadrature of a Circle 2

Reduces to the equation –

s 2 = π r 2 or s = r π

s

r

s2 = r2However, π is not an algebraic number and hence cannot be constructed.

x 2 - π = 0

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§14-2 Duplication ProblemReduces to the equation x 3 = 2 or x 3 – 2 = 0

But this has no rational roots and hence is not possible.

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§14-2 Angle TrisectionTrig Identity cos θ = 4 cos 3 (θ/3) – 3 cos (θ/3)

Let θ = 60º and x = cos (θ/3) then the identity becomes:

½ = 4 x 3 – 3x or 8x 3 – 6x – 1 = 0

But this has no rational roots and hence is not possible.

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§14 -3 Compass or Straightedge

Student Discussion.

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§14 -3 CompassLorenzo Mascheroni and Georg Mohr

All Euclidean constructions can be done by compass alone.

Need only show:

1. Intersection of two lines.

2. Intersection of one line and a circle.

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§14 -3 StraightedgeJean Victor Poncelet

All Euclidean constructions can be done by straight edge alone in the presence of one circle with center. Fully developed by Jacob Steiner later.

Need only show:

1. Intersection of one line and a circle.

2. Intersection of two circles.

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§14 -3 Compass or StraightedgeAbû’l-Wefâ proposed a straightedge and a rusty compass.

Yet others used a two-edged ruler with sides not necessarily parallel.

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§14- 4 Projective Geometry

Student Discussion.

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§14- 4 PonceletPrinciple of duality

Two points determine a line.

Two lines determine a point.

Principle of continuity – from a case proven in the real plane there is a continuation into the imaginary plane.

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14-5 Analytic Geometry

Student Discussion.

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14-5 Julius PlückerLine Coordinates

• A line is defined by the negative reciprocals of its x and y intercepts.

• A point now becomes a “linear” equation.

• A line becomes an ordered pair of real numbers.

More later.

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§14 - 6 N-Dimensional Geometry

Student Comment

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§14 - 6 N-Dimensional GeometryHyperspace for n dimensions and n > 3.

Emerged from analysis where analytic treatment could be extended to arbitrary many variables.

n dimensional space has -• Points as ordered n-tuples (x 1, x 2, . . . , x n)• Metric d (x, y) = [(x 1–y1) 2 + (x 2–y2) 2 + . . . +(x n–yn) 2]• Sphere of radius r and center (a 1, a 2, . . . , a n ) (x 1–a1) 2 + (x 2–a2) 2 + . . . +(x n–an) 2 = r2

• Line through (x 1, x 2, . . . , x n) and (y 1, y 2, . . . , y n) (k (y 1–x1) 2, k (y 2–x2) 2, . . . , k (y n–xn) 2 ) k 0.

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§14-7 Differential Geometry

Student Discussion.

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§14 – 8 Klein and theErlanger Program

Student Discussion.

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§14 – 9 Arithmetization of Analysis

Student Discussion.

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§14–10 Weierstrass and Riemann

Student Discussion.

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§14–11 Cantor, Kronecker, and Poincaré

Student Discussion.

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§14–12 Kovalevsky, Noether and Scott

Student Discussion.

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§14–13 Prime Numbers

Student Discussion.

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§14–13 How many Prime NumbersIs there a formula to calculate the number of primes less than some given number?

Consider the following:

n Number of primes < n

10 4

100 14

. . . . . .

10 9 50,847,534

10 10 455,052,511

n ? ? ?

Confirm this.

n / ln n

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§14–13 2 n - 1

2 n - 1 generates primes:

n 2 n - 1

1 1

2 3

3 7

4 31

5 63

. . . . . .

127 39 digit prime

521 How many digits?

216,091 64,828 digits

Composite

2 n # digits

10 4

20 7

30 10

40 13

. . .

10k 1 + 3k

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§14–13 Fermat thought that generated only primes:

n

1 5

2 17

3 257

4 65,537

5 4,294,967,297

Also composite for n = 145 and lots of others

Composite

12n2

12n2

12n2

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§14–13 Palindromic Primes11, 131, 151, . . . , 345676543, . . .

There are no four digit palindromic primes. WHY?

There are 5,172 five digit palindromic primes.

11 is the only palindromic primes with an even number of digits.

Homework – find the smallest five digit prime.

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Functions to Generate Primesf (n) = n 2 – n + 41 yields primes for n < 41.

41 43 47 53 61 71

n 1 2 3 4 5 6 . . .

f (n)

f (n) = n 2 – 79 n + 160 yields primes.

Homework – find a polynomial that yields all primes.

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Twin Primes2, 3 and 5, 7 and 11, 13 and 137, 139 and 1007, 1009 and infinitely many more.

My new phone number is 2 5 · 5 3 · 11 · 191

Goldbach’s Conjecture – Every even integer > 2 can be written as a sum of two prime numbers. 1000 = 3 + 997Homework – write 2002 as the sum of two primes.

My new phone number is 2 5 · 5 3 · 11 · 191 Note 2, 5, 11 and 191 are the first of twin primes.My new phone number is 2 5 · 5 3 · 11 · 191 Note 2, 5, 11 and 191 are the first of twin primes. Note 2, 5, 11, and 191 are all palindromic primes.

Goldbach Bingo

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Poincaré’s ModelHyperbolic Geometry

Normal points

Ideal points

Ultra-ideal points

l

m

no

P

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Assignment

Papers presented from Chapters 11 and 12.

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