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 PresentationTRANSCRIPT
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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.