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Modern Geometry Fall 2011. Portfolio By: Alexandria Croom Professor: Dr. David James. Description of project. - PowerPoint PPT Presentation
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MODERN GEOMETRYFALL 2011
P O R T F O L I OB Y : A L E X A N D R I A C R O O MP R O F E S S O R : D R . D A V I D J A M E S
DESCRIPT
ION OF
PROJECT
As a senior at Howard University, I enrolled in a directed reading course with Dr. David James. This project is a way to document my progress in the course. This document is a work in progress that will continued to be completed throughout the semester. I would like to especially thank Dr. James for being willing to teach this course. His enthusiasm has brought much joy to the learning process.
HOW WAS MODERN
GEOMETRY DEVELO
PED?
Check out this video!
TABLE OF CONTENTSThe Geometries
EuclideanHyperbolicSpherical
AxiomsDistanceIncidenceLinesReal Ray
Theorems & ProofsUnique Middle TheoremTriangle Inequality
THE GEO
METRIES
E X A M P L E S OF F
I VE S
Y S T E M S
EUCLIDEAN GEOMETRY
The Euclidean plane points and lines includes: Coordinates - set of points is an ordered pair Equation of lines
Nonvertical y= mx+ b Vertical x= a
Distance Formula e(AB)=
HYPERBOLIC GEOMETRYConsists of all points inside (but not on) the unit circle in the
Euclidean plane. That is all (x, y)Lines in H are the chords of the circleDistance in the Hyperbolic Plane is calculated:
Where M, N are endpoints of chords, and A, B are points on a line
SPHERICAL GEOMETRYS(r): notation for spherical plane surface of a sphere of radius rS- set of all (x, y, z) with Definition: Great circles- intersection of sphere with plane that cuts it in
halfHow is distance defined in S?The distance between two points is the length of the minor arc of the
great circle (line) through A and B
AXIOMS
A D E S C R I P T I ON O
F ON E O
R MO
R E MA T H S T R U C T U R E S C O
N S I S T I N G
O F T E R MS , D
E F I N I T I ON S , A S S U M
P T I ON S , T H E O
R E MS , A N D
P R OP O
S I T I ON S
AXIOMS OF DISTANCEFor all points P and Q
1. Positivity: PQ 2. Definiteness: iff 3. Symmetry:
AXIOMS OF INCIDENCEIn a system1. There are at least two different lines2. Each line contains at least two different points3. Each two different points lie in at least one line4. Each two different points P, Q, with lie in at most one line
THREE AXIOMS FOR THE LINEDefinition: Point B lies between points A and C (written A-B-C) provided
that A, B, C are different collinear points, and AB+BC = ACUnique Middle Theorem follows (see theorems and proofs)
Betweenness of Points AxiomIf A, B, and C are different collinear points and if then there exists a
betweenness relation among A, B, and C.
Triangle inequality theorem follows (see theorems & proofs)
THREE AXIOMS FOR THE LINEQuadrichotomy Axiom for Points
If A, B, C, X are distinct, collinear points, and if A-B-C, then at least one of following must hold:
X-A-B, A-X-B, B-X-C, OR B-C-X
THREE AXIOMS FOR THE LINENontriviality AxiomFor any point A on a line there exists a point n with
REAL RAY AXIOMDefinition: Let A, B, be points with The ray is the set
For any ray and any real number with there is a point X in with
Leads to Theorem 8.2 ( see theorems & proofs)
THEOREMS & PR
OOFS
K E Y T H E O R E MS A N D P O S T U L A T E S F O
R WO
R K I N G I N MO
D E R N
G E OM
E T R Y
UNIQUE MIDDLE THEOREMIf A-B-C, then both B-A-C and A-C-B are false.Proof:By definition of betweenness, A, B, C are distinct collinear point and AB + BC = AC.Assume B-A-C. Then,
BA + AC = BCBA + AB + BC = BC2AB + BC= BC (by symmetry axiom AB =BA)2AB=0
Thus, AB= 0. This implies that 0 + BC = AC By positive definiteness this would make A=B, contradicting the assumption that they are distinct points
Therefore, B-A-C is false
A-C-B is likewise false.
TRIANGLE INEQUALITY
If A, B, C are distinct and collinear then AB + BC ≥ AC
Proof of Triangle Inequality
Case 1: 𝐴𝐵+ 𝐵𝐶≥ 𝜔
Then by Betweenness of Points Axiom, there is some betweenness relationship between A, B, C.
Without loss of Generality consider two case: 1) A-B-C and 2) A-C-B
1) Since A-B-C by definition AB+BC = AC
Therefore it holds that 𝐴𝐵+ 𝐵𝐶≥ 𝐴𝐶
2) For A-C-B,
𝐴𝐶+ 𝐵𝐶= 𝐴𝐵
𝐴𝐶= 𝐴𝐵− 𝐵𝐶 ≤ 𝐴𝐵+ 𝐵𝐶
Therefore it holds that 𝐴𝐵+ 𝐵𝐶 ≥ 𝐴𝐶
Case 2: 𝐴𝐵+ 𝐵𝐶 > 𝜔
For E, G, H, and M, 𝜔= ∞. Do as longs as 𝐴𝐵+ 𝐵𝐶= 𝑥 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑥∈𝐑 there will always be a 𝜔 greater.
Thus ∃ a betweenness relation fro A, B, C such that by case 1 it holds that 𝐴𝐵+ 𝐵𝐶 ≥ 𝐴𝐶
For A,B, C ∈ H, if 𝐴𝐵+ 𝐵𝐶> 𝜔 then 𝐴𝐵+ 𝐵𝐶 > 𝜋𝑟 = 𝜔
AB + BC must also be the length of the major axis.
It follows AC must be the length of the major axis.
Therefore 𝐴𝐶≤ 𝜋𝑟 = 𝜔< 𝐴𝐵+ 𝐵𝐶
Therefore it holds that 𝐴𝐶≤ 𝐴𝐵+ 𝐵𝐶. Thus, the proposition is true.
THIS IS STILL A WORK IN PROGRESS!
Special Thanks to :Dr. David James
AndThe Howard University Mathematics Department