Download Review Problems from 2.2 through 2.4

Euclidean and Non-Euclidean Geometry – Fall 2007
Dr. Hamblin
Problems from 2.2 through 2.4
The following problems cover topics from Sections 2.2, 2.3, and 2.4.
1. Consider this axiomatic system.
Undefined terms: space, connected
Axiom 1: Space consists of a finite set of integers, called elements.
Axiom 2: Each element is connected to at least two others, and no element is ever connected to
itself.
Axiom 3: The character of an element is the sum of the elements it is connected to. Every
element has a positive character.
Axiom 4: There are at least four elements.
Axiom 5: No two elements have the same character.
a. Show that the following is a model for this axiomatic system:
Elements: 6, -5, 10, 14
Connections: 6  -5, 6  14, 6  10, -5  14, -5  10, 10  14
b. Construct a model for this system that has only two positive elements.
c. Demonstrate that “if every element is odd then every element has an even character” is
not a theorem in this system.
d. Prove that “if every element is even, then every element has an even character” is a
theorem in this system.
e. Show that Axiom 5 is independent from the others.
2. Which of the five incidence axioms I-1 through I-5 are satisfied in the following model?
Points: S = {1, 2, 3, 4, 5}
Lines: {1, 2, 3}, {2, 4}, {1, 4, 5}, {3, 4}, {2, 5}, {3, 5}
Planes: {1, 2, 3, 4}, {1, 2, 3, 5}, {3, 4, 5}
3. You are given that the points A, B, C, and D are all collinear and these distance measurements:
BC = 3, BD = 3, CD = 6, AB = 4, and AC = 4.
a. Prove that C-B-D.
b. Prove that the Ruler Postulate cannot be true in this situation. (Hint: Assign coordinates
and try to find the coordinate of A.)
Euclidean and Non-Euclidean Geometry – Fall 2007
Dr. Hamblin
Solutions
1. a.
All of the axioms are true.
b.
Answers vary. Try using two large positive numbers and two small negative numbers.
c.
Answers vary. Find a model where one element is connected to an odd number of odd
elements.
d.
Proof: The character of each element is the sum of the elements it is connected to. Since
each element is even, the character of each element is a sum of even numbers. From
Discrete Math, we know that any sum of even numbers is even.
e.
Answers vary.
2. Axiom I-3 is false, all others are true.
3. a.
b.
Clearly B, C, and D are collinear. We have CB + BD = BC + BD = 3 + 3 = 6 = CD.
One solution is to let C[0], B[3], and D[6]. Then AB = 4, so the coordinate of A is either -1 or
7, but neither of these is consistent with AC = 4.