Download MAT 141 Midterm I Solutions: Geometry Axioms and Models and more Exercises Analytical Geometry and Calculus in PDF only on Docsity!
MAT 141
Midterm I – solutions
Name: KEY
No notes. No calculators. All incidents of cheating or the appearance thereof will be brought to the attention of Student Judicial Affairs.
- (10 Points)
(a) State Euclid’s Postulate 4.
All right angles are congruent.
(b) State the negation of Euclid’s Postulate 4. There are
two right angles that are not congruent.
(c) Rewrite the following sentence using only the logic words if, then, and, either, or, not, is, there is, are, there are, for every and the words point, line, equal, circle, incident, be- tween, congruent and assuming Euclid’s first postulate:
The segment AB is equal to the segment BA.
The set of points whose members are A, B and all points
C incident to the line ↔ AB that are between A and B is equal to the set of points whose members are B, A and all points D incident to the line ↔ BA that are between B and A.
(d) Repeat the exercise above for the following sentence: M is the midpoint of the segment AB.
M is the point on the line ↔ AB that is between A and B and such that the set of points whose members are A, M and all points C incident to the line ↔ AM that are between A and M is congruent to the set of points whose members are M, B and all points D incident to the line ↔ M B that are between M and B.
- (10 Points) Use ruler and compass to erect a perpendic- ular to a given point on a given line. You can see this in a
separate file.
- (10 Points) State the axioms of incidence geometry and prove that if l and m are distinct lines that are not parallel, then l and m have a unique point in common.
I-1: For every pair of distinct points, there is a unique line incident to the two points. I-2: For every line, there are at least two distinct points incident to the line. I-3: There exist three distinct noncollinear points.
- (10 Points) Is the following axiomatized theory consis- tent? Substantiate your answer.
B-1: There are at most four points.
B-2: Every line is incident to at least 3 points.
B-3: There are at least four lines.
B-4: There are parallel lines.
The axiomatized theory is INCONSISTENT. By B-1, there are at most 4 points. By B-2 each line is incident to at least 3 points. Case 1: There are 3 points or fewer. In this case, there can at most one line. We contradict B-3. Case 2: There are 4 points. In this case, there can be at most 4 lines, each characterized by the point to which it is not incident. In Cases 1 there are no lines and hence to parallel lines. In Case 2 there is no pair of distinct lines and hence no par- allel lines. In Case 3, consider a pair of distinct lines. Each is characterized by the point to which it is not incident. This leaves two points to which both lines are incident. They are thus not parallel. In all cases, B-4 is contradicted. Thus the axiomatized theory is INCONSISTENT.
ADDENDUM: A more elegant argument: The axiomatized theory is INCONSISTENT. Assume that it is consistent. Then by B-4, there are parallel lines. By B-2, each of these
lines is incident to at least 3 points. By the definition of parallel, these 6 points must be distinct. This contradicts B-1, which tells us that there are at most 4 points. Con- tradiction.
