BRIEF ANSWERS TO TEST 1 (PDF file for Test 1)

1. • True. By using a truth table or a rule of logic one can show that the negation of (P=>~Q) is (P and Q).
   • False. The negation of "Every dog has its day" is "At least one dog does not have its day".

     (Remark: It may help to rewrite the statement symbolically: let "d" be a symbol for "dog", let "D" be a symbol for "Day", and let R(d,D) be a symbol whose meaning is "dog d has Day D". Then "Every dog has its day" is written

     (for all d)(there exists D)(R(d,D)),

     while its negation may be simplified to

     (there exists d)(for all D)(~R(d,D)).

     In words, this says "There exists a dog d such that, for every day D, day D is not dog d's." One can smooth this out to "At least one dog does not have its day".)

   • False. To disprove a statement is to prove its NEGATION (not its converse). For example, the negation of (P=>Q) is (P and ~Q), but the converse is (Q=>P) (which is different).

2. • An interpretation of a language is an assignment of meaning to the primitive concepts of the language. A model of a set of sentences in the language is an interpretation for which those sentences are true.
   • An affine plane is an incidence plane that satisfies the Euclidean parallel postulate. The smallest one has 4 points and 6 lines.

3. The Fano plane is the projective completion of the smallest affine plane. It has 7 points and 7 lines. If we define betweenness so that A*B*C means "A, B, and C are distinct collinear points", then axioms B1 and B2 hold. (Remark: in order to prove that axiom B2 holds, one must show that whenever B and D are distinct points on a line, then there exist A, C and E such that A*B*D, B*C*D, and B*D*E. One shows this by taking A = C = E = the unique third point on the line through B and D.)

4. To show that B2 is independent of I1, I2, I3, B1, and B3 we have to describe two models: one where I1-3 and B1, B2, B3 hold, and another where I1-3, B1, ~B2, B3 hold.

   We can take R^2 (the real plane) for the first model.

   For the other model we can take the 3 point incidence plane (a triangle) and define betweenness so that A*B*C NEVER HOLDS. Then I1-3 hold because the triangle is an incidence plane. B1 and B3 hold because there are no instances of betweenness and no lines with 3 distinct points on them. B2 fails because there exist a pair of distinct points B and D on the same line, but no point C such that B*C*D.

BRIEF ANSWERS TO TEST 2 (PDF file for Test 2)

1. • False. This is the Crossbar Theorem.
   • False. There is no reason for B and D to be on opposite sides of Line(A,C). (You should draw a picture illustrating why the claim is false.)

2. • Suppose that Angle(A) = Angle(CAD) and Angle(B) = Angle (EBF). Then Angle(CAD) is less than Angle (EBF) if there is a ray Ray(B,G) between Ray(B,E) and Ray(B,F) such that Angle(EBG) is congruent to Angle (CAD).

   • An ordered field is Euclidean if it is closed under square roots of positive elements.

3. • Assume that Ray(A,E) is opposite to Ray(A,D). Since Ray(A,D) and Ray(A,E) are between Ray(A,B) and Ray(A,C), points D and E are in Int(Angle(BAC)) by definition of "betweenness for rays". By Proposition 3.8(b), if D is in Int(Angle(BAC)), then no point on the ray opposite to Ray(A,D), i.e., no point of Ray(A,E), is in Int(Angle(BAC)). But point E is such a point, so we have a contradiction.

   • By the Crossbar Theorem, both Ray(A,D) and Ray(A,E) intersect segment BC. After renaming, we may assume that the intersection points are D and E. If any of the points B, C, D or E are equal, then it is easy to see that either the rays Ray(A,D) and Ray(A,E) are not distinct, or at least one of them is not between Ray(A,B) and Ray(A,C). Therefore, B, C, D and E are distinct points of segment BC. By Proposition 3.5, segment BC is the union of BE and EC. Hence D belongs to BE or EC and is not an endpoint, which implies that either B*D*E or E*D*C holds. In the first case, Ray(A,D) is between Ray(A,E) and Ray(A,B) (by the Proposition 3.7 and the definition of "betweenness for rays"), and in the second case Ray(A,D) is between Ray(A,E) and Ray(A,C) (same reasons).

   • Assume that we have the first case of the second part of this problem, namely that Ray(A,D) is between Ray(A,E) and Ray(A,B). (The other case is similar.) Then by definition of "less than", Angle(DAE) is less than Angle(BAE). Again by definition of "less than", Angle(BAE) is less than Angle(BAC). By transitivity, Angle(DAE) is less than Angle(BAC).