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MAT 141
Homework 5 – solutions
- Logic - Suppose that the following are true: Danyi likes Faulkner. Jorden likes London. Sean likes Morrison. Amandeep likes Welty.
Decide whether each of the following is true or false. a: If Danyi does not like Faulkner and Jorden does not like London, then Sean likes Morrison. true b: If Danyi does not like Faulkner or Jorden does not like London, then Sean does not like Morrison. true c: If Danyi does not like Faulkner and Jorden does not like London, then Sean likes Morrison, furthermore, Aman- deep does not like Welty. false d: If Danyi does not like Faulkner and Jorden does not like London, then Sean likes Morrison, furthermore, Aman- deep likes Welty or Jorden does not like London. true
- An ordered incidence plane is a model for the three in- cidence axioms and the four betweenness axioms. For each interpretation below, state whether or not it is an incidence plane. If it is an incidence plane, verify all 7
axioms. If it not an incidence plane, state an axiom that fails and provide an example of points, lines that fail the axiom.
Interpretation a: (The Cartesian Plane) Points are ordered pairs (v, w) of real numbers. Lines are expressions of the form x = a or y = mx + b. A point (v, w) is incident to a line x = a if and only if v = a. It is incident to a line y = mx + b if and only if w = mv + b. Given two points, use notation (c, d) < (e, f ) iff either c < e or c = e and d < f. The point (e, f) is between points (c, d) and (g, h) iff (c, d), (e, f), (g, h) are collinear and either (c, d) < (e, f ) < (g, h) or (g, h) < (e, f ) < (c, d).
This interpretation is and ORDERED INCIDENCE PLANE. Verification of Axioms:
I-1: To verify I-1, consider distinct points (c, d), (e, f ). If c = e, then the two points are incident to the unique line x = c. If c ≠ e, the two points are incident to the unique line
y =
f − d e − c
(x − c) + d
I-2: To verify I-2, consider a line l. If l is given by x = a, then (a, 0 ) and (a, 1 ) are incident to l. If l is given by y = mx + b, then ( 0 , b) and ( 1 , m + b) are incident to l.
I-3: To verify I-3, we consider the points ( 0 , 0 ), ( 0 , 1 ) and ( 1 , 0 ). The first two are incident to the unique line x = 0. The latter is not incident to this line. Thus the three points are not collinear.
l is given by x = a, Then c = e and e = g, hence c = g, whence (c, d) and (g, h) lie on the same side. If l is given by y = mx + b, then either d < mc + b and f < me + b or mc + b > d and me + b > f and f < me + b and h < mg + b or or me + b > g and mg + b > h. This can only happen if either d < mc + b and f < me + b and h < mg + b or or d > mc + b and f > me + b and mg + b > h. In both cases, all three points are on the same side of l. To verify B-4(ii), we argue similarly.
Interpretation b: Points are points on the 2-sphere, i.e., points (x, y, z) such that x^2 + y^2 + z^2 = 1. Lines are great circles. (A great circle is the intersection of a plane through the origin with the 2-sphere.) Incidence is as usual.
Interpretation b is NOT AN ORDERED INCIDENCE PLANE. To see this, consider I-1. The antipodal points (0,0,1) and (0, 0, -1), for instance, are incident to several distinct lines.
Interpretation c: Points are points inside the unit disk, i.e., points (x, y) such that x^2 + y^2 < 1. Lines chords, i.e., they are the portions of lines in 2-space that lie in the disk. Incidence is as usual. Betweenness is as usual.
This interpretation is and ORDERED INCIDENCE PLANE. Verification of Axioms:
I-1: To verify I-1, consider distinct points (c, d), (e, f ). The points (c, d), (e, f ) also lie in the Cartesian plane, where
they are incident to a unique line l. The portion ˆl of l that lies in the disk is the unique chord to which (c, d) and (e, f ) are incident in Interpretation c.
I-2: To verify I-2, consider a line ˆl. Then ˆl is the portion of a line l in the Cartesian plane that meets the disk. If l is given by x = a, then (a, 0 ) and (a,
1 − a^2 ~ 2 ) are incident to l. If l is given by y = mx + b, then we claim that there must be at least one point on ˆl. If there were no point incident to ˆl, then ˆl would not be a chord. Call this point P. Now consider (0, b). If (0, b), which is incident to l, is also incident to ˆl (because it is lies inside the unit disk), then we set Q = (0, b). If (0, b) is not incident to ˆl (because it lies outside the unit disk), consider the point R on the unit circle that is between P and (0, b) in the Cartesian plane and choose Q to be a point between P and R. In both cases, P, Q are points on ˆl. (We are tacitly using a Continuity Principle here to guarantee that point R exists.)
I-3: To verify I-3, we consider the points ( 0 , 0 ), ( 0 , 1 ~ 2 ) and ( 1 ~ 2 , 0 ). The first two are incident to the unique line x = 0. The latter is not incident to this line. Thus the three points are not collinear.
B-1: This axiom holds by the definition of betweenness given for the Cartesian plane.
B-2: To verify B-2, consider the two points (c, d) and (e, f). Proceed as in the verification of B-2 for the Cartesian plane but argue as in the end of the verification of I-2 for this interpretation.
B-3: To verify B-3, note that a point is between two others
Note that all points must be incident to l, for otherwise there would be another line incident to B, D, violating the uniqueness portion of I-1. Thus l is incident to the five points A, B, C, D, E. ∎
Claim: Every segment is incident to infinitely many points.
Proof: Let AB be a segment we prove by induction that the number of points incident to AB is unbounded. Base Case: The number of points incident to AB is greater than 0, because by Betweenness Axiom B-2, there is a point between A and B. Inductive Step: We wish to show that if n is not a strict upper bound for the number of points on AB, then neither is n+1. Suppose that n is not a strict upper bound for the num- ber of points incident to AB. Then there are at least n points incident to AB. By Betweenness Axiom B-3, there must be two among these n points that have none of the others between them. Choose such a pair of points and name them P and Q. By Betweenness Axiom B-2, there is point R between P and Q. The collection of n points incident n together with R is a collection of n+1 points incident to AB. Thus n+1 is not a strict upper bound for the number of points incident to AB. ∎.
18: Claim: A halfplane is convex.
Proof: Let HA be a half-plane. By definition, HA consists of all points that lie on the
same side of a given line l as point A. In particular, let B, C be points in HA. By definition, A and B are on the same side of l and A and C are on the same side of l. By Betweenness Axiom B-4, B and C are on the same side of l. By definition, BC is contained in HA. ∎
Claim: The interior of an angle is convex.
Proof: The interior of an angle is the intersection of two halfplanes. Let B, C be points in the interior of an angle. Then BC lies in each of the halfplanes and hence in the interior of the angle. ∎
Claim: The interior of a triangle is convex.
Proof: Similarly.
No, the exterior of a triangle is not convex.
