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Solutions to Greenberg's "Euclidean and Non-Euclidean geometries"
Typology: Exercises
Uploaded on 12/11/2016
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Homework 8 – solutions
The copyright for this text rests with the author.
Decide whether each of the following are true, false or not determined by the information given:
a: Everybody in the group is on a team. true b: Nobody in the group plays chess. not determined by the information given c: If somebody in the group plays soccer, then somebody else plays golf. not determined by the information given
Negate statements a, b, c.
Negation of a: Somebody in the group is not on a team. Negation of b: Somebody in the group plays chess. Negation of c: Somebody in the group plays soccer and nobody else plays golf.
Proof: We prove two implications. (“⇒”) Suppose that Hilbert’s Parallel Postulate holds, i.e., suppose that, given any line l and a point P not on l, there is at most one line parallel to l that is incident to P. Let l, l′^ be a parallel lines and and let m be a line incident to l in a point P. Suppose, to derive a contradiction, that m is disjoint from l′. Then both l and m are parallels to l′^ incident to P. This contradicts Hilbert’s Parallel Postulate. Thus m intersects l′. (“⇐”) Suppose that every line intersecting one of two parallel lines also intersects the other. Let n be a line and let Q be a point not on n. Suppose, to derive a contradiction, that there is more than one line parallel to n and incident to Q. Let n′, m′^ be two such lines. Then m′^ intersects n′^ in the point Q and n′^ is parallel to n. Hence by hypothesis, m′^ also intersects the line n. Thus m′^ is not parallel to n, a contradiction. Hence Hilbert’s Parallel Postulate holds. ∎
n through Q. Then t′^ is a transversal for n, m and also for n, m′. By hypothesis, alternate interior angles are congruent in both cases. By Congruence Axiom C-4 there is a unique ray ema- nating from Q on a given side of t′^ congruent to an interior angle formed by t′^ and n. Hence m and m′^ must coincide. Thus Hilbert’s Parallel Postulate holds. ∎.
Proposition 4.9: Hilbert’s Parallel Postulate holds if and only if whenever a transversal t meets a pair of parallel lines l, l′, if t ⊥ l, then t ⊥ l′.
Proof: We prove two implications. (“⇒”) Suppose that Hilbert’s Parallel Postulate holds, i.e., suppose that, given any line l and a point P not on l, there is at most one line parallel to l that is incident to P. Let l, l′^ be parallel lines and let t be a transversal to l and l′. Furthermore, suppose that t ⊥ l. By the converse to AIA, alternate interior angles are congruent. Hence, by the definition of perpendicular and right an- gle, all four interior angles for t, l, l′^ are right angles. Thus t ⊥ l′. (“⇐”) Suppose that whenever a transversal t meets a pair of parallel lines l, l′, if t ⊥ l, then t ⊥ l′. Let n be a line and Q a point not on n. Suppose that m, m′^ are parallels to n incident to Q.
By Proposition 3.16, there is a line t perpendicular to n and incident to Q. By hypothesis, t ⊥ m and t ⊥ m′. Thus m and m′^ must coincide. Therefore Hilbert’s Parallel Postulate holds. ∎.
Proposition 4.10: Hilbert’s Parallel Postulate holds if and only if for every pair of parallel lines k, l and lines m ⊥ k, n ⊥ l, either m = n or m is parallel to n.
Proof: We prove two implications. (“⇒”) Suppose that Hilbert’s Parallel Postulate holds, i.e., suppose that, given any line l and a point P not on l, there is at most one line parallel to l that is incident to P. Let k, l be parallel lines and m, n lines such that m ⊥ k and n ⊥ l. Case 1: m = n. In this case the conclusion follows. Case 2: m ≠ n. In this case, by Proposition 4.9, m ⊥ l. In particular, l is a transversal for the pair of lines m, n and has congruent alternate interior angles. By AIA, m, n are parallel. (“⇐”) Suppose that for every pair of parallel lines k, l and lines m ⊥ k, n ⊥ l, either m = n or m is parallel to n. Let l′^ be a line and P ′^ a point not on l′. Suppose that m′, m” are parallels to l′^ incident to P ′. By Proposition 3.16, there is a line t ⊥ l′^ incident to P ′. Denote the point at which t and l′^ meet by Q. By Proposition 3.16, there is a line t′^ ⊥ m′^ through Q.