Math 201-B Introduction to Proofs Instructor: Alex Roitershtein Iowa State University Department of Mathematics Fall 2015 Homework #5 (solutions) 1. Exercise 2 from Section 11.1 of the textbook BP. Solution: R is not reflexive because (a, a) 6∈ R. R is not symmetric because (a, b) ∈ R but (b, a) 6∈ R. R is transitive. 2. Exercise 8 from Section 11.1 of the textbook BP. Solution: The relation can be described in words as “the distance between x and y is less than one”. It is reflexive, symmetric, but not transitive. Counterexample: 0.1R0.9 and 0.9R1.8 but clearly (0, 1; 1.8) 6∈ R. 3. Exercise 4 from Section 11.2 of the textbook BP. Solution: aRd implies a and d in the same equivalence class. Then eRa implies that e is in the same class. Finally, cRe establishes that c is alo in thisequivalence class, while bRc implies that so is b. Thus R consists of just one equivalence class. 4. Exercise 6 from Section 11.2 of the textbook BP. Solution: To describe the equivalence relation it suffices to describe the equivalence classes it induces. The classes must form a partition of the original set. Furtehrmore, any such partition can be viewed as induced by some equivalence relation. Thus the number of possible equivalence relations coincides with the number of partitions of the set. In teh case under consideration we have the following 5: Three classes: {a}, {b}, {c}. Two classes: {a}, {b, c}. Two classes: {a, b}, {c}. Two classes: {a, c}, {b}. One class: {a, b, c}. 5. Exercise 12 from Section 11.2 of the textbook BP. Solution: False. Consider, for instance the following relations S and R on Z : S = {(1, 1), (2, 2), (1, 2), (2, 1)} and R = {(1, 1), (3, 3), (1, 3), (3, 1)}. Then (2, 1) ∈ S ∪ R, (1, 3) ∈ S ∪ R, but (2, 3) 6∈ S ∪ R violating the transitivity property of the relation. 1 6. Exercise 4 from Section 11.3 of the textbook BP. Solution: The relation induces the following four equivalence classes: {4k : k ∈ Z}, {4k + 1 : k ∈ Z}, {4k + 2 : k ∈ Z}, {4k + 3 : k ∈ Z}. 7. Exercise 2 from Section 11.4 of the textbook BP. Solution: We have: [0] × [0] = [0], [0] × [1] = [0], [0] × [2] = [0], [1] × [1] = [1], [1] × [2] = [2], [2] × [2] = [1], [0] + [0] = [0], [0] + [1] = [1], [0] + [2] = [2], [1] + [1] = [2], [1] + [2] = [0], [2] + [2] = [1]. The rest can be determined using the above identities and the symmetry properties [a] × [b] = [b] × [a] and [a] + [b] = [b] + [a]. 2