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.
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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].
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