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Appendix 1: Solutions to Selected Exercises 343
Section 4.3
1.

a
b bad
c bed
d cab
e

3.

12 3 4

5. S ◦ R ={(a, y), (a, z), (b, x), (c, y), (c, z)}.
7. (→) Suppose R is reflexive. Let (x, y) be arbitrary element of i A . Then by
the definition of i A , x = y ∈ A. Since R is reflexive, (x, y) = (x, x) ∈ R.
Since (x, y) was arbitrary, this shows that i A ⊆ R.
(←) Suppose i A ⊆ R. Let x ∈ A be arbitrary. Then (x, x) ∈ i A ,so
since i A ⊆ R,(x, x) ∈ R. Since x was arbitrary, this shows that R is reflex-
ive.
10. Suppose (x, y) ∈ i D . Then x = y ∈ D = Dom(S), so there is some z ∈ A
−1
such that (x, z) ∈ S. Therefore (z, x) ∈ S ,so(x, y) = (x, x) ∈ S −1 ◦ S.
Thus, i D ⊆ S −1 ◦ S. The proof of the other statement is similar.
13. (a) Yes. To prove it, suppose R 1 and R 2 are reflexive, and suppose a ∈ A.
Since R 1 is reflexive, (a, a) ∈ R 1 ,so(a, a) ∈ R 1 ∪ R 2 .
(b) Yes. To prove it, suppose R 1 and R 2 are symmetric, and suppose
(x, y) ∈ R 1 ∪ R 2 . Then either (x, y) ∈ R 1 or (x, y) ∈ R 2 .If(x, y) ∈
R 1 then since R 1 is symmetric, (y, x) ∈ R 1 ,so(y, x) ∈ R 1 ∪ R 2 . Sim-
ilar reasoning shows that if (x, y) ∈ R 2 then (y, x) ∈ R 1 ∪ R 2 .
(c) No. Counterexample: A ={1, 2, 3}, R 1 ={(1, 2)}, R 2 ={(2, 3)}.
17. First note that by part 2 of Theorem 4.3.4, since R and S are symmetric,
−1
R = R −1 and S = S . Therefore
R ◦ S is symmetric iff R ◦ S = (R ◦ S) −1 (Theorem 4.3.4, part 2)
iff R ◦ S = S −1 ◦ R −1 (Theorem 4.2.5, part 5)
iff R ◦ S = S ◦ R.