Exercise 1: Checking Properties of R = {(1,1),(2,2),(3,3),(1,2)} — Possible Solution ==================================================================== REFLEXIVE — HOLDS ------------------------------ Reflexive requires (a,a) to be in R for every a in {1,2,3}. Checking: (1,1) is present, (2,2) is present, (3,3) is present. All three diagonal pairs are in R, so reflexive holds. SYMMETRIC — DOES NOT HOLD ------------------------------ Symmetric requires that whenever (a,b) is in R, (b,a) must be too. (1,2) is in R. Checking whether (2,1) is also in R: it is not listed anywhere in R. This single missing pair is enough to break symmetry - R is NOT symmetric. TRANSITIVE — HOLDS ------------------------------ Transitive requires that whenever (a,b) and (b,c) are both in R, (a,c) must be too. Checking every possible chain: (1,1) and (1,1) -> need (1,1): present. (1,1) and (1,2) -> need (1,2): present. (1,2) and (2,2) -> need (1,2): present. (2,2) and (2,2) -> need (2,2): present. (3,3) and (3,3) -> need (3,3): present. No chain produces a missing pair, so transitive holds. ANTISYMMETRIC — HOLDS ------------------------------ Antisymmetric requires that whenever BOTH (a,b) and (b,a) are in R, a must equal b. The only pairs in R where this "both directions" situation could apply are the diagonal ones ((1,1), (2,2), (3,3)), where a already equals b trivially. For the one off-diagonal pair, (1,2), the reverse (2,1) is not in R at all - so the "both present" condition never actually triggers for it, meaning there's nothing to violate. Antisymmetric holds (vacuously, for the only pair where it could have mattered). SUMMARY ------------------------------ R is reflexive, transitive, and antisymmetric, but NOT symmetric. A BONUS OBSERVATION ------------------------------ Per this chapter, "reflexive + antisymmetric + transitive" is exactly the definition of a partial order - and R happens to satisfy all three. R is, in fact, a (very small) partial order on {1,2,3}, with 1 "below" 2 in that ordering, and 3 unrelated to either. WHY THIS WORKS AS AN ANSWER ------------------------------ Each of the four properties is checked systematically against R's own explicit list of pairs, citing the specific pair (or absence of one) that confirms or breaks each property, rather than asserting the answer without justification.