quiz2sol - Yes We must show if (a,b) R and (b,c) R, then...

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COT 3100 Quiz #2 Solutions Date: 10/19/00 1) Consider the following relation: R = {(a,b) | a Z + b Z + 5 c | a = b + 2c, where c Z} (Remember that Z + = {1, 2, 3, . ..} and Z = {0, 1, -1, 2, -2, . ..}.) a) (4 pts) Is R reflexive? Yes For all positive integers a, we have a = a + 2(0). Thus, for all positive integers a (a,a) R, meaning that R is reflexive. b) (4 pts) Is R irreflexive? No Because (1,1) R, R can not be irreflexive. c) (4 pts) Is R symmetric? Yes We must show in all cases that if (a,b) R, then (b,a) R. if (a,b) R, that means there exists a c such that a = b + 2c. b = a – 2c b = a + 2(-c) If c is an integer, so is –c. Thus, the last line shows, by definition of R that (b,a) R. d) (4 pts) Is R antisymmetric? No (1,3) R and (3,1) R, but 1 and 3 are not equal. e) (4 pts) Is R transitive?
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Unformatted text preview: Yes We must show if (a,b) R and (b,c) R, then (a,c) R. Our assumption leads to the following two equations, based on the definition of R: a = b + 2d , d is an integer b = c + 2e , e is an integer Substitute for b in the first equation to yield: a = (c + 2e) + 2d a = c + 2e + 2d a = c + 2(d+e) Since d and e are both integers, their sum is as well. This means that (a,c) R. Bonus Question(4 pts): What kind of relation is this? Does this relation partition the set Z + ? If so, how? A equivalence relation. All equivalence relations partition a set. This one has two equivalence classes, [1] and [2]. [1] is the set of positive odd integers while [2] is the set of positive even integers....
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