hw10sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 10 (due April 20, 2009) 8.1, #5: a) Reflexive, symmetric, not antisymmetric, not transitive. b) Not reflexive, symmetric, not antisymmetric, not transitive. c) Not reflexive, symmetric, not antisymmetric, not transitive. The relation is not reflexive because there exist webpages which have no links at all. d) Not reflexive, symmetric, not antisymmetric, not transitive. Simi- larly, there are webpages which are not linked to. 8.1, #32: a) R 6 b) R 2 c) R 5 d) e) f) R 5 g) R 6 h) R 6 8.1, #34: a) R 1 b) R 1 c) R 2 d) R 2 e) R 1 f) R 2 g) R 2 h) R 3 8.1, #46: a) 2. Call the single element of S 1 and then a relation on S is either { (1 , 1) } or , both of which are transitive. b) 13. Suppose S = { 1 , 2 } . There are 4 elements in S S , which means that there are 2 4 = 16 relations on S . Note that the transitivity condition, that ( a, b ) , ( b, c ) R implies ( a, c ) R , is automatically satisfied if a = b or b = c . Thus, the relations where are not transitive are those which contain both (1 , 2) and (2 , 1), but dont contain (1 , 1) or dont contain (2 , 2). Thus, including the possibility of containing neither, we have to rule out 3 relations, which leaves 16 3 = 13 transitive relations. c) 171. 1 Let S = { 1 , 2 , 3 } . First, well count the transitive relations on S which happen to be antisymmetric. If R is a transitive, antisymmetric rela- tion on S , then let R be its reflexive closure, which is still transitive and antisymmetric, and so a partial ordering. There are 5 possible un- labelled partial orderings on a set of 3 elements, shown by their Hasse diagrams: The number of ways of labelling these diagrams with the elements of S is 1, 6, 3, 3, and 6 respectively. Thus, there are 19 possible partial orderings R on S . Because the not necessarily reflexive relation R is antisymmetric, it can either contain or not contain each relation (1 , 1), (2 , 2) and (3 , 3) and still be transitive. Thus, there are 2 3 = 8 possi- ble relations R whose reflexive closure is a given partial ordering R . Therefore, there are 19 8 = 152 antisymmetric, transitive relations on S . Second, well count the transitive relations on S which are not anti- symmetric. This means there are two distinct integers a, b S such that ( a, b ) and ( b, a ) are both in R . As the first subcase, assume this pair of integers a and b are the only pair of integers where this is true. By transitivity, ( a, a ) and ( b, b ) must also be in S . Let c be the ele- ment of S other than a and b . Then R contains ( a, c ) if and only if it contains ( b, c ) and it contains ( c, a ) if and only if it contains ( c, b )....
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hw10sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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