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hw8solutions - EE 369 Homework#8 Solutions(Sixth Edition...

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EE 369 Homework #8 Solutions (Sixth Edition) Section 4.1 ----------- 10) Let S={0,1,2,4,6} Test the following relations on S for reflexivity, symmetry, transitivity, and antisymmetry. a) \rho={(0,0),(1,1),(2,2),(4,4),(6,6),(0,1),(1,2),(2,4),(4,6)} Solution: reflexive,antisymmetric b) \rho={(0,1),(1,0),(2,4),(4,2),(4,6),(6,4)} Solution: symmetric c) \rho={(0,1),(1,2),(0,2),(2,0),(2,1),(1,0),(0,0),(1,1),(2,2)} Solution: symmetric, transitive d) \rho={(0,0),(1,1),(2,2),(4,4),(6,6),(4,6),(6,4)} Solution: reflexive, symmetric, transitive e) \rho=\empty Solution: symmetric, antisymmetric, transitive 17) Find the reflexive, symmetric and transitive closures of each of the relations in (8) above. Solution: a) reflexive closure = \rho itself symmetric closure - add (1,0), (2,1), (4,2), (6,4) transitive closure - add (0,2), (1,4), (2,6), (0,4), (0,6), (1,6) b) reflexive closure - add (0,0), (1,1), (2,2), (4,4), (6,6) symmetric closure - \rho itself transitive closure - add (0,0), (1,1), (2,2), (2,6) (4,4), (6,6), (6,2) c) reflexive closure - add (4,4), (6,6) symmetric closure = transitive closure = \rho itself d) \rho is its own closure with respect to all three properties e) reflexive closure - add (0,0), (1,1), (2,2), (4,4), (6,6) symmetric closure = transitive closure = \rho itself 20) Does it make sense to look for the closure of a relation with respect to the following properties? Why or why not? a) irreflexive property Solution: No- if the relation is irreflexive, it is its own irreflexive closure. If the relation is not irreflexive, there
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must be some x \in S with (x,x) in the relation; extending the relation will not remove this pair,so no extension can be irreflexive. b) asymmetric property Solution: No- if the relation is asymmetric, it is its own asymmetric closure. If the relation is not asymmetric, there must be two pairs (x,y) and (y,x) or one pair (x,x) in the relation; extending the relation will not remove these pairs,so no extension can be asymmetric. 24) Given the partial orderings in Exercise 20, name any least elements, minimal elements, greatest elements, and maximal elements. a. S= {a,b,c} p={(a,a),(b,b),(c,c),(a,b),(b,c),(a,c)} Solution: a is minimal and least c is maximal and greatest b. S= {a,b,c,d} \rho={(a,a),(b,b),(c,c),(d,d),(a,b),(a,c)} Solution: a and d are minimal b, c, d are maximal c. S={\empty,{a},{a,b},{c},{a,c},{b}} A p B <-> A \subseteq B Solution: \empty is minimal and least {a,c} and {a,b} are maximal 30e) Let \rho be a binary relation on a set S. Then a binary relation called the inverse of \rho, denoted \rho^-1 is defined x \rho^-1 <-> y \rho x. Prove that if \rho is a transitive relation on a set S, then \rho^-1 is transitive. Solution : Let x \rho^-1 y and y \rho^-1 z. Then y \rho x and z \rho y. By the transitivity of \rho , z \rho x and therefore x \rho^-1 z. 40) a) Given the partition {1,2} and {3,4} of the set S={1,2,3,4}, list the ordered pairs in the corresponding equivalence relation.
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