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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/24/2009 for the course ECE 369 taught by Professor Bagchi during the Spring '08 term at Purdue University-West Lafayette.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online