EE 369 Homework #8 Solutions (Sixth Edition)
Section 4.1
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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