exam2pracsol - COT3100 Spring2001 Practice problems for the...

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Spring’2001 Practice problems for the test#2 Relations. 1. Let A ={1, 2, 3}, B ={ w , x , y , z }, and C = {4, 5, 6}. Define the relation R A × B , S B × C , and T B × C , where R = {(1, w ), (3, w ), (2, x ), (1, y )}, S ={( w , 5), ( x , 6), ( y , 4), ( y , 6)}, and T ={( w , 4), ( w , 5), ( y , 5)}. a) Determine R ( S T ) and ( R S ) ( R T ). S T = {( w , 4), ( w , 5), ( x , 6), ( y , 4), ( y , 5), ( y , 6)}, R ( S T ) = {(1, 4), (1, 5), (1, 6), (2, 6), (3, 4), (3, 5)} R S = {(1, 4), (1, 5), (1, 6), (2, 6), (3, 5)}, R T = {(1, 4), (1, 5), (3, 4), (3, 5)} ( R S ) ( R T ) = {(1, 4), (1, 5), (1, 6), (2, 6), (3, 4), (3, 5)}. b) Determine R ( S T ) and ( R S ) ( R T ). S T ={( w , 5)}; R ( S T ) ={(1, 5), (3, 5)}; ( R S ) ( R T ) = {(1, 4), (1, 5), (3, 5)}. 2. Consider the relation T over real numbers: T = {( a , b ) | a 2 + b 2 =1}. Define all properties of this relation (reflexive, irreflexive, symmetric, anti-symmetric, and transitive). Not reflexive (for example, a =0.5, ( a , a ) T ). Not irreflexive ( a =1/ 2, ( a , a ) T ) Symmetric (for any real a , b , ( a , b ) T ( b , a ) T , because a 2 + b 2 =1 b 2 + a 2 =1). Not anti-symmetric (counterexample: (0, 1) T and (1, 0) T ) Not transitive (counterexample: (0, 1) T , (1, 0) T , but (0, 0) T ) 3. Let R A × A be a symmetric relation. Prove that t ( R ) is symmetric. Proof. Assume R A × A is symmetric. To prove that t ( R ) is symmetric, we need to show that if ( a , b ) t ( R ) then ( b , a ) t ( R ). So, pick arbitrary ( a , b ) t ( R ) (1). By the definition of transitive closure, (1) implies that there is a path from a to b in R . By the definition of a path it means, that there are some elements a 0 = a , a 1 , a 2 , … , a n = b A such that ( a 0 , a 1 ), ( a 1 , a 2 ), … ( a n - 1 , a n ) R . But since R is symmetric by assumption, all inverted pairs belong to R as well, i. e. ( a n , a n - 1 ), …, ( a 2 , a 1 ), ( a 1 , a 0 ) R . It signifies that R contains the path from a n = b to a 0 = a . By the definition of the transitive closure it implies that ( b , a ) t ( R ). 4. Prove or disprove the following propositions about arbitrary relations on A : a) R 1 , R 2 symmetric R 1 R 2 is symmetric. Proof. Assume R 1 and R 2 are symmetric. To prove that R 1 R 2 is symmetric we need to show that if ( a , b ) R 1 R 2 , then ( b , a ) R 1 R 2 . So, take arbitrary ( a , b ) R 1 R 2 . It implies that either ( a , b ) R 1 or ( a , b ) R 2 . In the first case ( a , b ) R 1 implies ( b , a ) R 1 by symmetric property of R 1 and ( b , a ) R 1 R 2 by the union property. In the second case, ( a , b ) R 2 implies ( b , a ) R 2 by symmetric property of R 2 , and then ( b , a ) R 1 R 2 by the union property. So, in both cases we showed that (
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exam2pracsol - COT3100 Spring2001 Practice problems for the...

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