key3sp00 - COT3100C-01 Spring 2000 S Lang Solution Key to...

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COT3100C-01, Spring 2000 S. Lang Solution Key to Assignment #3 (40 pts.) 3/19/2000 1. (15 pts.) Let A denote an arbitrary non-empty set, and R and T denote arbitrary binary relations defined over A . Determine if each of the following statements is true or false, and give a brief explanation of your answer. In the cases of a false statement, use A = { a , b , c } and appropriate relations over A for your counter-example. (a) If R is symmetric, then R {( a , a ) | a A } is symmetric. Proof: In order to prove the symmetric property, let ( x , y ) R {( a , a ) | a A } ---- (1), we need to prove ( y , x ) R {( a , a ) | a A } ---- (2). From (1), ( x , y ) R or ( x , y ) {( a , a ) | a A }. Thus, there are two cases: (Case 1) Suppose ( x , y ) R . Since R is symmetric by assumption, so ( y , x ) R R {( a , a ) | a A }, so (2) is proved in this case. (Case 2) Suppose ( x , y ) {( a , a ) | a A }. In this case, x = y . Thus, ( y , x ) = ( x , y ) {( a , a ) | a A } R {( a , a ) | a A }, so (2) is proved in this case. Therefore, we proved (2) in both cases. (b) If R is transitive and R T , then T is transitive. Disproof: Let A = { a , b , c }, R = {( a , b ), ( b , c ), ( a , c )}, T = {( a , b ), ( b , c ), ( a , c ), ( b , a )}. Then, R is transitive and R T , but T is not transitive because both ( a , b ) and ( b , a ) T but ( a , a ) T . (c) ( R T ) –1 = R –1 T –1 . Proof: We will prove ( x , y ) ( R T ) –1 ( x , y ) R –1 T –1 . Note that ( x , y ) ( R T ) –1 ( y , x ) ( R T ) , by the definition of ( R T ) –1 ( y , x ) R or ( y , x ) T , definition of union ( x , y ) R
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key3sp00 - COT3100C-01 Spring 2000 S Lang Solution Key to...

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