key6fa02 - COT3100.01, Fall 2002 S. Lang Solution Keys to...

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COT3100.01, Fall 2002 S. Lang Solution Keys to Assignment #6 11/26/2002 1. (12 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 T , then s ( R ) s ( T ). Proof: Note that R T implies R –1 T –1 , by the definition of inverse Thus, s ( R ) = R R –1 , by the definition of s ( R ) T T –1 = s ( T ). (b) If R is transitive and T is transitive, then R T is transitive. Disproof: We give a counter-example as follows. Let A = {1, 2, 3}. Define two relations R = {(1, 2)}, and T = {(2, 3)}. Then both R and T are transitive (since there are no pairs to be composed together in both relations). However, R T ={(1, 2), (2, 3)}; it is not transitive since the pair (1, 3) R T . (c) If R = T –1 , then T = R –1 . Proof: We prove T R –1 and R –1 T simultaneously as follows. Note that ( a , b ) T ( b , a ) T –1 (definition of inverse) ( b , a ) R (since R = T –1 by assumption) ( a , b ) R –1 (definition of inverse) (d) If R T is irreflexive, then R is irreflexive. Proof: We prove by contradiction. That is, we assume R is not irreflexive. Thus, there exists some ( a , a ) R . Since R R T , so ( a , a ) R T . Thus, R T is not irreflexive, a contradiction to the assumption. 2. (4 pts.) Let A , B , and C denote 3 sets. Prove that A × ( B C ) = ( A × B ) ( A × C ). Proof:
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key6fa02 - COT3100.01, Fall 2002 S. Lang Solution Keys to...

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