# hw3_key - COT3100 Summer’2001 Assignment#3(Solution...

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Unformatted text preview: COT3100 Summer’2001 Assignment #3 (Solution) Assigned: 06/12, due: 06/26 on lecture. Total 100 pts. 1. Suppose A = {1, 2, 3}, B ={ a , b , c }, R = {(1, a ), (1, b ), (2, b ), (3, c )} is a relation from A to B and S = {( a , b ), ( a , c ), ( b , a ), ( c , c )} is a relation from B to B . Find the following relations: a) (2pts) R S R S = { (1, 4), (1, 5), (1, 6), (2, 4), (3, 6)} b) (2pts) S S S S = {(4, 4), (4, 6), (5, 5), (5, 6), (6, 6)} c) (3pts) R S —1 R S —1 = {(1, 4), (1, 5), (2, 4), (3, 4), (3, 6)} d) (3pts) S R –1 S R –1 = { (4, 1), (4, 2), (4, 3), (5, 1), (6, 3)} 2. Suppose R and S are two relations from A to B , R ⊆ A × B , S ⊆ A × B . Prove or disprove each of the following statements: a) (5pts) If R ⊆ S , then R- 1 ⊆ S- 1 . Proof . Assume R ⊆ S . To prove R- 1 ⊆ S- 1 , take arbitrary ( x , y ) ∈ R- 1 to show that ( x , y ) ∈ S- 1 . If ( x , y ) ∈ R- 1 , then ( y , x ) ∈ R in accordance with definition of inverse relation. But R ⊆ S by assumption, so ( y , x ) ∈ S by subset definition. If ( y , x ) ∈ S , then ( x , y ) ∈ S- 1 by the definition of inverse relation. b) (8pts) ( R ∪ S )- 1 = R- 1 ∪ S- 1 Proof . We need to prove two subset relations, i) ( R ∪ S )- 1 ⊆ R- 1 ∪ S- 1 and ii) R- 1 ∪ S- 1 ⊆ ( R ∪ S )- 1 . i) Take arbitrary ( x , y ) ∈ ( R ∪ S )- 1 to prove that ( x , y ) ∈ R- 1 ∪ S- 1 . If ( x , y ) ∈ ( R ∪ S )- 1 , then ( y , x ) ∈ R ∪ S by the definition of inverse relation. Then there might be two cases: either ( y , x ) ∈ R or ( y , x ) ∈ S . If ( y , x ) ∈ R , then ( x , y ) ∈ R- 1 by the definition of inverse relation, and ( x , y ) ∈ R- 1 ∪ S- 1 , because R- 1 ⊆ R- 1 ∪ S- 1 . If ( y , x ) ∈ S , then ( x , y ) ∈ S- 1 , and ( x , y ) ∈ R - 1 ∪ S- 1 , since S- 1 ⊆ R- 1 ∪ S- 1 . ii). Take arbitrary ( x , y ) ∈ R- 1 ∪ S- 1 to prove that ( x , y ) ∈ ( R ∪ S )- 1 . If ( x , y ) ∈ R- 1 ∪ S- 1 , then either ( x , y ) ∈ R- 1 or ( x , y ) ∈ S- 1 by the definition of the union. In case ( x , y ) ∈ R- 1 , ( y , x ) ∈ R , that implies that ( y , x ) ∈ R ∪ S . But if ( y , x ) ∈ R ∪ S , then ( x , y ) ∈ ( R ∪ S )- 1 . In another case, ( x , y ) ∈ S- 1 , we may imply, that ( y , x ) ∈ S , and then that ( y , x ) ∈ R ∪ S ....
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hw3_key - COT3100 Summer’2001 Assignment#3(Solution...

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