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Unformatted text preview: MATH 290 – EXAM 4 – SOLUTIONS 1. Define the relation S on R × R by ( x,y ) S ( z,w ) if and only if x ≤ z and y ≤ w . (a) (8 pts.) Prove that S is a partial order on R × R . (b) (4 pts.) Is S a linear order ? Why or why not? (c) (4 pts.) Find an upper bound for the rectangle R = [1 , 2] × [3 , 4] = { ( x,y ) ∈ R × R :1 ≤ x ≤ 2 , 3 ≤ y ≤ 4 } , and find sup( R ). You need not verify that your answers are correct. Solution: (a). First we need to show that S is reflexive. Let ( x,y ) ∈ R × R . Since x ≤ x and y ≤ y , ( x,y ) S ( x,y ) and S is reflexive. Now we will show that S is antisymmetric. Let ( x,y ) , ( z,w ) ∈ R × R . Assume that ( x,y ) S ( z,w ) and that ( z,w ) S ( x,y ). We must show that ( x,y ) = ( z,w ), that is, that x = z and y = w . Since ( x,y ) S ( z,w ), x ≤ z and since ( z,w ) S ( x,y ), z ≤ x . Hence x = z . Also, since ( x,y ) S ( z,w ), y ≤ w and since ( z,w ) S ( x,y ), w ≤ y . Hence y = w ....
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 Fall '08
 Alligood,K
 Math, Advanced Math, Order theory, Equivalence relation, upper bound, 4 pts, 8 pts, 3 k

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