Homework 3 Solutions - Solutions to Homework Three CS 191 :...

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Solutions to Homework Three CS 191 : Discrete Structures I Feb. 20 , Winter 2007 Due Date: Mar. 6, Tuesday, before the class meeting. Section 3.1, p123 In Exercises 9-12, draw the diagraph of the relation. Exercise (11) The relation R = {(1,2), (2,3), (3,4), (4,1) } on X = {1, 2, 3, 4}. Answer: 1 2 3 4 In Exercises 13-16, write the relation as a set of ordered pairs. Exercise (14) 1 2 3 4 5 Answer: {(1,1), (2,2), (3,3), (3,5), (4,3), (4,4), (5,5), (5,4)}. 1
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R on the set {1, 2, 3, 4, 5} defined by the rule ( x , y ) R if 3 divides x y . Exercise (20) List the elements of R -1 . Answer : {(1,1), (4,1), (2,2), (5,2), (3,3), (1,4), (4,4), (2,5), (5,5)} Exercise (27) Is the relation of Exercise 25 reflexive, symmetric, antisymmetric, transitive, and/or a partial order? Note, the relation R of Exercise 25 is defined on the set {1, 2, 3, 4, 5} by the rule ( x , y ) R if x + y 6.. Answer: It is symmetric. It is not reflexive, not antisymmetric, not transitive, and not a partial order. In Exercises 29-34, determine whether each relation defined on the set of positive integers is reflexive, symmetric, antisymmetric, transitive, and/or a partial order. Exercise (30) ( x , y ) R if x > y . Answer: It is not reflexive. It is antisymmetric, transitive. It is not a partial order. Exercise (34) ( x , y ) R if 3 divides x + 2 y . Answer: It is reflexive because x + 2 x = 3 x is divisible by 3. It is symmetric because if x + 2 y is divisible by 3, then y + 2 x = 2( x + 2 y ) – 3 y will be also divisible by 3. It is transitive because if x + 2 y is divisible by 3 and y + 2 z is divisible by 3, then x + 2 z = ( x + 2 y ) + ( y + 2 z ) -3 y will be also divisible by 3. It is not antisymmetric obviously and not a partial order.
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Homework 3 Solutions - Solutions to Homework Three CS 191 :...

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