hw2solsp09

# hw2solsp09 - Math 55: Discrete Mathematics UC Berkeley,...

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Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 2 (due February 2, 2009) 2.1, #5: a) yes b) no c) yes d) no e) no f) no 2.1, #8: a) true b) true c) false d) true e) true f) true g) true 2.1, #21: a) 3 b) 4 c) 2 2.1, #28: a) { ( a, x, 0) , ( a, x, 1) , ( a, y, 0) , ( a, y, 1) , ( b, x, 0) , ( b, x, 1) , ( b, y, 0) , ( b, y, 1) , ( c, x, 0) , ( c, x, 1) , ( c, y, 0) , ( c, y, 1) } b) { (0 , x, a ) , (1 , x, a ) , (0 , y, a ) , (1 , y, a ) , (0 , x, b ) , (1 , x, b ) , (0 , y, b ) , (1 , y, b ) , (0 , x, c ) , (1 , x, c ) , (0 , y, c ) , (1 , y, c ) } c) { (0 , a, x ) , (1 , a, x ) , (0 , a, y ) , (1 , a, y ) , (0 , b, x ) , (1 , b, x ) , (0 , b, y ) , (1 , b, y ) , (0 , c, x ) , (1 , c, x ) , (0 , c, y ) , (1 , c, y ) } d) { ( x, x, x ) , ( x, x, y ) , ( x, y, x ) , ( x, y, y ) , ( y, x, x ) , ( y, x, y ) , ( y, y, x ) , ( y, y, y ) } 2.1, #38: a) Suppose that S is a member of S . Since S consists of those elements which do not contain themselves, then S is not an element of S . This is a contradiction. b) Suppose that S does not a member S . Since S constists of all elements which do not contain themselves, then S is an element of S . This is a contradiction. 2.2, #4: a) A B = { a, b, c, d, e, f, g, h } b) A B = { a, b, c, d, e } 1 c) A B = d) B A = { f, g, h } 2.2, #14: A = { 1 , 3 , 5 , 6 , 7 , 8 , 9 } , B = { 2 , 3 , 6 , 9 , 10 } 2.2, #24: Let A , B , and C be sets....
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## hw2solsp09 - Math 55: Discrete Mathematics UC Berkeley,...

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