# hw2sol - Math 55 - Discrete Mathematics UC Berkeley -...

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Unformatted text preview: Math 55 - Discrete Mathematics UC Berkeley - Spring 2010 Solutions to Homework #2 (due February 3) 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 2 . 1, #21 Recall that |P ( A ) | = 2 | A | . Then: a. 8 b. 16 c. 2 2 . 1, #28 a. A B C = { ( 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. C B A = { (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. C A B = { (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. B B B = { ( x,x,x ) , ( x,y,x ) , ( y,x,x ) , ( y,y,x ) , ( x,x,y ) , ( x,y,y ) , ( y,x,y ) , ( y,y,y ) } . 2 . 1, #38 a. Assume S S . By definition of the set, this implies that S / S , a contradiction. b. Assume S / S . Followint the definition of the set, this implies that S S , since it is not an element of the set S . Again, we obtain a contradiction. 2 . 2, #4 a. A B = B = { a,b,c,d,e,f,g,h } b. A B = A = { a,b,c,d,e } c. A r B = d. B r A = { f,g,h } 2 . 2, #14 Recall that for any two sets X,Y we have X = ( X r Y ) ( X Y ). Thus, A = { 1 , 5 , 7 , 8 } { 3 , 6 , 9 } = { 1 , 3 , 5 , 6 , 7 , 8 , 9 } , B = { 2 , 10 } { 3 , 6 , 9 } = { 2 ,...
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## This note was uploaded on 04/05/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.

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hw2sol - Math 55 - Discrete Mathematics UC Berkeley -...

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