ps06 - MAT 300 Mathematical Structures Problem Set 6...

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MAT 300, Mathematical Structures Dr. L. Mantini Problem Set 6 Solutions Spring 2010 4.1 : 3, 4, 5, 6, 7, 10, 12; 6.1 : 2, 5, 6, 9, 12 4.1.3 Describe the truth sets of the following statements as subsets of R 2 (pictures omitted). (a) y = x 2 - x - 2: region is graph of the parabola opening up with x -intercepts at (2 , 0) and ( - 1 , 0) (b) y < x : region is the open half-plane below the line y = x (c) either y = x 2 - x - 2 or y = 3 x - 2: region is union of given parabola and line (d) y < x , and either y = x 2 - x - 2 or y = 3 x - 2: region is the portion of the graph from (c) that lies strictly below the line y = x 4.1.5 Prove parts 2 and 3 of Theorem 4.1.3. Solution : We prove part 2 as a two-part proof showing both set containments sepa- rately, and we prove part 3 using a string of equivalent statements. 2. Prove that A × ( B C ) = ( A × B ) ( A × C ). Proof : ( ) Let ( x, y ) A × ( B C ). Then x A and y B C . So either y B or y C . If y B , then ( x, y ) A × B . If y C , then ( x, y ) A × C . In either case, ( x, y ) ( A × B ) ( A × C ). ( ) Now let ( x, y ) ( A × B ) ( A × C ). So either ( x, y ) A × B or ( x, y ) A × C . If ( x, y ) A × B , then x A and y B B C , so ( x, y ) A × ( B C ). If ( x, y ) A × C , then x A and y C B C , so ( x, y ) A × ( B C ). In either case, ( x, y ) A × ( B C ), which was what we had to prove. 3. Prove that ( A × B ) ( C × D ) = ( A C ) × ( B D ). Proof : Let x and y be arbitrary. Then ( x, y ) ( A × B ) ( C × D ) ( x, y ) A × B ( x, y ) C × D x A y B x C y D ( x A x C ) ( y B y D ) x A C y B D ( x, y ) ( A C ) × ( B D ) . 4.1.6 What’s wrong with the proof given in the book that for any sets A , B , C , and D , ( A C ) × ( B D ) ( A × B ) ( C × D )?
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