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

This preview shows pages 1–2. Sign up to view the full content.

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 )?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern