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Unformatted text preview: Math 55. Sample Answers to Final Exam 1. (15 points) Let A and B be sets. (a). Show that if A is a subset of B then the power set of A is a subset of the power set of B . (b). Suppose that A and B are sets such that the power set of A is a subset of the power set of B . Does it follow that A is a subset of B ? a . If A B , then all S P ( A ) are subsets of A , hence subsets of B (since S A B , hence elements of P ( B ). Therefore P ( A ) P ( B ). b . Yes. We have A A , so A P ( A ) P ( B ), therefore A B . 2. (20 points) Find real numbers a and b such that n X k =1 k 2 (1 . 1) k = ( n a b n ) for positive integers n . Prove your answer. The only numbers that work are a = 2 and b = 1 . 1. Discarding all but the last term, we have n X k =1 k 2 (1 . 1) k n 2 (1 . 1) n , so n k =1 k 2 (1 . 1) k is ( n 2 (1 . 1) n ). To get an upper bound, we have n X k =1 k 2 (1 . 1) k n 2 ( 1 . 1 + (1 . 1) 2 + + (1 . 1) n ) = n 2 1 . 1 (1 . 1) n +1 1 1 . 1 = 10 n 2 ( (1 . 1) n +1 1 . 1 ) 11 n 2 (1 . 1) n , and therefore n k =1 k 2 (1 . 1) k is O ( n 2 (1 . 1) n ). Since n k =1 k 2 (1 . 1) k is both ( n 2 (1 . 1) n ) and O ( n 2 (1 . 1) n ), it is ( n 2 (1 . 1) n ). 1 2 3. (35 points) Use the fact that 35 89 18 173 = 1 to solve the following problems. (a). Find an integer x such that 89 x 1 (mod 173). (b). Find an integer y such that 18 y 3 (mod 89). (c). Find an integer z such that z 3 (mod 173) and z 4 (mod 89). a . x = 35. b . We have 18( 173) 1 (mod 89), so multiplying by 3 gives an answer: y = 3 173. c . We have 35 89 1 (mod 173) and 35 89 0 (mod 89) and 18 173 0 (mod 173) and 18 173 1 (mod 89) , so multiplying the first number by 3 and the second by 4 gives z = 3 35 89 4 18 173 ....
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This note was uploaded on 03/08/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.
 Spring '08
 STRAIN
 Sets

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