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Unformatted text preview: Answer key No.1 to selected homework problems: Math. 61 1.7 1. Suppose the formula for n and try to prove the formula for n + 1: 1(1!) + 2(2!) + Β·Β·Β· + n ( n !) + ( n + 1)(( n + 1)!) = ( n + 1)! 1 + ( n + 1)(( n + 1)!) = ( n + 1)! + ( n + 1)(( n + 1)!) 1 = ( n + 1)!(1 + n + 1) 1 = ( n + 2)! 1 . 9. First check the equality: 1 a 2 1 = 1 2( a 1) 1 2( a + 1) . And use this formula for a = n + 2 in the induction process. 13. Write S n for the fraction of the LHS (the left hand side). Then S n +1 = S n Γ (2 n + 1) 2( n + 1) β€ ( n + 1) 1 / 2 Γ (2 n + 1) 2( n + 1) β€ ( n + 2) 1 / 2 . The last inequality: ( n + 1) 1 / 2 Γ (2 n +1) 2( n +1) β€ ( n + 2) 1 / 2 is equivalent to (2 n + 1) 2( n + 1) β€ ( n + 1) 1 / 2 ( n + 2) 1 / 2 , which in turn equivalent to: (2 n + 1) 2 4( n + 1) 2 = ( n + 1 2 ) 2 ( n + 1) 2 β€ ( n + 1) ( n + 2) , which can be checked by computation. Indeed ( n + 1 2 ) 2 ( n + 1) 2 β€ ( n + 1) ( n + 2) is equivalent to n 3 + 3 n 2 + 9 4 n + 1 2 = ( n + 1 2 ) 2 ( n + 2) β€ ( n + 1) 3 = n 3 + 3 n 2 + 3 n + 1 and RHSLHS = 3 4 n + 1 2 β₯ 0 if n β₯ 0. 14. Check the starting point of the induction by your self. Now suppose: 2 n + 1 β€ 2 n and prove: 2( n + 1) β€ 2 n +1 as follows: 2( n + 1) + 1 = 2 + 2 n + 1 β€ 2 + 2 n β€ 2 n + 2 n = 2 Γ 2 n = 2 n +1 . 2.1 56. Let P ( X ) denote the set of all subsets in a set X . Then by Theorem 2.1.6, we know that P ( X )  = 2 n if  X  = n . The only nonproper subset in P ( X ) is X itself. Thus the answer is 2 n 1 . 78. Suppose X, A, B, C to be subsets of a universe U . We prove the statement A C = B C β A = B . By definition, we have X C = ( X βͺ C ) ( X β© C ) = ( X C ) βͺ ( C X ) . Pick an arbitrary element a in X . Then there are two possibilities: (i) a β C or (ii) a β C. 1 2 If a β C , then a β X C . If a β C , then a β X C . Since (i) and (ii) never happen at the same time, we see that for a β X (a) a β C β a β X C and a β C β a β X C. Pick an arbitrary element x outside X . Then (b) x β X C β x β C and x β X C β x β C. Suppose a β A β© C . Then by (a) applied to X = A , a β B C = A C . Then by (b) applied to X = B , a β B β a β C , a contradiction because a β A β© C . Thus a β B . This shows that A β© C β B β© C . Since the roles of A and B are interchangeable, we conclude A β© C β B β© C . This shows that A β© C = B β© C . Now suppose that a β A C . Then by (a) applied to X = A , we see a β A C = B C . Since a β C , we know a β B C because B C = ( B C ) βͺ ( C B ). Thus A C β B C . Interchanging the role of A and B , we get A C β B C . This shows A C = B C ....
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 Spring '09
 SUGANO
 Equivalence relation, X, ak bk, 8bit strings, sn bn+1

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