Exam s 3-4 - n [ k =1 A k ! = n [ k =1 ( B A k ) . 4. (a)...

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MATH 290 – 3 JUNE 2010 – EXAM 3 Answer all of the following questions on the answer sheets provided. Show all work, as partial credit may be given. This test will be counted out of a total of 60 points. 1. (16 pts.) Prove by induction that for all natural numbers n , n X k =1 ( k · k !) = ( n + 1)! - 1. 2. (16 pts.) Prove by induction that for all natural numbers n 4, 2 n < n !. 3. (16 pts.) Suppose that { A n : n N } is an indexed family of sets indexed by the natural numbers, and let B be a set. Prove by induction that for all natural numbers n , B
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Unformatted text preview: n [ k =1 A k ! = n [ k =1 ( B A k ) . 4. (a) (12 pts.) For all sets A , B , C , D , prove that ( A B ) ( C D ) ( A C ) ( B D ) . (b) (4 pts.) Find an example of sets A , B , C , and D where the inclusion in part (a) is proper, that is, where ( A B ) ( C D ) 6 = ( A C ) ( B D ). You do not have to provide a proof that your example works, a picture or similar justication will be sucient. (Hint: Try intervals on the real line.)...
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