fpsolns

# fpsolns - Mathematics 55 Spring 2005 Solutions for Final...

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Mathematics 55, Spring 2005 – Solutions for Final Exam Review Problems 1 Here are solutions, answers, hints, and references for solutions for some of the review problems. (A1) Prove the identity n i = k ( i k ) = ( n +1 k +1 ) by induction. (Here n is any positive integer.) Solution. We prove this by induction on n (the instructions didn’t say whether to argue by induction on n , or on k , so it’s up to us). If n < k then both sides are zero; there are then no terms in the sum on the left, while ( n +1 k +1 ) = 0 by definition since n + 1 < k + 1. Basis step: n = k . Then the left-hand side is n i = n ( i n ) = ( n n ) = 1. The right-hand side is ( n +1 n +1 ) = 1, so they’re equal. Inductive step: Assume true for n , prove for n + 1: n +1 i = k i k = n i = k i k + n + 1 k = n + 1 k + 1 + n + 1 k = n + 1 k + 1 by Pascal’s identity. (A2) Identify each of the following sets as finite, countably infinite, or uncountable. (No proofs are required.) (A2a) The set Z of all integers. (A2b) N (A2c) The set R of all real numbers. (A2d) The set of all rational numbers. (A2e) The set of all subsets of Z . (A2f) The set of all infinite sequences ( a 1 , a 2 , a 3 , · · · ), where each a k ∈ { 0 , 1 } . Answers. Countably infinite: Z , the set of all rationals, and the set of all infinite sequences of the type described. Uncountable: R (see Example 20 page 235 of Rosen) and the set of all subsets of Z (certainly not finite; and we showed earlier in the semester that the power set of a set S can never be placed into one-to-one correspondence with S , so the only remaining possibility is that the power set of Z be uncountable). (A3) Suppose that a 0 = 1 and that for every positive integer n , a n = na n - 1 + ( - 1) n . Show that a n = n ! n k =0 ( - 1) k k ! for every n 0. Solution. We prove this by induction on n . Basis step n = 1: the recurrence relation gives 1 a 0 + ( - 1) 1 = 1 - 1 = 0, while on the other hand 1! 1 k =0 ( - 1) k k ! = 1 0! - 1 1! = 1 - 1 = 0. So the formula holds for n = 1. Inductive step: Assume true for n . a n +1 = ( n + 1) a n + ( - 1) n +1 = ( n + 1) n ! n k =0 ( - 1) k k ! + ( - 1) n +1 = ( n + 1)! n k =0 ( - 1) k k ! + ( - 1) n +1 = ( n + 1)! n +1 k =0 ( - 1) k k ! - ( n + 1)! ( - 1) n +1 ( n + 1)! + ( - 1) n +1 = ( n + 1)! n +1 k =0 ( - 1) k k ! - ( - 1) n +1 + ( - 1) n +1 = ( n + 1)! n +1 k =0 ( - 1) k k ! . (A4) A game between two players A, B has these rules: There is a pile of n tokens. Player A goes first, and removes some number of tokens from the pile. Player B then does the same, and they take turns. At each turn, the player must remove at least one token, and no more than 4 tokens. The last player to remove a token is the loser. 1 c 2005 by Michael Christ. All rights reserved. 1

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Prove that if n is congruent to 1 modulo 5 then the second player will always win if he/she plays correctly, but that in all other cases the first player will always win if she/he plays correctly. Hint. The winning strategy: A player should play so that the number of tokens remaining after her/his turn is congruent to 1 modulo 5. n may be written as 5 q + r for some unique nonnegative integer q and some r ∈ { 0 , 1 , 2 , 3 , 4 } . Prove by induction on q that if 5 q + r tokens are in the pile, then if the second player follows this strategy, then she/he will always win.
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