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Unformatted text preview: ICS 6D ( Discrete Mathematic): Solution to Homework #5 Section 4.2 4. a) It’s true for 18 , 19 , 20 , 21 because 18 = 7+7+4, 19 = 4+4+7+4, 20 = 4+4+4+4+4, 21 = 7+7+7. b) Inductive hypothesis: Assume that for some k we have that P ( i ) is true for all i s.t. 18 ≤ i ≤ k . (Note that “usually” the inductive hypothesis in a strong induction proof would be that P ( i ) is true for all i ≤ k , but that actually only applies to proofs of theorems where the base case is k = 0. Here the base case is k = 18 and hence we can only assume that P ( i ) is true for i between 18 and k .) c) What we need to prove is that P ( k + 1) holds, i.e. that k + 1 can be represented as a sum of 7’s and 4’s, assuming the inductive hypothesis on k , as stated above. d) The proof is simple: Since k + 1 = k + 4 if k = k 3 then if you can represent k = k 3 as a sum of 7’s and 4’s then you can do the same with k + 1. In other words P ( k 3) implies P ( k + 1). But is P ( k 3) implied by the inductive assumption stated in part (b)? Yes, because k ≥ 21 and therefore k 3 ≥ 18, and at the same time obviously k 3 ≥ k , so therefore the inductive assumption, that P ( i ) holds for all i between 18 and k , covers the case i = k 3, and therefore it implies P ( k 3). Since we’ve seen that P ( k 3) implies P ( k + 1), this implies P ( k + 1), as needed. e) This finishes the proof because part (a) covered the cases n = 18 , 19 , 20 , 21, and part (d) showed that for any k ≥ 21, if P ( i ) holds for all i between 18 and k then it holds for k + 1, and this implies that P ( n ) holds for all n ≥ 21. Thus P ( n ) holds for all n ≥ 18. 10. The answer is that exactly n 1 breaks are needed. Let P ( n ) state that every nsquared rectangular bar needs exactly n 1 breaks to be broken to single pieces. The base case: P (1) is obviously true. The (strong) inductive case: Let k be any integer k ≥ 1. Assume that P ( i ) holds for all i s.t. 1 ≤ i ≤ k . We’ll show that P ( k + 1) holds as well. Why? Assume you have a ( k + 1)squared rectangular bar. Assume you broke it somehow using a single break, and therefore now you have two rectangular bars, with n 1 and n 2 pieces each s.t. n 1 + n 2 = k + 1. Moreoever we have 1 ≤ n 1 , n 2 ≤ k . Therefore the inductive assumption kicks in for both n 1 and n 2 , and therefore we know that it takes exactly n 1 1 steps to break the first bar and exactly n 2 1 steps to break the second bar. (Note that you couldn’t do this step that easily using standard induction!!) Therefore it took ( n 1 1) + ( n 2 1) + 1 = ( n 1 + n 2 ) 1 = ( k + 1) 1 steps to break the main bar, which was what we needed to show.the main bar, which was what we needed to show....
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This homework help was uploaded on 01/30/2008 for the course ICS 6D taught by Professor Jarecki during the Fall '07 term at UC Irvine.
 Fall '07
 Jarecki

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