Discrete Mathematics and Its Applications with MathZone

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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 n -squared 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.
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