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C__work_280sol2

# C__work_280sol2 - Solutions to Homework 2 February 9 2007...

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Solutions to Homework 2 February 9, 2007 General Comments: This problem set will probably end up being the hardest in the course for many of you. Your score will, in all likelihood, be significantly lower than on the previous homeworks. If you are frustrated with your performance, don’t give up. Many people are new to this kind of methodology and it will take some time for it to sink in. There are lots of induction problems in the practice exercises for you to try. See me or a TA or grader if you’re having trouble with them. Also, read these solutions carefully . They are an example of good mathematical exposition and may help you to write better, clearer, and more concise proofs next time around. 2.2: 24 Let P ( n ) be the statement that n 3 - n is divisible by 3. (a) We prove P ( n ) for n = 0 , 1 , 2 , 3 , . . . by induction. Base case: P (0) says that 0 3 - 0 = 0 is divisible by 3, which is clearly true. Inductive step: Assume P ( n ). We want to show P ( n +1): ( n +1) 3 - ( n + 1) is divisible by 3. Note that ( n + 1) 3 - ( n + 1) = n 3 + 3 n 2 + 3 n + 1 - ( n + 1) = ( n 3 - n ) + 3 n 2 + 3 n By the induction assumption, n 3 - n is divisible by 3. Clearly 3 n 2 + 3 n is divisible by 3. Hence, their sum, which is ( n +1) 3 - ( n +1) is divisible by 3. 1

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2.2: 26 Let P ( n ) be the statement: 13 n + 6 is divisible by 7 if n is even. We prove P ( n ) by strong induction for n 0. Basis: P (0) is 13 0 + 6 is divisible by 7. Since 13 0 + 6 = 7, this is clearly true. Inductive step: Suppose P (0) , . . . , P ( n ) are true. We prove P ( n + 1), which is the statement that 13 n +1 + 6 is divisible by 7 if n + 1 is even. If n +1 is odd, there is nothing to prove. If n +1 is even, then so is n - 1. Moreover, n - 1 0 (since if n + 1 is even, n + 1 is at least 2). By P ( n - 1), 13 n - 1 + 6 is divisible by 7. Note that 13 n +1 + 6 = 13 2 (13 n - 1 + 6) - 13 2 (6) + 6 = 13 2 (13 n - 1 + 6) - 1008 . Since 13 n - 1 + 6 is divisible by 7 (by P ( n - 1)), so is 13 2 (13 n - 1 + 6). It is easy to check that 1008 is also divisible by 7. Thus, 13 2 (13 n - 1 + 6) - 1008 is divisible by 7, which is what we want to show. 2.3: 12 The obvious generalization is A 1 . . . A n = A 1 . . . A n for n 2. Let’s prove this by strong induction. Let P ( n ) be the statement A 1 . . . A n = A 1 . . . A n . We prove P ( n ) for n 2. You were told you could assume the base case. (You should be able to prove it though.) Assume P (2) , . . . , P ( n ). Let’s prove P ( n + 1): A 1 . . . A n A n +1 = ( A 1 . . . A n ) A n +1 [note the grouping here] = A 1 . . . A n A n +1 [by P (2)] = A 1 . . . A n A n +1 [by P ( n )] 2.3: 19 (a) You also have to show that P (1), P (2), P (3), and P (4) are true. (These become the base cases.) If you can show that P (0), P (1), P (2), and P (3) are true, then you can’t use the induction hypothesis as stated, because it says that P ( n +1) is true for n > 3 (which means that n +1 will be at least 5!) if P ( n ), P ( n - 1), P ( n = 2), and P ( n - 3) are true. You lost 1 if you did this. If you prove that P (4), P (5), P (6), and P (7) are true, then you can use the induction hypothesis as stated, but you’ll end up proving that P ( n ) holds for all n 4, not for all positive n , which is what the problem asks you to do. You lost 1 if you did this.
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C__work_280sol2 - Solutions to Homework 2 February 9 2007...

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