This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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, dont 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 youre 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 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 + 6 is divisible by 7. Since 13 + 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. Lets 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 ). Lets 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 cant 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!) ifwill be at least 5!...
View
Full
Document
This note was uploaded on 09/04/2008 for the course CS 280 taught by Professor  during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 

Click to edit the document details