Mathematical Induction

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1 Mathematical Induction William Cherry February 2011 These notes provide some additional examples to supplement the section of the text on mathematical induction. Inequalities. It happens that often in mathematics, the more freedom one has in creating a solution, the more difficult it is to solve a problem. Often the easiest problems to solve are those where there is really only one way to get to the solution. In particular, this means that it is often more difficult to prove an inequality than an equality. Because your textbook does not work through examples of how to use induction to prove inequalities and yet these can be some of the more difficult exercises, these notes are intended to provide some examples of using induction to prove inequalities. Proposition 1. For every n N, we have n 2 + 6n + 7 < 20n 2. Let P (n) be the proposition n 2 + 6n + 7 < 20n 2. Base Step. We check P (1), which says (1) + 7 < 20(1) 2. The left-hand-side is 14 and the right hand side is 20, so P (1) is true and the base step is complete. Assume: n 2 + 6n + 7 < 20n 2. Show: (n + 1) 2 + 6(n + 1) + 7 < 20(n + 1) 2. We will begin with the left-hand-side of the inequality we want to show because this is the more complicated looking side: (n + 1) 2 + 6(n + 1) + 7 = n 2 + 2n n [multiply out] = (n 2 + 6n + 7) + (2n + 7) [group terms to make use of our assumption] < 20n 2 + (2n + 7) [using our induction assumption] = 20n 2 + 2n + 7 < 20n n + 7 [since 2n < 40n] < 20n n + 20 [since 7 < 20] = 20(n 2 + 2n + 1) [factor out 20] = 20(n + 1) 2. [factor] Thus, (n + 1) 2 + 6(n + 1) + 7 < 20(n + 1) 2, which is what we needed to show. Remark. When we do a proof like this, it is important that all of our inequalities go the same way. In this case, they were all <. We may not mix < and >. 1

2 Mathematical Induction 2 Proposition 2. For every natural number n 12, we have 5 n < n!. Base Step. In this case, we only claim the inequality is true for n 12, so that makes our base step n = 12. Thus, we need to check whether 5 12 < 12!. Using a calculator or computer (or a lot of patience), we determine that and thus we see that the base step is true. Assume: 5 n < n!. Show: 5 n+1 < (n + 1)! = 244, 140, 625 and 12! = 479, 001, 600, We ll start with the left-hand-side of what we are trying to show: Hence, 5 n+1 < (n + 1)! as required. 5 n+1 = 5 5 n [re-write so we can use our assumption] < 5 n! [since 5 n < n! by our induction assumption] < (n + 1) n! [since n + 1 > 5 (remember n 12)] = (n + 1)!. [simplifying] Remark. This last example also shows the necessity of the base step. Notice that the only thing we needed in the induction step was that n + 1 > 5, so the induction step works as long as n > 5. However, 5 6 = 3125 and 6! = 720, so the proposition is not true for n = 6. We really need the base step too for our proof to be valid. Sometimes we might have to prove several inequalities in order to get to the one we want. For example, suppose we want to prove n 3 < 2 n. If we take a look at the induction step, it would go something like this: Assume: n 3 < 2 n. Show: (n + 1) 3 < 2 n+1. If we now start working with what we want to show, we get something like (n + 1) 3 = n 3 + 3n 2 + 3n + 1 [multiply out] < 2 n + 3n 2 + 3n + 1. But, now we are kind of stuck. If, however, we somehow knew that 3n 2 + 3n + 1 < 2 n, we could continue: 2 n + 3n 2 + 3n + 1 < 2 n + 2 n = 2 2 n = 2 n+1. Well, we should then try to prove 3n 2 + 3n + 1 < 2 n, which we can also do by induction, after taking another detour.

3 Mathematical Induction 3 Proposition 3. For each naturual number n 6, we have 6n + 6 < 2 n. Base Step. When n = 6, we have Assume: 6n + 6 < 2 n. Show: 6(n + 1) + 6 < 2 n+1. Now, as was to be shown. 6(n + 1) + 6 = 6n (6) + 6 = 42 < 64 = 2 6. < 2 n + 6 [using our induction assumption] < 2 n + 2 n [since 6 < 2 n when n 3] = 2 2 n = 2 n+1, Proposition 4. For each natural number n 8, we have 3n 2 + 3n + 1 < 2 n. Base Step. When n = 8, we have and so the base step is true. 3(8) 2 + 3(8) + 1 = 217 < 256 = 2 8, Assume: 3n 2 + 3n + 1 < 2 n. Show: 3(n + 1) 2 + 3(n + 1) + 1 < 2 n+1. Starting with the left-hand-side of what we need to show, we have 3(n + 1) 2 + 3(n + 1) + 1 = 3(n 2 + 2n + 1) + 3n as required. = 3n 2 + 6n n = (3n 2 + 3n + 1) + (6n + 6) [Re-group to apply induction assumption] < 2 n + (6n + 6) [By our induction assumption] < 2 n + 2 n [By Proposition 3] = 2 2 n = 2 n+1, We are now finally able to get to the inequality we wanted. Proposition 5. For each natural number n 10, we have n 3 < 2 n. Base Step. When n = 10, we have 10 3 = 1000 < 1024 = 2 10,

4 Mathematical Induction 4 and so the base step holds. Assume: n 3 < 2 n. Show: (n + 1) 3 < 2 n+1. Starting with the left-hand-side, and so we are done. (n + 1) 3 = n 3 + 3n 2 + 3n + 1 [multiply out] < 2 n + 3n 2 + 3n + 1 [using our induction assumption] < 2 n + 2 n [by Proposition 4] = 2 2 n = 2 n+1, Generalized Strong Induction. The method of proof by induction can be generalized as follows. Suppose we have a proposition P (n) and suppose we want to prove that P (n) is true for all integers n such that n m for some integer m. Suppose we also prove the following: Generalized base step: There exists an l Z such that P (j) is true for j = m, m + 1,..., l. Strong induction step: For each k Z such that k l, we have (P (m) P (m + 1) P (m + 2) P (k)) P (k + 1). In analogy with my stairstep explanation in class, this goes as follows. For the generalized base step, suppose we can show that we can get to any of the first five stairs on the infinite stair case. Now suppose that we can show that if we can get all the stairs before the k + 1-st step, then we can also get to k + 1-st step. Then we can climb all the stairs. Imagine, for example, that in order to go up a stair, we need to both stand on the stair before, and use the assistance of a pole resting on the stair before that. In practice, strong induction often works as follows. We do a double base step, for example, we show P (1) and P (2) are true. Then, for the induction step, we show that P (n 1) and P (n) together imply P (n + 1). That is, we use the assistance of two previous stairs to help us climb to the next step. Here is an example. Proposition 6. For each natural number n, there exist natural numbers a and b such that 5 n = a 2 + b 2. Proof by strong induction. Base step. We will do a double base step, allowing us to go two steps back in our induction step. For n = 1, we can choose a = 1 and b = 2 and we have 5 1 = 5 = = For n = 2, we can choose a = 3 and b = 4, since 5 2 =

5 Mathematical Induction 5 For this proof, instead of showing that P (n) implies P (n + 1), we will show that P (n 1) implies P (n + 1). Because we are skipping over a step each time, we need the double base step above. Assume: There exist a and b in N such that 5 n 1 = a 2 + b 2. Show: There exist c and d in N such that 5 n+1 = c 2 + d 2. We start with the left-hand-side, and write 5 n+1 = n 1. Then, we use our induction assumption that 5 n 1 = a 2 + b 2. So, we then have 5 n+1 = n 1 = 5 2 (a 2 + b 2 ) = 5 2 a b 2 = (5a) 2 + (5b) 2. Thus, we can choose c = 5a and d = 5b to get and our proof is complete. 5 n+1 = c 2 + d 2, Fibonacci numbers. The Fibonacci numbers are defined recursively as follows: F 0 = 0, F 1 = 1, and F n = F n 1 + F n 2 for n 2. So, for instance, the Fibonacci numbers start out 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,... where the first two are 0 and 1 and each successive number is obtained by adding the previous two. Problems involving Fibonacci numbers are natural candidates for going backward two steps. Proposition 7. For each n = 0, 1, 2, 3,..., we have F n < 2 n. Proof by strong induction. Base Step. We will verify the inequality for n = 0 and n = 1 for our base step. For n = 0, we have F 0 = 0 < 1 = 2 0. For n = 1, we have F 1 = 1 < 2 = 2 1, and so we have verified the inquality for n = 0 and n = 1. Induction step. Because we have completed a double base step, we are allowed to use two induction assumptions. Assumption 1. F n 1 < 2 n 1. Assumption 2. F n < 2 n. Show. F n+1 < 2 n+1. Starting with the definition of the Fibonacci numbers, we have F n+1 = F n + F n 1 [definition of F n ] < 2 n + 2 n 1 [Using our two induction assumptions] < 2 n + 2 n [since 2 n 1 < 2 n.] = 2 2 n = 2 n+1. Hence, F n+1 < 2 n+1 as was to be shown.

6 Mathematical Induction 6 The method of strong induction can sometimes be used to shorten an induction proof of an inequalty. Earlier in these notes we proved that n 3 < 2 n for all natural numbes n 10, but our proof was rather tedius. Using strong induction, we can shorten our proof. Proposition 8. For each natural number n 10, we have n 3 < 2 n. Proof by strong induction. Base Step. First, we will do an extended base step: Induction step. Assume: n 10 and n 3 < 2 n. Show: (n + 2) 3 < 2 n+2. n n 3 2 n (n + 2) 3 = n 3 + 6n n + 8 [multiply out] < n 3 + n n + 8 [since 6 < n] < n 3 + n 3 + n [since 12 < n 2 ] < n 3 + n 3 + n 3 + n 3 [since 8 < n 3 ] = 4n 3 < 4 2 n [by our induction assumption] = 2 n+2.

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