A guide to Proof by Induction

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Gordon Parsons
5 years ago
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1 A guide to Proof by Induction Adapted from L. R. A. Casse, A Bridging Course in Mathematics, The Mathematics Learning Centre, University of Adelaide, Inductive reasoning is where we observe of a number of special cases and then propose a general rule. For example, if we observe five or six times that it rains as soon as we hang out the washing, then we might propose that hanging out the washing causes it to rain. Obviously, inductive reasoning has a basic flaw. Lots of special cases might fit the general proposal but that still does not prove that all cases will fit. Here is a mathematical example: Special case: = 4 Special case: = 9 Special case: = 16 Special case: =? There seems to be a pattern here. The next answer in this sequence looks like it should be 25 (or 5 2 ). To propose a general rule we need to relate the right-hand side to the left. Perhaps the relationship is This is a proposition P (n). The sum of the first 2 odd counting numbers = 4 = 2 2 The sum of the first 3 odd counting numbers = 9 = 3 2 The sum of the first 4 odd counting numbers = 16 = 4 2 The sum of the first n odd counting numbers = n 2. P (n) : nth odd counting number = n 2 It would help to replace the phrase nth odd counting number with something more mathematical. 2nd odd number (n = 2) is 3 3rd odd number (n = 3) is 5 4th odd number (n = 4) is 7 5th odd number (n = 5) is 9 G Coates 1 September 2014

2 In each case, the nth odd number is one less than double n, so nth odd number is 2n 1. Our proposition is now P (n) : (2n 1) = n 2 So, how do we prove that P (n) is true for all cases (ie. n 1)? The Principle of Mathematical Induction uses the structure of propositions like this to develop a proof. What we do is assume we know that the proposition is true for an arbitrary special case (call it n = k) and then use this assumption to show that the proposition is true for the next special case (ie. n = k + 1). In our example, we know that the proposition is true for n = 2, 3 and 4. (In fact, it s clearly also true for n = 1.) So, we know it s true for k = 4. If we have shown that the statement above in italics is true, this automatically proves the proposition is true for the next one, k = 5. Now we know it s true for k = 5, it must also be true for k = 6, and so on. In mathematical notation, here is the definition of Mathematical Induction: The Principle of Mathematical Induction Suppose P(n) is a proposition defined for every integer n a. If (1) P (a) is true, and (2) P (k + 1) is true assuming P (k) is true, where k a, then P (n) is true for all integers n a. To prove (2) in our example, start with what we are allowed to assume: Now state what we need to prove: P (k) : (2k 1) = k 2. P (k + 1) : (2(k + 1) 1) = (k + 1) 2. How can we use P (k) to prove P (k + 1)? Well, the left-hand side (LHS) of the formula for P (k + 1) is just the LHS of the formula for P (k) plus an extra term: P (k + 1) : (2k 1) + (2(k + 1) 1) = (k + 1) 2. If we replace these terms with the assumed right-hand side of P (k) we get P (k + 1) : k 2 + (2(k + 1) 1) = (k + 1) 2. G Coates 2 September 2014

3 We now manipulate this new LHS to make it equal to the right-hand side (RHS): k 2 + (2(k + 1) 1) = k 2 + 2k = k 2 + 2k + 1 = (k + 1) 2 = RHS We now know that P (1) is true (by direct examination) and P (k + 1) is true if P (k) is true for k 1. Hence we have proved the proposition by induction. There are two other broad proposition structures that can be proved by induction, divisibility and inequality propositions. Divisibility: Prove P (n) : 3 2n 1 is divisible by 8 (1) The smallest value of n is 1 so P (1) claims that = 8 is divisible by 8. Clearly this is true. (2) We assume that P (k) is true, so 3 2k 1 is assumed to be divisible by 8. Now, P (k + 1) states that 3 2(k+1) 1 is divisible by 8. To prove it, re-arrange this expression to include 3 2k 1: 3 2(k+1) 1 = 3 2k+2 1 = 9 3 2k 1 = 9 ( 3 2k 1 ) + 8 We now have two terms. The first is divisible by 8 because we assume the factor in brackets is divisible by 8. The second is just 8 so that is also divisible by 8. Hence, by the principle of mathematical induction, P (n) is true for all integers n 1. Inequality: Prove P (n) : 2 n > n + 4 for n 3. (1) The smallest value of n is 3 so P (3) claims that 2 3 = 8 is greater than = 7. Clearly this is true. (2) We assume that P (k) is true, so 2 k > k + 4 is assumed to be true. Now, P (k + 1) states that 2 k+1 > (k + 1) + 4. With inequalities it helps to have zero on one side: 2 k+1 (k + 1) 4 > 0 G Coates 3 September 2014

4 Now, this can be re-arranged to include 2 k : 2 2 k k 5 > 0 We are assuming that 2 k > k + 4 so, if we replace 2 k in the above expression with the smaller number k + 4, we produce a smaller result. So: Now, 2(k + 4) k 5 = k + 3 so Since k 3, we know that k + 3 > 0, so 2 2 k k 5 > 2(k + 4) k k k 5 > k k k 5 > k + 3 > 0. Hence, by the principle of mathematical induction, P (n) is true for all integers n 1. G Coates 4 September 2014

5 Exercises Prove the following propositions using mathematical induction: (3n + 1) = (n + 1)(3n + 2) 2 for n n 1 = 5n n = n(n + 1) n 2 = n(n + 1)(2n + 1) n 3 = n2 (n + 1) r + r r n = 1 rn+1 1 r for n 0 and any r n > n 8.* 4 n > n n(n + 1) = n n (1 + a) n 1 + na for n 1 where a > 1 is a fixed real number. 11. (a) 7 2n 1 is divisible by 48 (b)* n 3 + 5n is divisible by 6 G Coates 5 September 2014

CSC 344 Algorithms and Complexity. Proof by Mathematical Induction

CSC 344 Algorithms and Complexity Lecture #1 Review of Mathematical Induction Proof by Mathematical Induction Many results in mathematics are claimed true for every positive integer. Any of these results