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Unformatted text preview: 1 Induction 1.1 Welcome Welcome to the twelfth module in our course on proofs. Before going on, please do the assigned readings. 1.2 Introduction Induction is a common and powerful technique and should be your first choice whenever you en counter a statement of the form for every integer n ≥ 1, P ( n ) is true where P ( n ) is a statement that depends on n . Here are two examples of propositions in this form. Proposition 1. For every integer n ≥ 1 n X i =1 i 2 = n ( n + 1)(2 n + 1) 6 . Often the clause “For every integer n ≥ 1” is implied and does not actually appear in the proposition, as in the following version of the same theorem. Proposition 2. The sum of the first n perfect squares is n ( n +1)(2 n +1) 6 . The second example uses sets, not equations. Proposition 3. Every set of size n has exactly 2 n subsets. 1.3 How To Use Simple Induction Our technique relies on the Principle of Mathematical Induction (POMI) sometimes called Simple Induction . We will see variations later on. Principle of Mathematical Induction (POMI) Let P ( n ) be a statement that depends on n ∈ P . If 1. P (1) is true, and 2. P ( k ) is true implies P ( k + 1) is true then P ( n ) is true for all n ∈ P . There are three parts in a proof by induction. Base Case Verify that P (1) is true . This is usually easy. In condensed proofs you will often see the statement “It is easy to see that the statement is true for n = 1.” It is best to write this step out completely. 1 Inductive Hypothesis Assume that P ( k ) is true, k ≥ 1. It is best to write out the statement P ( k ). Inductive Conclusion Using the assumption that P ( k ) is true, show that P ( k +1) is true . Again, it is usually best to write out the statement P ( k + 1) before trying to prove it. 1.4 Why Does Induction Work? The basic idea is simple. We show that P (1) is true. We then use P (1) to show that P (2) is true. And then we use P (2) to show that P (3) is true and continue indefinitely. That is P (1) ⇒ P (2) ⇒ P (3) ⇒ ... ⇒ P ( i ) ⇒ P ( i + 1) ⇒ ... 1.5 Two Examples of Simple Induction Our first example is very typical and uses an equation containing the integer n , and we have already seen it. Proposition 4. n X i =1 i 2 = n ( n + 1)(2 n + 1) 6 . Proof. We begin by formally writing out our inductive statement P ( n ) : n X i =1 i 2 = n ( n + 1)(2 n + 1) 6 . We must establish our Base Case. Base Case We verify that P (1) is true where P (1) is the statement P (1) : 1 X i =1 i 2 = 1(1 + 1)(2 × 1 + 1) 6 . As in most base cases involving equations, we can make our way from the lefthand side of the equation to the right side of the equation with just a little algebra. 1 X i =1 i 2 = 1 2 = 1 = 1(1 + 1)(2 × 1 + 1) 6 ....
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This note was uploaded on 10/13/2011 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS

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