Page 275 - HOW TO PROVE IT: A Structured Approach, Second Edition
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                                              Proof by Mathematical Induction          261
                              We’ll say more about the justification of the method of mathematical induc-
                            tion later, but first let’s look at an example of a proof that uses mathematical
                            induction. The following list of calculations suggests a surprising pattern:
                                                    0
                                                            1
                                                   2 = 1 = 2 − 1
                                                0
                                                                   2
                                                    1
                                               2 + 2 = 1 + 2 = 3 = 2 − 1
                                                    2
                                            0
                                                1
                                                                       3
                                           2 + 2 + 2 = 1 + 2 + 4 = 7 = 2 − 1
                                                                           4
                                       0
                                                2
                                                    3
                                            1
                                      2 + 2 + 2 + 2 = 1 + 2 + 4 + 8 = 15 = 2 − 1
                            The general pattern appears to be:
                                                   1
                                                            n
                                               0
                                              2 + 2 +· · · + 2 = 2 n+1  − 1.
                            Will this pattern hold for all values of n? Let’s see if we can prove it.
                                                                          0   1        n
                            Example 6.1.1. Prove that for every natural number n, 2 + 2 +· · · + 2 =
                            2 n+1  − 1.
                            Scratch work
                            Our goal is to prove the statement ∀n ∈ NP(n), where P(n) is the statement
                                          n
                                 1
                             0
                            2 + 2 + ··· + 2 = 2 n+1  − 1. According to our strategy, we can do this by
                            proving two other statements, P(0) and ∀n ∈ N(P(n) → P(n + 1)).
                                                                                 0
                                                                                     1
                              Plugging in 0 for n, we see that P(0) is simply the statement 2 = 2 − 1,
                            the first statement in our list of calculations. The proof of this is easy – just do
                            the arithmetic to verify that both sides are equal to 1. Often the base case of an
                            induction proof is very easy, and the only hard work in figuring out the proof
                            is in carrying out the induction step.
                              For the induction step, we must prove ∀n ∈ N(P(n) → P(n + 1)). Of
                            course, all of the proof techniques discussed in Chapter 3 can be used in
                            mathematical induction proofs, so we can do this by letting n be an arbitrary
                            natural number, assuming that P(n) is true, and then proving that P(n + 1) is
                            true. In other words, we’ll let n be an arbitrary natural number, assume that
                                                                            1
                                 1
                             0
                                                                        0
                                          n
                            2 + 2 + ··· + 2 = 2 n+1  − 1, and then prove that 2 + 2 +· · · + 2 n+1  =
                            2 n+2  − 1. This gives us the following givens and goal:
                                         Givens                        Goal
                                                                 1
                                                             0
                               n ∈ N                        2 + 2 +· · · + 2 n+1  = 2 n+2  − 1
                                0
                                    1
                                             n
                               2 + 2 +· · · + 2 = 2 n+1  − 1
                              Clearly the second given is similar to the goal. Is there some way to start
                            with the second given and derive the goal using algebraic steps? The key to the
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