 ## Presentation on theme: "Copyright © 2007 Pearson Education, Inc. Slide 8-1."— Presentation transcript:

Copyright © 2007 Pearson Education, Inc. Slide 8-2 Chapter 8: Further Topics in Algebra 8.1Sequences and Series 8.2Arithmetic Sequences and Series 8.3Geometric Sequences and Series 8.4The Binomial Theorem 8.5Mathematical Induction 8.6Counting Theory 8.7Probability

Copyright © 2007 Pearson Education, Inc. Slide 8-3 8.5 Mathematical Induction Mathematical induction is used to prove statements claimed true for every positive integer n. For example, the summation rule is true for each integer n > 1.

Copyright © 2007 Pearson Education, Inc. Slide 8-4 8.5 Mathematical Induction Label the statement S n. For any one value of n, the statement can be verified to be true.

Copyright © 2007 Pearson Education, Inc. Slide 8-5 8.5 Mathematical Induction To show S n is true for every n requires mathematical induction.

Copyright © 2007 Pearson Education, Inc. Slide 8-6 8.5 Mathematical Induction Principle of Mathematical Induction Let S n be a statement concerning the positive integer n. Suppose that 1. S 1 is true; 2. For any positive integer k, k < n, if S k is true, then S k+1 is also true. Then, S n is true for every positive integer n.

Copyright © 2007 Pearson Education, Inc. Slide 8-7 8.5 Mathematical Induction Proof by Mathematical Induction Step 1 Prove that the statement is true for n = 1. Step 2 Show that for any positive integer k, if S k is true, then S k+1 is also true.

Copyright © 2007 Pearson Education, Inc. Slide 8-8 8.5 Proving an Equality Statement Example Let S n be the statement Prove that S n is true for every positive integer n. Solution The proof uses mathematical induction.

Copyright © 2007 Pearson Education, Inc. Slide 8-9 8.5 Proving an Equality Statement Solution Step 1 Show that the statement is true when n = 1. S 1 is the statement which is true since both sides equal 1.

Copyright © 2007 Pearson Education, Inc. Slide 8-10 8.5 Proving an Equality Statement Solution Step 2 Show that if S k is true then S k+1 is also true. Start with S k and assume it is a true statement. Add k + 1 to each side

Copyright © 2007 Pearson Education, Inc. Slide 8-11 8.5 Proving an Equality Statement Solution Step 2

Copyright © 2007 Pearson Education, Inc. Slide 8-12 8.5 Proving an Equality Statement Solution Step 2 This is the statement for n = k + 1. It has been shown that if S k is true then S k+1 is also true. By mathematical induction S n is true for all positive integers n.

Copyright © 2007 Pearson Education, Inc. Slide 8-13 8.5 Mathematical Induction Generalized Principle of Mathematical Induction Let S n be a statement concerning the positive integer n. Suppose that Step 1 S j is true; Step 2 For any positive integer k, k > j, if S k implies S k+1. Then, S n is true for all positive integers n > j.

Copyright © 2007 Pearson Education, Inc. Slide 8-14 8.5 Using the Generalized Principle Example Let S n represent the statement Show that S n is true for all values of n > 3. Solution Since the statement is claimed to be true for values of n beginning with 3 and not 1, the proof uses the generalized principle of mathematical induction.

Copyright © 2007 Pearson Education, Inc. Slide 8-15 8.5 Using the Generalized Principle Solution Step 1 Show that S n is true when n = 3. S 3 is the statement which is true since 8 > 7.

Copyright © 2007 Pearson Education, Inc. Slide 8-16 8.5 Using the Generalized Principle Solution Step 2 Show that S k implies S k+1 for k > 3. Assume S k is true. Multiply each side by 2, giving or or, equivalently

Copyright © 2007 Pearson Education, Inc. Slide 8-17 8.5 Using the Generalized Principle Solution Step 2 Since k > 3, then 2k >1 and it follows that or which is the statement S k+1. Thus S k implies S k+1 and, by the generalized principle, S n is true for all n > 3.