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following statement. For every integer n 2 1, 3" - 2 is even. g is a proposed proof by mathematical induction for the statement.

following statement. For every integer n 2 1, 3" - 2 is even. g is a proposed proof by mathematical induction for the statement.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A proof by mathematical induction is supposed to show that a given property is true for every integer greater than or equal to an initial value. In order for it to be valid, the property must be true for the initial value, and the argument in the inductive step must be correct for every integer greater than or equal to the initial value. Consider the following statement. For every integer n 2 1, 3'" – 2 is even. The following is a proposed proof by mathematical induction for the statement. Since the property is true for n = 1, the basis step is true. Suppose the property is true for an integer k, where k > 1. That is, suppose that 3K – 2 is even. We must show that 3K + 1 - 2 is even. Observe that %3D 3k. 3 – 2 = 3k(1 + 2) – 2 (3k – 2) + 3k . 2. 3k + 1 2 Now 3K – 2 is even by inductive hypothesis, and 3K · 2 is even by inspection. Hence the sum of the two quantities is even by (Theorem 4.1.1). It follows that 3* + - - 2 is even, which is what we needed to show. Identify the error(s) in the proof. (Select all that apply.) 3k + 1 - 2 + (3k – 2) + 3k . 2 The inductive hypothesis is assumed to be true. The property is not true for n = 1. 3k 2 is odd by the inductive hypothesis. O (3k – 2) + 3k. 2 + 3*(1 + 2) – 2
Transcribed Image Text:A proof by mathematical induction is supposed to show that a given property is true for every integer greater than or equal to an initial value. In order for it to be valid, the property must be true for the initial value, and the argument in the inductive step must be correct for every integer greater than or equal to the initial value. Consider the following statement. For every integer n 2 1, 3'" – 2 is even. The following is a proposed proof by mathematical induction for the statement. Since the property is true for n = 1, the basis step is true. Suppose the property is true for an integer k, where k > 1. That is, suppose that 3K – 2 is even. We must show that 3K + 1 - 2 is even. Observe that %3D 3k. 3 – 2 = 3k(1 + 2) – 2 (3k – 2) + 3k . 2. 3k + 1 2 Now 3K – 2 is even by inductive hypothesis, and 3K · 2 is even by inspection. Hence the sum of the two quantities is even by (Theorem 4.1.1). It follows that 3* + - - 2 is even, which is what we needed to show. Identify the error(s) in the proof. (Select all that apply.) 3k + 1 - 2 + (3k – 2) + 3k . 2 The inductive hypothesis is assumed to be true. The property is not true for n = 1. 3k 2 is odd by the inductive hypothesis. O (3k – 2) + 3k. 2 + 3*(1 + 2) – 2
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