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Prove the following statement by mathematical induction. For every integer n ≥ 0, 7"-2" is divisible by 5. Proof (by mathematical induction): Let P(n) be the following sentence. 7n2n is divisible by 5. We will show that P(n) is true for every integer n ≥ 0. Show that P(0) is true: Select P(0) from the choices below. (7⁰-20) 15 051 (7⁰-20) O 5 is a multiple of 7⁰ - 20 O 70-20 < 5 The truth of the selected statement follows from the definition of divisibility and the fact that 7⁰ ? ✓ = 0. Show that for each integer k ≥ 0, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below. O 7k-2k <5 O5 is a multiple of 7k - 2k O 5 is divisible by 7k - 2k O 7k 2k is divisible by 5 [This is P(k), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below. 5 is a multiple of 7k+ 1 - 2k +1 O 7k+12k+1 is divisible by 5 O7k+12k+1 < 5 O 5 is divisible by 7k+1_2k + 1 By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 2k = 5r. Then 7k+1_2k+1=7.7k-2.2k= (5 + 2)-7k 2.2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k+1_2k+ 1 = 5. ( This quantity is an integer because k and r are integers. Thus 7k + 1-2k + 1 ---Select---divisible by 5, and hence P(k + 1) is ---Select---, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

Prove the following statement by mathematical induction. For every integer n ≥ 0, 7"-2" is divisible by 5. Proof (by mathematical induction): Let P(n) be the following sentence. 7n2n is divisible by 5. We will show that P(n) is true for every integer n ≥ 0. Show that P(0) is true: Select P(0) from the choices below. (7⁰-20) 15 051 (7⁰-20) O 5 is a multiple of 7⁰ - 20 O 70-20 < 5 The truth of the selected statement follows from the definition of divisibility and the fact that 7⁰ ? ✓ = 0. Show that for each integer k ≥ 0, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below. O 7k-2k <5 O5 is a multiple of 7k - 2k O 5 is divisible by 7k - 2k O 7k 2k is divisible by 5 [This is P(k), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below. 5 is a multiple of 7k+ 1 - 2k +1 O 7k+12k+1 is divisible by 5 O7k+12k+1 < 5 O 5 is divisible by 7k+1_2k + 1 By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 2k = 5r. Then 7k+1_2k+1=7.7k-2.2k= (5 + 2)-7k 2.2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k+1_2k+ 1 = 5. ( This quantity is an integer because k and r are integers. Thus 7k + 1-2k + 1 ---Select---divisible by 5, and hence P(k + 1) is ---Select---, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Prove the following statement by mathematical induction. For every integer n ≥ 0, 7" - 2" is divisible by 5. Proof (by mathematical induction): Let P(n) be the following sentence. 7" - 2n is divisible by 5. We will show that P(n) is true for every integer n ≥ 0. Show that P(0) is true: Select P(0) from the choices below. (70 - 20) | 5 5 | (7⁰ - 20) 5 is a multiple of 7⁰ -2⁰ 7⁰ -2⁰ <5 The truth of the selected statement follows from the definition of divisibility and the fact that 7⁰ ? V = 0. Show that for each integer k≥ 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below. 7K-2k < 5 5 is a multiple of 7k - 2k 5 is divisible by 7k - 2k O 7k - 2k is divisible by 5 [This is P(K), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+1) from the choices below. 5 is a multiple of 7k + 1 - 2k + 1 7k + 1 2k +1 is divisible by 5 O 7k+ 2k + 1 <5 5 is divisible by 7k + 1 2k + 1 By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 2k = 5r. Then 2 7k + 12k + 1 = 7.7k - 2. 2k = (5 + 2) · 7k − 2.2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k + 1 _ - 2k + 1 = 5. This quantity is an integer because k and r are integers. Thus 7k + 12k + 1 ---Select---divisible by 5, and hence P(k + 1) is ---Select--- ✓ which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Transcribed Image Text:Prove the following statement by mathematical induction. For every integer n ≥ 0, 7" - 2" is divisible by 5. Proof (by mathematical induction): Let P(n) be the following sentence. 7" - 2n is divisible by 5. We will show that P(n) is true for every integer n ≥ 0. Show that P(0) is true: Select P(0) from the choices below. (70 - 20) | 5 5 | (7⁰ - 20) 5 is a multiple of 7⁰ -2⁰ 7⁰ -2⁰ <5 The truth of the selected statement follows from the definition of divisibility and the fact that 7⁰ ? V = 0. Show that for each integer k≥ 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below. 7K-2k < 5 5 is a multiple of 7k - 2k 5 is divisible by 7k - 2k O 7k - 2k is divisible by 5 [This is P(K), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+1) from the choices below. 5 is a multiple of 7k + 1 - 2k + 1 7k + 1 2k +1 is divisible by 5 O 7k+ 2k + 1 <5 5 is divisible by 7k + 1 2k + 1 By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 2k = 5r. Then 2 7k + 12k + 1 = 7.7k - 2. 2k = (5 + 2) · 7k − 2.2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k + 1 _ - 2k + 1 = 5. This quantity is an integer because k and r are integers. Thus 7k + 12k + 1 ---Select---divisible by 5, and hence P(k + 1) is ---Select--- ✓ which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
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