ng statement by mathematical induction. For every integer n ≥ 0, 7 - 1 is divisible by 6. Proof (by mathematical induction): Let P(n) be the following sentence. 7- 1 is divisible by 6. We will show that P(n) is true for every integer n 20. Show that P(0) is true: Select P(0) from the choices below. 061(7⁰-1) 6 is a multiple of 70 - 1 (7⁰-1) 16 O 1 is a factor of 70 - 1 X The truth of the selected statement follows from the definition of divisibility and the fact that 70-1=-1 X 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 2 0, and suppose that P(k) is true. Select P(K) from the choices below. 6 is a multiple of 7k - 1 O (7k-1) is divisible by 6 O 6 is divisible by (7k - 1) O 1 is a factor of 7k- 1 X [This is P(k), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below. 6 is a multiple of 7k +1 -1. O 6 is divisible by (7k+1-1) O1 is a factor of 7k +1 -1 O (7k+1-1) is divisible by 6

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 - 1 is divisible by 6.
Proof (by mathematical induction): Let P(n) be the following sentence.
7-1 is divisible by 6.
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.
0 61 (7⁰1)
O 6 is a multiple of 70 - 1
(7⁰-1) 16
O 1 is a factor of 70 - 1
The truth of the selected statement follows from the definition of divisibility and the fact that 70-1 = -1
X
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 2 0, and suppose that P(k) is true. Select P(K) from the choices below.
6 is a multiple of 7k - 1
O (7k 1) is divisible by 6
O 6 is divisible by (7k - 1)
O 1 is a factor of 7k - 1
X
Now
[This is P(K), the inductive hypothesis.]
We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below.
Ⓒ6 is a multiple of 7k +1 -1.
O 6 is divisible by (7k+11)
O 1 is a factor of 7k +1 -1.
O (7k+11) is divisible by 6
X
6.67
X
By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 1 = 6r, and so 7k = 6r + 1.
4
7k+1 -1 =7k. 7-1.
When 6r + 1 is substituted for 7K in the above equation, and the right-hand side is simplified, the result can be expressed in terms of k and r as follows.
7k+1 -1
Transcribed Image Text:.webassign.net/web/Student/Assignment-Responses/submit?dep=33264788&tags=autosave#question4603719_0 Prove the following statement by mathematical induction. For every integer n ≥ 0, 7 - 1 is divisible by 6. Proof (by mathematical induction): Let P(n) be the following sentence. 7-1 is divisible by 6. 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. 0 61 (7⁰1) O 6 is a multiple of 70 - 1 (7⁰-1) 16 O 1 is a factor of 70 - 1 The truth of the selected statement follows from the definition of divisibility and the fact that 70-1 = -1 X 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 2 0, and suppose that P(k) is true. Select P(K) from the choices below. 6 is a multiple of 7k - 1 O (7k 1) is divisible by 6 O 6 is divisible by (7k - 1) O 1 is a factor of 7k - 1 X Now [This is P(K), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below. Ⓒ6 is a multiple of 7k +1 -1. O 6 is divisible by (7k+11) O 1 is a factor of 7k +1 -1. O (7k+11) is divisible by 6 X 6.67 X By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 1 = 6r, and so 7k = 6r + 1. 4 7k+1 -1 =7k. 7-1. When 6r + 1 is substituted for 7K in the above equation, and the right-hand side is simplified, the result can be expressed in terms of k and r as follows. 7k+1 -1
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