Prove the following statement by mathematical induction. For every integer n 2 0, 7" - 1 is divisible by 6. Proof (by mathematical induction): Let P(n) be the following sentence. 7n - 1 is divisible by 6. We will show that P(n) is true for every integer n 2 0. Show that P(0) is true: Select P(0) from the choices below. O (7° - 1) | 6 O 61 (7° – 1) O 1 is a factor of 7° - 1 O 6 is a multiple of 70 - 1 The truth of the selected statement follows from the definition of divisibility and the fact that 7° - 1 = Show that for each integer k 2 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. O 6 is a multiple of 7k - 1 1 is a factor of 7* - 1 O 6 is divisible by (7* - 1) O (7k - 1) is divisible by 6 [This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. Select P(k + 1) from the choices below. O (7k * 1 - 1) is divisible by 6 O 1 is a factor of 7k + 1 _ 1 6 is a multiple of 7k *1 - 1 O 6 is divisible by (7k +1 - 1) By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 1 = 6r, and so 7k - 6r+ 1. Now 7k+1 - 1 = 7*.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. zk+1 - 1 = 6· This quantity is an integer because k and r are integers. Select the final sentence from the choices below. Hence, 6 is a multiple of (7**1 - 1), and so P(k+1) is false, which completes the inductive step. O Hence, 6 is divisible by (7**1 - 1), and so P(k+1) is true, which completes the inductive step. O Hence, (7k+1 1) is divisible by 6, and so P(k+1) is false, which completes the inductive step. Hence, (7**1 - 1) is divisible by 6, and so P(k+1) is true, which completes the inductive step. O Hence, 1 is a factor of (7**1 - 1), and so P(k+1) is false, which completes the inductive step. O Hence, 1 is a factor of (7k+1 - 1), and so P(k+1) is true, which completes the inductive step.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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## Mathematical Induction Example

### Prove the following statement by mathematical induction:

For every integer \( n \geq 0 \), \( 7^n - 1 \) is divisible by 6.

### Proof (by mathematical induction):

Let \( P(n) \) be the following sentence:

\( 7^n - 1 \) is divisible by 6.

We will show that \( P(n) \) is true for every integer \( n \geq 0 \).

#### Show that \( P(0) \) is true:

Select \( P(0) \) from the choices below.

- \( 7^0 - 1 = 6 \)
- 6 is a multiple of \( 7^0 - 1 \)
- 1 is a factor of \( 7^0 - 1 \)
- 6 is divisible by \( 7^0 - 1 \)
- 6 is a multiple of \( 7^0 - 1 \)

The truth of the selected statement follows from the definition of divisibility and the fact that \( 7^0 - 1 = \) __.

#### Show that for each integer \( k \geq 0 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:

Let \( k \) be any integer with \( k \geq 0 \), and suppose that \( P(k) \) is true. Select \( P(k) \) from the choices below.

- 6 is a multiple of \( 7^k - 1 \)
- 1 is a factor of \( 7^k - 1 \)
- 6 is divisible by \( 7^k - 1 \)
- \( 7^k - 1 \) is divisible by 6

_[This is \( P(k) \), the inductive hypothesis.]_

We must show that \( P(k + 1) \) is true. Select \( P(k + 1) \) from the choices below.

- \( 7^{k + 1} - 1 \) is divisible by 6
- 1 is a factor of \( 7^{k + 1} - 1 \)
- 6 is a multiple of \( 7^{k + 1} - 1 \)
- 6 is
Transcribed Image Text:## Mathematical Induction Example ### Prove the following statement by mathematical induction: For every integer \( n \geq 0 \), \( 7^n - 1 \) is divisible by 6. ### Proof (by mathematical induction): Let \( P(n) \) be the following sentence: \( 7^n - 1 \) is divisible by 6. We will show that \( P(n) \) is true for every integer \( n \geq 0 \). #### Show that \( P(0) \) is true: Select \( P(0) \) from the choices below. - \( 7^0 - 1 = 6 \) - 6 is a multiple of \( 7^0 - 1 \) - 1 is a factor of \( 7^0 - 1 \) - 6 is divisible by \( 7^0 - 1 \) - 6 is a multiple of \( 7^0 - 1 \) The truth of the selected statement follows from the definition of divisibility and the fact that \( 7^0 - 1 = \) __. #### Show that for each integer \( k \geq 0 \), if \( P(k) \) is true, then \( P(k + 1) \) is true: Let \( k \) be any integer with \( k \geq 0 \), and suppose that \( P(k) \) is true. Select \( P(k) \) from the choices below. - 6 is a multiple of \( 7^k - 1 \) - 1 is a factor of \( 7^k - 1 \) - 6 is divisible by \( 7^k - 1 \) - \( 7^k - 1 \) is divisible by 6 _[This is \( P(k) \), the inductive hypothesis.]_ We must show that \( P(k + 1) \) is true. Select \( P(k + 1) \) from the choices below. - \( 7^{k + 1} - 1 \) is divisible by 6 - 1 is a factor of \( 7^{k + 1} - 1 \) - 6 is a multiple of \( 7^{k + 1} - 1 \) - 6 is
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