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Prove that 7" –- 1 is a multiple of 6 for all n e N.

Prove that 7" –- 1 is a multiple of 6 for all n e N.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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**Title:** Prove that \(7^n - 1\) is a Multiple of 6 for All \(n \in \mathbb{N}\). **Objective:** Demonstrating that \(7^n - 1\) is divisible by 6 for any natural number \(n\). **Overview:** The statement asserts that for any natural number \(n\), the expression \(7^n - 1\) can be divided by 6 without leaving a remainder. The task involves using mathematical proof techniques to confirm this proposition. **Approach:** To show \(7^n - 1\) is a multiple of 6, it suffices to demonstrate: 1. \(7^n - 1\) is divisible by 2. 2. \(7^n - 1\) is divisible by 3. Since 6 is the product of 2 and 3, proving divisibility by both numbers will establish the required result. **Potential Mathematical Techniques:** - Mathematical induction - Modular arithmetic **Details of Proof:** 1. **Divisibility by 2:** - Consider the parity (odd or even nature) of \(7^n\) and \(7^n - 1\). 2. **Divisibility by 3:** - Analyze \(7 \equiv 1 \pmod{3}\), and extend to \(7^n\). The combination of these cases will show \(7^n - 1\) meets the criteria for being a multiple of 6, effectively concluding the proof.
Transcribed Image Text:**Title:** Prove that \(7^n - 1\) is a Multiple of 6 for All \(n \in \mathbb{N}\). **Objective:** Demonstrating that \(7^n - 1\) is divisible by 6 for any natural number \(n\). **Overview:** The statement asserts that for any natural number \(n\), the expression \(7^n - 1\) can be divided by 6 without leaving a remainder. The task involves using mathematical proof techniques to confirm this proposition. **Approach:** To show \(7^n - 1\) is a multiple of 6, it suffices to demonstrate: 1. \(7^n - 1\) is divisible by 2. 2. \(7^n - 1\) is divisible by 3. Since 6 is the product of 2 and 3, proving divisibility by both numbers will establish the required result. **Potential Mathematical Techniques:** - Mathematical induction - Modular arithmetic **Details of Proof:** 1. **Divisibility by 2:** - Consider the parity (odd or even nature) of \(7^n\) and \(7^n - 1\). 2. **Divisibility by 3:** - Analyze \(7 \equiv 1 \pmod{3}\), and extend to \(7^n\). The combination of these cases will show \(7^n - 1\) meets the criteria for being a multiple of 6, effectively concluding the proof.
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