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Prove that for any integer a, 9/(a – 3).

Prove that for any integer a, 9/(a – 3).

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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I try do to the prove using the quotient remainder theorem and division into cases and my teacher said it was wrong . So can you please help me prove this by the quotient remainder theorem and division into case where a^2 = 9q + 3
**Title: Mathematical Proof Exercise** **Objective:** Prove that for any integer \( a \), 9 divides \( a^2 - 3 \). **Exercise Explanation:** In this exercise, you need to demonstrate a divisibility property involving integers. The goal is to show that the expression \( a^2 - 3 \) is divisible by 9 for every integer \( a \). This proof can be approached using techniques such as modular arithmetic, mathematical induction, or direct algebraic manipulation. **Steps for Solution:** 1. **Using Modular Arithmetic:** - Investigate values of \( a \) modulo 3, as \( a^2 \mod 9 \) is influenced heavily by \( a \mod 3 \). - For \( a \equiv 0, 1, 2 \pmod{3} \), calculate \( a^2 \mod 9 \) and verify that \( a^2 - 3 \equiv 0 \pmod{9} \). 2. **Using Mathematical Induction:** - Base Case: Prove for \( a = 0 \) and possibly other small integers. - Inductive Step: Assume it’s true for \( a = k \) and prove for \( a = k+1 \). 3. **Direct Algebraic Manipulation:** - Explore factorizations or algebraic identities that simplify the expression. By applying these methods, you can effectively demonstrate the divisibility condition for all integer values of \( a \).
Transcribed Image Text:**Title: Mathematical Proof Exercise** **Objective:** Prove that for any integer \( a \), 9 divides \( a^2 - 3 \). **Exercise Explanation:** In this exercise, you need to demonstrate a divisibility property involving integers. The goal is to show that the expression \( a^2 - 3 \) is divisible by 9 for every integer \( a \). This proof can be approached using techniques such as modular arithmetic, mathematical induction, or direct algebraic manipulation. **Steps for Solution:** 1. **Using Modular Arithmetic:** - Investigate values of \( a \) modulo 3, as \( a^2 \mod 9 \) is influenced heavily by \( a \mod 3 \). - For \( a \equiv 0, 1, 2 \pmod{3} \), calculate \( a^2 \mod 9 \) and verify that \( a^2 - 3 \equiv 0 \pmod{9} \). 2. **Using Mathematical Induction:** - Base Case: Prove for \( a = 0 \) and possibly other small integers. - Inductive Step: Assume it’s true for \( a = k \) and prove for \( a = k+1 \). 3. **Direct Algebraic Manipulation:** - Explore factorizations or algebraic identities that simplify the expression. By applying these methods, you can effectively demonstrate the divisibility condition for all integer values of \( a \).
**Prove that for any integer \( a \), \( 9 \nmid (a^2 - 3) \)** **Proof (by contradiction):** Suppose not. That is, suppose \( a \in \mathbb{Z} \) and \( 9 \mid (a^2 - 3) \). \[ \exists q \in \mathbb{Z} \Rightarrow a^2 - 3 = 9q \quad \text{by definition of divisibility.} \] \( a = 3k + 1 \) or \( a = 3k + 2 \) for some integer \( k \), by the definition of the quotient-remainder theorem. **Case 1:** \( a = 3k + 1 \) Then, \[ a^2 - 3 = 9q \] \[ (3k+1)^2 - 3 = 9q \quad \text{by substitution.} \] \[ 9k^2 + 6k - 2 = 9q \quad \text{by algebra.} \] \[ 3(3k^2 + 2k - 3q) = 2 \quad \text{by algebra.} \] \[ (3k^2 + 2k - 3q) = \frac{2}{3} \quad \text{by algebra.} \] Let \( \ell = 3k^2 + 2k - 3q \). \( \ell \in \mathbb{Z} \) since \( 3, 2, k^2, k, 9 \in \mathbb{Z} \), by the closure of integers under additions and multiplications. But \( \ell = \frac{2}{3} \notin \mathbb{Z} \), therefore this leads to the contradiction that \( 9 \nmid (a^2 - 3) \). **Case 2:** \( a = 3k + 2 \) Then, \[ a^2 - 3 = 9q \] \[ (3k+2)^2 - 3 = 9q \quad \text{by substitution.} \] \[ 9k^2 + 12k + 1 = 9q \
Transcribed Image Text:**Prove that for any integer \( a \), \( 9 \nmid (a^2 - 3) \)** **Proof (by contradiction):** Suppose not. That is, suppose \( a \in \mathbb{Z} \) and \( 9 \mid (a^2 - 3) \). \[ \exists q \in \mathbb{Z} \Rightarrow a^2 - 3 = 9q \quad \text{by definition of divisibility.} \] \( a = 3k + 1 \) or \( a = 3k + 2 \) for some integer \( k \), by the definition of the quotient-remainder theorem. **Case 1:** \( a = 3k + 1 \) Then, \[ a^2 - 3 = 9q \] \[ (3k+1)^2 - 3 = 9q \quad \text{by substitution.} \] \[ 9k^2 + 6k - 2 = 9q \quad \text{by algebra.} \] \[ 3(3k^2 + 2k - 3q) = 2 \quad \text{by algebra.} \] \[ (3k^2 + 2k - 3q) = \frac{2}{3} \quad \text{by algebra.} \] Let \( \ell = 3k^2 + 2k - 3q \). \( \ell \in \mathbb{Z} \) since \( 3, 2, k^2, k, 9 \in \mathbb{Z} \), by the closure of integers under additions and multiplications. But \( \ell = \frac{2}{3} \notin \mathbb{Z} \), therefore this leads to the contradiction that \( 9 \nmid (a^2 - 3) \). **Case 2:** \( a = 3k + 2 \) Then, \[ a^2 - 3 = 9q \] \[ (3k+2)^2 - 3 = 9q \quad \text{by substitution.} \] \[ 9k^2 + 12k + 1 = 9q \
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