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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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What is r
![**Prove the following statement by contradiction:**
*If \(a\) and \(b\) are rational numbers, \(b \neq 0\), and \(r\) is an irrational number, then \(a + br\) is irrational.*
**Proof by contradiction:** Select an appropriate statement to start the proof.
- Suppose not. That is, suppose there exist irrational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is a rational number, and \(a + br\) is rational.
- Suppose not. That is, suppose there exist irrational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is an irrational number, and \(a + br\) is rational.
- Suppose not. That is, suppose there exist rational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is an irrational number, and \(a + br\) is rational.
- **Suppose not. That is, suppose there exist rational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is an irrational number, and \(a + br\) is rational.** ✔️
**Then by definition of rational:**
\[a = \frac{c}{d}, \quad b = \frac{i}{j}, \quad \text{and } a + br = \frac{m}{n} \]
where \(c, d, i, j, m, \) and \(n\) are **integers** ✔️ and \([d \neq 0, j \neq 0, \text{and } n \neq 0]\) ✔️. Since \(b \neq 0\), we also have that \(i \neq 0\). By substitution,
\[ \frac{c}{d} + \frac{i}{j}r = \frac{m}{n} \]
Solving this equation for \(r\) and representing the result as a single quotient in terms of \(c, d, i, j, m, \) and \(n\) gives that
\[ r = \]
Note that \(r\) **is** ✔️ a ratio of two integers because products and differences of integers **are** ✔️](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96c2bf0d-d0d0-4024-bc9b-590e3c9ad5ff%2F18f3d163-15f8-4a55-8860-7cc254533fb9%2F7erd41q_processed.png&w=3840&q=75)
Transcribed Image Text:**Prove the following statement by contradiction:**
*If \(a\) and \(b\) are rational numbers, \(b \neq 0\), and \(r\) is an irrational number, then \(a + br\) is irrational.*
**Proof by contradiction:** Select an appropriate statement to start the proof.
- Suppose not. That is, suppose there exist irrational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is a rational number, and \(a + br\) is rational.
- Suppose not. That is, suppose there exist irrational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is an irrational number, and \(a + br\) is rational.
- Suppose not. That is, suppose there exist rational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is an irrational number, and \(a + br\) is rational.
- **Suppose not. That is, suppose there exist rational numbers \(a\) and \(b\) such that \(b \neq 0\), \(r\) is an irrational number, and \(a + br\) is rational.** ✔️
**Then by definition of rational:**
\[a = \frac{c}{d}, \quad b = \frac{i}{j}, \quad \text{and } a + br = \frac{m}{n} \]
where \(c, d, i, j, m, \) and \(n\) are **integers** ✔️ and \([d \neq 0, j \neq 0, \text{and } n \neq 0]\) ✔️. Since \(b \neq 0\), we also have that \(i \neq 0\). By substitution,
\[ \frac{c}{d} + \frac{i}{j}r = \frac{m}{n} \]
Solving this equation for \(r\) and representing the result as a single quotient in terms of \(c, d, i, j, m, \) and \(n\) gives that
\[ r = \]
Note that \(r\) **is** ✔️ a ratio of two integers because products and differences of integers **are** ✔️
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