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20. Prove: Let x and y be real numbers. If x is rational and y is irrational, then x + y is irrational.

20. Prove: Let x and y be real numbers. If x is rational and y is irrational, then x + y is irrational.

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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**Problem Statement:** 20. Prove: Let \( x \) and \( y \) be real numbers. If \( x \) is rational and \( y \) is irrational, then \( x + y \) is irrational. **Explanation:** This problem requires a proof in mathematical logic. The objective is to demonstrate that the sum of a rational number \( x \) and an irrational number \( y \) always results in an irrational number. Key Points to Consider: - A **rational number** can be expressed as the quotient of two integers (i.e., \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \)). - An **irrational number** cannot be expressed as a simple fraction and has a non-repeating, non-terminating decimal expansion. Steps for Proof: 1. Assume \( x \) is rational and can be written as \( \frac{m}{n} \) where \( m \) and \( n \) are integers, \( n \neq 0 \). 2. Assume \( y \) is irrational and cannot be expressed as a fraction. 3. Suppose \( x + y \) is rational. Then \( x + y = \frac{p}{q} \) for some integers \( p \) and \( q \). 4. Rearranging gives \( y = \frac{p}{q} - \frac{m}{n} \). 5. The right side of this equation is a difference of two rational numbers, which must be rational. 6. This contradicts the assumption that \( y \) is irrational. Conclusion: The assumption that \( x + y \) is rational leads to a contradiction, hence \( x + y \) must be irrational. This completes the proof.
Transcribed Image Text:**Problem Statement:** 20. Prove: Let \( x \) and \( y \) be real numbers. If \( x \) is rational and \( y \) is irrational, then \( x + y \) is irrational. **Explanation:** This problem requires a proof in mathematical logic. The objective is to demonstrate that the sum of a rational number \( x \) and an irrational number \( y \) always results in an irrational number. Key Points to Consider: - A **rational number** can be expressed as the quotient of two integers (i.e., \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \)). - An **irrational number** cannot be expressed as a simple fraction and has a non-repeating, non-terminating decimal expansion. Steps for Proof: 1. Assume \( x \) is rational and can be written as \( \frac{m}{n} \) where \( m \) and \( n \) are integers, \( n \neq 0 \). 2. Assume \( y \) is irrational and cannot be expressed as a fraction. 3. Suppose \( x + y \) is rational. Then \( x + y = \frac{p}{q} \) for some integers \( p \) and \( q \). 4. Rearranging gives \( y = \frac{p}{q} - \frac{m}{n} \). 5. The right side of this equation is a difference of two rational numbers, which must be rational. 6. This contradicts the assumption that \( y \) is irrational. Conclusion: The assumption that \( x + y \) is rational leads to a contradiction, hence \( x + y \) must be irrational. This completes the proof.
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