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Prove that for each irrational number r, there exists a unique integer n such that |r – n| < 1/2.

Prove that for each irrational number r, there exists a unique integer n such that |r – n| < 1/2.

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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To prove that for each irrational number \( r \), there exists a unique integer \( n \) such that \( |r - n| < 1/2 \), consider the following reasoning: ### Proof: 1. **Locate the Integer:** - Given any real number \( r \), to find the integer \( n \) such that \( |r - n| < 1/2 \), we can start by considering the floor function, which is defined as the greatest integer less than or equal to \( r \). Hence, let \( n = \lfloor r \rfloor \). 2. **Evaluate the Distance:** - The inequality \( |r - n| < 1/2 \) implies that the distance between \( r \) and \( n \) should be less than half a unit. - By definition, \( \lfloor r \rfloor \leq r < \lfloor r \rfloor + 1 \), which translates to \( 0 \leq r - n < 1 \). 3. **Choose the Correct Integer:** - Now consider \( r - n \). - If \( n = \lfloor r \rfloor \), then \( 0 \leq r - n < 1 \). - For \( |r - n| < 1/2 \) to hold, it must be that \( r - n < 1/2 \). Consequently, the only case where \( r - n \geq 1/2 \) and \( r < n + 1 \) must be avoided. Thus, \( n \) is indeed the integer we seek. 4. **Uniqueness:** - Assume there exists another integer \( m \neq n \) where \( |r - m| < 1/2 \). This would mean \( |r - n| < 1/2 \) and \( |r - m| < 1/2 \), which implies \( |n - m| < 1 \). The only integers \( n \) and \( m \) with \( |n - m| < 1 \) are such that \( n = m \), thus proving the uniqueness. This simple proof leverages the properties of the floor function and real numbers to establish the existence and uniqueness of such an integer \( n \). ### Explanation: - The goal of
Transcribed Image Text:To prove that for each irrational number \( r \), there exists a unique integer \( n \) such that \( |r - n| < 1/2 \), consider the following reasoning: ### Proof: 1. **Locate the Integer:** - Given any real number \( r \), to find the integer \( n \) such that \( |r - n| < 1/2 \), we can start by considering the floor function, which is defined as the greatest integer less than or equal to \( r \). Hence, let \( n = \lfloor r \rfloor \). 2. **Evaluate the Distance:** - The inequality \( |r - n| < 1/2 \) implies that the distance between \( r \) and \( n \) should be less than half a unit. - By definition, \( \lfloor r \rfloor \leq r < \lfloor r \rfloor + 1 \), which translates to \( 0 \leq r - n < 1 \). 3. **Choose the Correct Integer:** - Now consider \( r - n \). - If \( n = \lfloor r \rfloor \), then \( 0 \leq r - n < 1 \). - For \( |r - n| < 1/2 \) to hold, it must be that \( r - n < 1/2 \). Consequently, the only case where \( r - n \geq 1/2 \) and \( r < n + 1 \) must be avoided. Thus, \( n \) is indeed the integer we seek. 4. **Uniqueness:** - Assume there exists another integer \( m \neq n \) where \( |r - m| < 1/2 \). This would mean \( |r - n| < 1/2 \) and \( |r - m| < 1/2 \), which implies \( |n - m| < 1 \). The only integers \( n \) and \( m \) with \( |n - m| < 1 \) are such that \( n = m \), thus proving the uniqueness. This simple proof leverages the properties of the floor function and real numbers to establish the existence and uniqueness of such an integer \( n \). ### Explanation: - The goal of
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