Let b be a positive integer and \( R=\ (a, b) ∈ Z × Z a-b=nm \) for some \( m ≠ 0 ∈ Z\ \)

\[ Step 1:\] Define the Relation The relation \( R \) is defined as: \[ R = \ (a, b) ∈ Z × Z a - b = n · m for some m ≠ 0 ∈ Z \ \] This means \( (a, b) ∈ R \) if \( a - b \) is a multiple of \( n \) for some integer \( m ≠ 0 \). \[Step 2:\] Check Reflexivity A relation \( R \) is reflexive if \( (a, a) ∈ R \) for all \( a ∈ Z \). For \( (a, a) \), we have: \[ a - a = 0 = n · m \] This implies: \[...

Let b be a positive integer and \( R=\... - Solved

Question 54

Mathematics Function and Relation Hard

Let b be a positive integer and \( \mathrm{R}=\{(\mathrm{a}, \mathrm{b}) \in \mathrm{Z} \times \mathrm{Z} \mid \mathrm{a}-\mathrm{b}=\mathrm{nm} \) for some \( \mathrm{m} \neq 0 \in \mathrm{Z}\} \)

(A) Reflexive on \( Z \)

(B) Symmetric

(C) Transitive

(D) Equivalence relation of \( Z \)

View Dynamic Solution & Explanation ▼

Correct Solution: Option B

Step-by-step Solution:

\[ Step 1:\] Define the Relation The relation \( R \) is defined as: \[ R = \{(a, b) \in \mathbb{Z} \times \mathbb{Z} \mid a - b = n \cdot m \text{ for some } m \neq 0 \in \mathbb{Z}\} \] This means \( (a, b) \in R \) if \( a - b \) is a multiple of \( n \) for some integer \( m \neq 0 \). \[Step 2:\] Check Reflexivity A relation \( R \) is reflexive if \( (a, a) \in R \) for all \( a \in \mathbb{Z} \). For \( (a, a) \), we have: \[ a - a = 0 = n \cdot m \] This implies: \[ n \cdot m = 0 \] Since \( m \neq 0 \), the only way this equation holds is if \( n = 0 \). However, if \( n \neq 0 \), there is no \( m \neq 0 \) such that \( n \cdot m = 0 \). Therefore, \( (a, a) \notin R \) unless \( n = 0 \). Conclusion: \( R \) is not reflexive unless \( n = 0 \). \[Step 3:\] Check Symmetry A relation \( R \) is symmetric if \( (a, b) \in R \) implies \( (b, a) \in R \). If \( (a, b) \in R \), then \( a - b = n \cdot m \) for some \( m \neq 0 \). For \( (b, a) \), we have: \[ b - a = -n \cdot m \] Since \( -n \cdot m \) is also a multiple of \( n \) (with \( -m \neq 0 \)), \( (b, a) \in R \). Conclusion: \( R \) is symmetric. Step 4: Check Transitivity A relation \( R \) is transitive if \( (a, b) \in R \) and \( (b, c) \in R \) imply \( (a, c) \in R \). If \( (a, b) \in R \), then \( a - b = n \cdot m_1 \) for some \( m_1 \neq 0 \). If \( (b, c) \in R \), then \( b - c = n \cdot m_2 \) for some \( m_2 \neq 0 \). Adding these two equations: \[ a - c = (a - b) + (b - c) = n \cdot m_1 + n \cdot m_2 = n \cdot (m_1 + m_2) \] For \( (a, c) \in R \), \( m_1 + m_2 \neq 0 \) must hold. However, \( m_1 + m_2 = 0 \) is possible (e.g., \( m_1 = 1 \) and \( m_2 = -1 \)), in which case \( a - c = 0 \), and \( (a, c) \notin R \) unless \( n = 0 \). Conclusion: \( R \) is not transitive unless \( n = 0 \). \[Step 5:\] Final Conclusion - \( R \) is symmetric. - \( R \) is not reflexive unless \( n = 0 \). - \( R \) is not transitive unless \( n = 0 \). Thus, the relation \( R \) is only symmetric. Correct Answer: \(\boxed{B}\)