Exercise Problems 1. Determine whether each of the following relations are reflexive, symmetric and transitive. (i) Relation \( R \) in the set \( A=\{1,2,3, \ldots \ldots \ldots 12,13\} \) defined as \( R=\{(x, y) \mid 3 x-y=0\} \) (ii) Relation \( R \) in the set \( N \) of natural numbers defined as \( R=\{(x, y) \mid y=x+5 \) and \( x<4\} \) (iii) Relation \( R \) in the set \( A=\{1,2,3,4,5,6\} \) as \( R=\{(x, y) \mid y \) is divisible by \( x\} \)

To determine whether each of the given relations is reflexive, symmetric, and transitive, we will analyze each relation one by one. ### (i) Relation \( R \) in the set \( A=\{1,2,3,\ldots,13\} \) defined as \( R=\{(x, y) \mid 3x - y = 0\} \) 1. **Reflexive**: A relation is reflexive if for every element \( x \) in the set, \( (x, x) \) is in the relation. Here, \( 3x - x = 2x \neq 0 \) for \( x \neq 0 \). Therefore, \( (x, x) \) is not in \( R \) for any \( x \) in \( A \). Thus, \( R \) is **not reflexive**. 2. **Symmetric**: A relation is symmetric if whenever \( (x, y) \) is in \( R \), then \( (y, x) \) is also in \( R \). If \( (x, y) \in R \), then \( 3x = y \). For \( (y, x) \) to be in \( R \), we would need \( 3y = x \). However, if \( y = 3x \), then \( x = \frac{y}{3} \), which does not hold for all \( x, y \) in \( A \). Thus, \( R \) is **not symmetric**. 3. **Transitive**: A relation is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \). If \( (x, y) \in R \) implies \( y = 3x \) and \( (y, z) \in R \) implies \( z = 3y = 3(3x) = 9x \). For \( (x, z) \) to be in \( R \), we need \( z = 3x \), which is not true in general. Thus, \( R \) is **not transitive**. ### Summary for (i): - Reflexive: No - Symmetric: No - Transitive: No --- ### (ii) Relation \( R \) in the set \( N \) of natural numbers defined as \( R=\{(x, y) \mid y = x + 5 \text{ and } x < 4\} \) 1. **Reflexive**: For \( R \) to be reflexive, \( (x, x) \) must be in \( R \) for all \( x \). However, \( y = x + 5 \) cannot equal \( x \) for any \( x \). Thus, \( R \) is **not reflexive**. 2. **Symmetric**: For \( R \) to be symmetric, if \( (x, y) \in R \) (i.e., \( y = x + 5 \)), then \( (y, x) \) must also be in \( R \). If \( y = x + 5 \), then \( x = y - 5 \), which does not satisfy \( y = x + 5 \) in reverse. Thus, \( R \) is **not symmetric**. 3. **Transitive**: For \( R \) to be transitive, if \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \) must also be in \( R \). If \( y = x + 5 \) and \( z = y + 5 = (x + 5) + 5 = x + 10 \), then \( (x, z) \) would require \( z = x + 5 \), which is not true. Thus, \( R \) is **not transitive**. ### Summary for (ii): - Reflexive: No - Symmetric: No - Transitive: No --- ### (iii) Relation \( R \) in the set \( A=\{1,2,3,4,5,6\} \) as \( R=\{(x, y) \mid y \text{ is divisible by } x\} \) 1. **Reflexive**: A relation is reflexive if \( (x, x) \) is in \( R \) for all \( x \). Since every number is divisible by itself, \( (x, x) \) is