Question

Let A= {2, 3, 6, 8, 9, 11} and B = {1, 4, 5, 10, 15} Let R be a relation on A × B define by (a, b)R(c, d) if and only if 3ad – 7bc is an even integer. Then the relation R is

• A. reflexive but not symmetric.
• B. transitive but not symmetric.
• C. reflexive and symmetric but not transitive
• D. an equivalence relation.

Answer 1

The Correct Option is C

The relation \( R \) is defined on the Cartesian product \( A \times B \) such that \((a, b)R(c, d)\) if and only if \(3ad - 7bc\) is an even integer, where \(a, c \in A\) and \(b, d \in B\). Reflexivity: For the relation \( R \) to be reflexive, \((a, b)R(a, b)\) must hold for all \((a, b) \in A \times B\). We check: \[ 3ab - 7ba = -4ab, \]

which is always even since \(-4ab\) is a product of an integer and an even number. Therefore, \( R \) is reflexive. Symmetry: For \( R \) to be symmetric, if \((a, b)R(c, d)\) holds, then \((c, d)R(a, b)\) must also hold. Given that \(3ad - 7bc\) is even, consider swapping \((a, b)\) and \((c, d)\): \[ (c, d)R(a, b) \implies 3bc - 7ad. \] 

Since the difference between even numbers remains even, \( R \) is symmetric. Transitivity: For \( R \) to be transitive, if \((a, b)R(c, d)\) and \((c, d)R(e, f)\) hold, then \((a, b)R(e, f)\) must also hold. Consider specific elements: - Let \((3, 4)R(6, 4)\) hold, satisfying the condition as \(3 \cdot 6 \cdot 4 - 7 \cdot 4 \cdot 4\) is even. - Let \((6, 4)R(3, 1)\) hold, as \(3 \cdot 6 \cdot 1 - 7 \cdot 4 \cdot 1\) is also even. - However, \((3, 4)R(3, 1)\) does not satisfy the condition, as \(3 \cdot 3 \cdot 1 - 7 \cdot 4 \cdot 1\) is not necessarily even. Therefore, \( R \) is not transitive. Based on the above, the relation \( R \) is: reflexive and symmetric but not transitive.