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

Answer: (C)

reflexive and symmetric but not transitive

Answer 2

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.