Question

Let R be a relation on the set N of natural numbers defined by $nRm \Leftrightarrow {\text{n}}$ is a factor of m(i.e. n|m).Then R is-
A. Reflexive and symmetric
B. Transitive and symmetric
C. Equivalence
D. Reflexive, transitive but not symmetric

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.

Complete step-by-step answer:
For the relation to be reflexive, nRn should be satisfied, that is n is a factor of n. We know that every number is a factor of itself. Hence, nRn is satisfied. The relation R is reflexive.

For the relation to be symmetric, if nRm is satisfied then mRn should also be satisfied. That is if n is a factor of m then m is also a factor of n. We can use an example to check this-
(2, 6) implies that 2 is a factor of 6, which is true. But 6 is not a factor of 2, this means that the given relation is not symmetric.

For a relation to be transitive, if nRm and mRk are satisfied then nRk is also satisfied. For example,
(2, 6) => 2 is a factor of 6 which is true.
(6, 12) => 6 is a factor of 12 which is true.
Now we will check the condition for (2, 12). We know clearly that 2 is a factor of 12. Hence, the condition for transitivity is satisfied. The given relation is transitive.

The relation R is reflexive and transitive. The correct option is D.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.