Question

What is the type of the relation \[R\] defined on \[N\] as \[aRb \Leftrightarrow b\] is divisible by \[a\]?
A. Reflexive but not symmetric
B. Symmetric but not transitive
C. Symmetric and transitive
D. None of the above

Answer 1

Hint: In the given question, the relation on a set of natural numbers is given. By using the definitions of reflexive, symmetric, and transitive relation, we will find the type of relation \[R\].

Formula Used:
Let \[R\] be the relation on a set \[A\]. Then,
\[R\] is reflexive if for all \[x \in A\], \[xRx\].
\[R\] is symmetric if for all \[x,y \in A\], if \[xRy\], then \[yRx\].
\[R\] is transitive if for all \[x,y,z \in A\], if \[xRy\] and \[yRz\], then \[xRz\].

Complete step by step solution:
The given relation on set \[N\] is, \[aRb \Leftrightarrow b\] is divisible by \[a\].
Let’s check whether the relation is reflexive, symmetric, or transitive.
Reflexive:
If \[aRa\], then
For any number of a set \[N\]:
Since every number is divisible by itself.
So, \[a\] is divisible by \[a\].
Hence, the relation is reflexive.

Symmetric:
If \[aRb\], then \[bRa\]
For \[\left( {2,4} \right)\] and \[\left( {4,2} \right)\]:
Since 4 is divisible by 2. But 2 is not divisible by 4.
Hence, the relation is not symmetric.

Transitive:
If \[aRb\], \[bRc\], then \[aRc\]
For \[\left( {1,2} \right),\left( {2,4} \right)\] and \[\left( {1,4} \right)\]:
Since 2 is divisible by 1 and 4 is divisible by 2.
So, 4 is divisible by 1.
Hence, the relation is transitive.

Hence the correct option is D.


Note:There is another property of relation. The name of the property is antisymmetric property. It states that \[a,b\] belong to set \[A\] and the relation \[R\] is defined on \[A\], then \[\left( {a,b} \right) \in R\] does not imply that \[\left( {b,a} \right) \in R\] when \[a \ne b\].