Question

Let Z be the set of all integers and let R be a relation on Z defined by $a\text{ }R\text{ }b\text{ }\Leftrightarrow \text{ }(a-b)$ is divisible by 3 . Then R is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation

Answer 1

Hint:Here, we will use the definitions of reflexive, symmetric and transitive relations to check whether the given relations are reflexive, symmetric or transitive.

Complete step-by-step answer:
A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x, y) is in the relation.
A relation R is reflexive if each element is related itself, i.e. (a, a) $\in$ R, where a is an element of the domain.
A relation R is symmetric in case if any one element is related to any other element, then the second element is related to the first, i.e. if (x, y) $\in$ R then (y, x) $\in$ R, where x and y are the elements of domain and range respectively.
A relation R is transitive in case if any one element is related to a second and that second element is related to a third, then the first element is related to the third, i.e. if (x, y) $\in$ R and (y, z) $\in$ R then (x, z) $\in$ R.
Here, the given relation is:
$a\text{ }R\text{ }b\text{ }\Leftrightarrow \text{ }(a-b)$ is divisible by 3.
If we take an ordered pair (a, a), then there exists in R as (a-a) = 0 is divisible by 3.
So, R is reflexive.
Now, let us suppose that (a, b) $\in$ R, so (a-b) is divisible by 3.
For R to be symmetric, we will have to show that (b, a) $\in$ R, i.e. (b-a) is divisible by 3.
Since (a-b) is divisible by 3. So, we can write:
$\begin{aligned} & a-b=3k \\\ & \Rightarrow -\left( b-a \right)=3k \\\ & \Rightarrow b-a=3\left( -k \right) \\\ \end{aligned}$
Therefore, (b-a) is divisible by 3.
This means that R is symmetric.
Now, let us suppose that a-b $\in$ R and b-c $\in$ R, then R would be transitive if a-c $\in$ R.
We can write:
$a-b=3p........\left( 1 \right)$
$b-c=3q.........\left( 2 \right)$
On adding equations (1) and (2), we get:
$\begin{aligned} & a-b+b-c=3p+3q \\\ & \Rightarrow a-c=3\left( p+q \right) \\\ & \Rightarrow a-c=3k' \\\ \end{aligned}$
Here, k’=p+q
This means that a-c is divisible by 3, i.e. a-c $\in$ R.
This implies that R is transitive.
Therefore, R is reflexive, symmetric and transitive.
We know that a relation which is reflexive, symmetric as well as transitive is an equivalence relation.
Hence, option (d) is the correct answer.

Note: Students should note here that for a relation to be a particular type of relation, i.e. reflexive, symmetric or transitive, all its elements must satisfy the required conditions. If we are able to find even a single counter example, i.e. if any of the elements doesn’t satisfy the criteria, then we can’t proceed further.