Question

Let A = {1, 2, 3} and R = {(1, 1), (2,2), (3, 3), (1, 3), (3, 2), (1, 2)}. 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:
R = {(1, 1), (2,2), (3, 3), (1, 3), (3, 2), (1, 2)}
The elements present in A are 1, 2 and 3.
We can clearly see that (1, 1), (2, 2) and (3, 3) exists in R.
So, R is reflexive.
Now, we can see that (1, 3) $\in $ R but (3, 1) $\notin $ R.
This means that R is not symmetric.
Now, (1, 3) $\in $ R, (3, 2) $\in $ R and also (1, 2) $\in $ R.
We can also take (3, 3) $\in $ R, (3, 2) $\in $ R and also (3, 2) $\in $ R.
This implies that R is transitive.
Therefore, R is reflexive and transitive but not symmetric.
Hence, option (b) 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.