Question

Let R and S be two non-void relations on a set A. Which of the following statements is false?
(a)R and S are transitive implies R $\cap $ S is transitive.
(b) R and S are transitive implies R $\cup $ S is transitive.
(c) R and S are symmetric implies R $\cup $ S is symmetric.
(d) R and S are reflexive implies R $\cap $ S is reflexive.

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.
Consider the relation R as:
R = { (a, b), (b, c), (c, a), (e, d) } and consider S as:
S = { (x, y), (y, z), (z, x), (d, f), (e, d) }
Here both R and S are transitive.
On taking union of R and S, we get:
R $\cup $ S = { (a, b), (b, c), (c, a), (e, d), (x, y), (y, z), (z, x), (d, f) }.
This is not transitive because (e, f) $\notin $ R $\cup $ S.
R $\cap $ S = {(e, d)}.
Thus R $\cap $ S is not transitive.
Now, consider R as:
R = { (a, b), (b, a), (c, d), (d, c) } and S = { (x, y), (y, x), (s, t), (t, s) }.
Here R and S both are symmetric.
R $\cup $ S = { (a, b), (b, a), (c, d), (d, c), (x, y), (y, x), (s, t), (t, s) }.
This is symmetric.
Since both R and S are defined on A, the reflexive relations R and S are the same.
If there are ‘n’ numbers of elements in A, then:
R = { (a, a), (b, b), ……., (n, n) } and S = { (a, a) , (b, b), ………, (n, n) }.
R $\cap $ S = { (a, a), (b, b),…….(n, n) }.
So, R $\cap $ S is reflexive.
Hence, option (c) and option (d) are the correct answers.

Note: Students should note that a relation on a set A is said to be reflexive if and only if all the elements of A are related to themselves, i.e. none of the elements should be left out.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.