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(10) If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (−1, 2) and (0, 1) are two elements of
S, then write the remaining elements of S.
Solution: n(A) = 4. Clearly from the two distinct elements of S, the elements of A =
{−1, 0, 1, 2}. The elements of S = {(−1, 0), (−1, 1), (−1, 2), (0, 1), (0, 2), (1, 2)}.

Exercise - 1.2

1. Discuss the following relations for reflexivity, symmetricity and transitivity:
(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.
Solution: Since each positive integer divides itself, R is reflexive;
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Clearly m divides n ; n divides m whenever m 6= n
Hence R is not symmetric.
Let p divides q. Then q = kp for some positive integer k
Let q divides r. Then r = mq for some positive integer m
Since r = mq = m(kp) = (mk)p we have p divides r
Hence R is transitive.
(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “`Rm if ` is
perpendicular to m”.
Solution: No straight line can be perpendicular to itself. Hence P is not reflexive.
Since ` is perpendicular to m ⇒ m is perpendicular to `. Hence R is symmetric.
No three lines will be perpendicular to each other in Cartesian Plane. Hence P is not transitive.
(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if
a is not a sister of b”.
Solution: We know that a is not a sister of a. Hence R is reflexive. But a is not a sister of
b ; b is not a sister of a.So R is not symmetric. a is not a sister of b and b is not a sister of
c ; a is not a sister of c. Hence R is not transitive.
(iv) Let A be the set consisting of all the female members of a family. The relation R defined by
“aRb if a is not a sister of b”.
Solution: We know that a is not a sister of a. Hence R is reflexive. But a is not a sister of
b ⇒ b is not a sister of a because both a and b are females. Hence R is symmetric. Let a is
not a sister of b and b is not a sister of c. As a, b, c are all females, it follows that a cannot be
a sister of c. Hence R is transitive.
(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.
Solution: If xRx then for some natural number 3x = 1 which is not possible in natural
numbers. Hence R is not reflexive.
If xRy then x + 2y = 1. If yRx then 2x + y = 1. Solving both equations we get x = y which
has no solution in natural numbers. Hence R is not symmetric.
If xRy and yRz then x + 2y = 1 and y + 2z = 1. Solving we get x + 2y = y + 2z. that is,
x + 2z cannot be equal to 1 when x, z are natural numbers. Hence R is not transitive.
2. Let X = {a, b, c, d} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered
pairs to be included to R to make it
(i) reflexive (ii) symmetric (iii) transitive (iv) equivalence
Solution: Since (c, c), (d, d) are not in R, it is not reflexive. It is enough to include (c, c), (d, d) to
make it reflexive. (a, c) ∈ R but (c, a) /∈ R. It is enough to include (c, a) to make it symmetric.
After including the above mentioned elements, R = {(a, a), (b, b), (c, c), (d, d), (a, c), (c, a)} we
see that R is transitive. Hence the minimum number of ordered pairs to be included to R to make
it equivalence relation is THREE.

3. Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered
pairs to be included to R to make it