(i) \(\rho\) is not reflexive because (3, 3) is not in \(\rho\). As (1, 1) and (2, 2) are in \(\rho,\) it is enough to include the pair (3, 3) to \(\rho\) so as to make it reflexive.
(ii) \(\rho\) is not symmetric because (1, 2) is in p, but (2, 1) is not in \(\rho.\)  It is enough to include the pair (2, 1) to \(\rho\) so as to make it symmetric. It is enough to remove the pair (1, 2) from p so as to make it symmetric.
(iii) \(\rho\) is not transitive because (3, 1) and (1,3) are in,\(\rho\) but (3, 3) is not in \(\rho.\) To make it transitive we have to include (3, 3) in \(\rho\) . Even after including (3, 3), the relation is not transitive because (3, 1) and (I, 2) are in \(\rho\) , but (3, 2) is not in \(\rho\). To make it transitive we have to include (3, 2) also in \(\rho\) , Now it becomes transitive. So (3, 3) and (3, 2) are to be included into so as to make \(\rho\) transitive. But if we remove (3, 1) from \(\rho\), then it becomes transitive.
(iv) We have seen that
i) to make \(\rho\) reflexive, we have to include (3, 3);
ii) to make \(\rho\) symmetric, we have to include (2, 1);
iii) and to make \(\rho\) transitive, we have to include (3, 3) and (3, 2).
To make \(\rho\) as an equivalence relation we have to include all these pairs. So after including the pairs the relation becomes {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (3, 2)}.
But this relation is not symmetric because (3,2) is in the relation and (2,3) is not in the relation. So we have to include (2,3) also. Now the new relation becomes {(1, 1); (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (3, 2), (2, 3)}.
It can be seen that this relation is reflexive, symmetric and transitive and hence it is an equivalence relation. Thus we have to include (3, 3), (2, 1), (3, 2) and (2, 3) to \(\rho\) so as to make it an equivalence relation.

Let a, b, c are people in Chennai
Reflexivity: "a" is a friend of "a" \(\Rightarrow\)  a R    a \(\Rightarrow\) R is not reflexive.
Symmetric: a is friend of b \(\Rightarrow\) b is the friend of a.
\(\therefore\) aRb \(\Rightarrow\) bRa \(\Rightarrow\) R is symmetric
Transitive: a is the friend of b and b is the friend of c \(\Rightarrow\) a need not be the friend of c.
\(\therefore\) aRb \(\Rightarrow\) bRc \(\neq \) aRc \(\Rightarrow\) R is not transitive
Hence, the relation "friendship" is not equivalent.

(B\(\cup \)C) = {3, 4, 5, 6 ,7, 9}
Now, A\(\times\)(B\(\cup \)C) = {1, 2, 3} \(\times\){3, 4, 5, 6, 7, 9}
= {(1,3)(1,4)(1,5)(1,6)(1,7)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,9)(3,3)(3,4)(3,5)(3,6)(3,7)(3,9)} .....(1)
Now A\(\times\)B = {1,2,3} \(\times\) {4,5,6,7}
= {(1,4)(1,5)(1,6)(1,7)(2,4)(2,5)(2,6)(2,7)(3,4)(3,5)(3,6)(3,7)}
A\(\times\)C = {1,2,3} \(\times\) {3,4,5,9}
= {(1,3)(1,4)(1,5)(1,9)(2,3)(2,4)(2,5)(2,9)(3,3)(3,4)(3,5)(3,9)}
RHS(A\(\times\)B)\(\cup \)(A\(\times\)C) = {(1,3)(1,4)(1,5)(1,6)(1,7)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,9)(3,3)(3,4)(3,5)(3,6)(3,7)(3,9)} .....(2)
From (1) & (2), LHS = RHS
Hence verified

A \(\times\) (B\(\cap \)C) = (A\(\times\)B) \(\cap \) (A\(\times\)C)
Let A = {1,2,3}, B = (4,5,6,7} C = {4,3,5,9} and \(\cup\) = {1,2,3,4,5,6,7,8,9}
LHS = A x (B\(\cap \)C)
= A \(\times\) {4,5}    [∴B\(\cap \)C = {4, 5}]
= {1,2,3} \(\times\) {4,5}
= {(1,4) (1,5) (2,4) (2,5) (3,4) (3,5)}......(1)
A x B = {1,2,3} x {4,5,6,7}
= {(1,4) (1,5) (1,6) (1,7) (2,4) (2,5) (2,6) (2,7) (3,4) (3,5) (3,6) (3,7)} 
A \(\times\) C = {1,2,3} \(\times\) {3,4,5,6,9}
= {(1,3) (1,4) (1,5) (1,9) (2,3) (2,4) (2,5) (2,6) (2,9) (3,3) (3,4) (3,5) (3,9)} 
RHS = (A \(\times\) B) \(\cap \) (A \(\times\) C) = {(1,4) (1,5) (2,4) (2,5) (3,4) (3,5) }
From (1) and (2), LHS = RHS. Hence Verified....(2)

(i) As above we get the relation {(0, 0),(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (2, 1), (3, 2)} that is reflexive, symmetric and not transitive.
(ii) We have the relation {(0, 0), (1, 1), (2, 2), (3, 3)} which is reflexive, symmetric and transitive.

(i) For a relation on {0, 1, 2, 3} to be reflexive, it must have the pairs (0, 0), (1, 1), (2, 2), (3, 3). Fortunately, it becomes symmetric and transitive. Therefore, as in (i) if we insert (1, 2) and (2, 3) we get the required one. Thus {(0, 0), (1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive; it is not symmetric and it is not transitive. 
(ii) Proceeding like this we get the relation {(0, 0), (1, 1), (2, 2), (3, 3), (1, 2)} that is reflexive, transitive and not symmetric.

(i) Let us start with the pair (1, 2). Since we need symmetricity, we have to include the pair (2, 1). At this stage as (1, 1), (2, 2) are not here, the relation is not transitive.  Thus {(1, 2), (2, 1)} is not reflexive; it is symmetric; and it is not transitive.
(ii) If we include the pairs (1, 1) and (2, 2) to the relation discussed in (iii), it will become transitive. Thus {(1, 2), (2, 1), (1, 1), (2, 2)} is not reflexive; it is symmetric and it is transitive.

(i) Let us use the pair (1,2) to make the relation "not symmetric" and consider the relation {(1, 2)}. It is transitive. If we include (2, 3) and not include (1, 3), then the relation is not transitive. So the relation {(1, 2), (2, 3)} is not reflexive, not symmetric and not transitive.
(ii) Just now we have seen that the relation {(1,2)} is transitive, not reflexive and not symmetric.

(i) (c, c)
(ii) (c, a)
(iii) nothing
(iv) (c, c) and (c, a)