(i) Let subset A = {2, 4, 6, 8, …} ∈ N
A relation given in N


From equation (i) putting the value of y in equation (ii) we have

Hence, R is transitive relation.
Hence, from (i), (a), (b) and (c) the given relation is a partial ordred relation.

Hence, the second rule is not followed then the given relation is not a total ordered relation.
For this it is clear that to prove a relation is a total ordered relative must be reflexive, symmetric and transitive and also every a, b ∈ A, (a, b) ∈ R or (b, a) ∈ R or a = b is true.
Hence, the given relation and their results are:
Relation |
Conclusion |
(i) (2, 4, 6, 8,…} |
No |
(ii) {0, 2, 4, 6,…} |
No |
(iii) {3, 9, 5, 15,…} |
No |
(iv) {5, 15, 30} |
Yes |
(v) {1, 2, 3, 4} |
No |
(vi) {a, b, ab} ∀ a, b ∈ R |
No |