Answer -> C
S = (i1j1) ⊑ (i2j2) iff (i1<i2) or ((i1=i2)∧(j1≤j2))
1. (m,n) R (m,n) ?
yes, here m !< n , so we go at second criteria.
Now m=n & n =n. So This is reflexive.
(1,2) R (2,3)
Is (2,3) R (1,2) ? No as 2 < 1.
If you see the defination, it is clear that other than diagonal element no other element is related to itself. So antisymmetric.
3. Transitive ->
(1,2) R (2,3) & (2,3) R (2,4) (It is easy to prove)
(1,2) R (2,4) ? Yes. It can be seen easily from following property
S = (i1j1) ⊑ (i2j2) iff (i1<i2) or ((i1=i2)∧(j1≤j2)).
Not going to prove this formally.
4. It is Not reflexive (1,2) R (2,3) but (2,3 ) ~R ( 3,2)
5. This is well ordered. We do not had infinite descending chain. As we have least element (0,0) we our chain stops there.