Answer -> C
S = (i_{1}j_{1}) ⊑ (i_{2}j_{2}) 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.
2. Antisymmetric
(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 = (i_{1}j_{1}) ⊑ (i_{2}j_{2}) 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.
Ref :-
https://en.wikipedia.org/wiki/Infinite_descending_chain
https://books.google.co.in/books?id=OR5KAAAAQBAJ&pg=PA17&lpg=PA17&dq=does+natural+no+have+infinite+descending+chain&source=bl&ots=dDJdZ26vWP&sig=WMJUx-35YMNbgkR5twmKDMwL2bo&hl=en&sa=X&ved=0ahUKEwje9of1-rfJAhUSC44KHRqGDI4Q6AEIKjAC#v=onepage&q=does%20natural%20no%20have%20infinite%20descending%20chain&f=false
https://en.wikipedia.org/wiki/Well-order#Examples_and_counterexamples