The relation R on the set N × N defined by (a, b) R (c, d) ⟺a+d=b+c ∀ (a,b) ϵ N×N is, where N is set of natural Number:

Correct option is D

We have (a,b)R(a,b) for all (a,b)∈N×N Since a+b=b+a. hence, R is reflexive.
R is symmetric for we have (a,b)R(c,d)=>a+d=b+c
=>d+a=b+c=>c+b=d+a=>(c,d)R(a,b)
Hence R is symmetric
If (a,b)R(c,d) & (c,d)R(e,f)
Then by definition of R, we have
a+d=b+c" and " c+f=d+e
Hence by addition, we get
a+d+c+f=b+c+d+e" or " a+f=b+e
Hence (a,b)R(e,f)
Thus (a,b)R(c,d) and (c,d)R(e,f)=>(a,b)R(e,f)
Hence R is transitive.