Final answer:
The relation R is an equivalence relation on N2 because it is reflexive, symmetric, and transitive.
Step-by-step explanation:
The relation R is an equivalence relation if it satisfies three properties: reflexive, symmetric, and transitive.
Reflexive: For any ordered pair (m,n) in N2, (m,n) R (m,n) if m+n = n+m, which is true. So, the relation R is reflexive.
Symmetric: For any two ordered pairs (m,n) and (j,k) in N2, if (m,n) R (j,k), then m+k = n+j. But this can be rewritten as j+k = m+n, which means (j,k) R (m,n). So, the relation R is symmetric.
Transitive: For any three ordered pairs (m,n), (j,k), and (p,q) in N2, if (m,n) R (j,k) and (j,k) R (p,q), then m+k = n+j and j+q = k+p. Adding these equations together, we get m+k+j+q = n+j+k+p, which simplifies to (m,q) R (n,p). So, the relation R is transitive.
Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.