Final answer:
The binary relation R on N² is an equivalence relation because it is reflexive (any element is related to itself), symmetric (if one element is related to another, the reverse is also true), and transitive (if the first element is related to a second, and the second to a third, then the first is also related to the third).
Step-by-step explanation:
The student has asked to prove that the binary relation R on N² defined by (m, n) R (j, k) if m + k = n + j is an equivalence relation. To do so, we must show that the relation is reflexive, symmetric, and transitive.
For any element (m, n) in N², we have m + n = n + m, by the commutative property of addition, which holds for ordinary numbers. Therefore, (m, n) R (m, n), and the relation is reflexive
If (m, n) R (j, k), then m + k = n + j. By rearranging the equation, we also get j + n = k + m, which implies (j, k) R (m, n), thus the relation is symmetric.
If (m, n) R (j, k) and (j, k) R (p, q), then m + k = n + j and j + q = k + p. Adding these two equations gives m + k + j + q = n + j + k + p. After canceling common terms, we have m + q = n + p, implying (m, n) R (p, q). Therefore, the relation is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation on N².