Assertion (A) The relation R on the set $N\times N$ , defined by (a, b) R (c, d) $\Leftrightarrow $ a+d = b+c for all (a, b), (c, d) $\in \text{ N}\times \text{N}$ is an equivalence relation.
Reason (R) Any relation R is an equivalence relation, if it is reflexive, symmetric and transitive.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.

Hint:Here, we will use the definitions of reflexive, symmetric and transitive relations to check whether the given relations are reflexive, symmetric or transitive.

Complete step-by-step answer:
A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x, y) is in the relation.
A relation R is reflexive if each element is related itself, i.e. (a, a) $\in $ R, where a is an element of the domain.
A relation R is symmetric in case if any one element is related to any other element, then the second element is related to the first, i.e. if (x, y) $\in $ R then (y, x) $\in $ R, where x and y are the elements of domain and range respectively.
A relation R is transitive in case if any one element is related to a second and that second element is related to a third, then the first element is related to the third, i.e. if (x, y) $\in $ R and (y, z) $\in $ R then (x, z) $\in $ R.
Here, the given relation is:
(a, b) R (c, d) $\Leftrightarrow $ a+d = b+c
Since, we know that (a +d)= (a + d).
Therefore, (a+d, a+d) $\in $ R.
So, R is reflexive.
Now, suppose that (a + d, b + c) $\in $ R.
Then we can write:
 a+d=b+c
Or, b+c = a+d
So, (b+c, a+d) $\in $ R.
This means that R is symmetric.
Now, let us take an ordered pair (a + d, b + c) $\in $ R and (b + c, e+f) $\in $ R.
We can write:
$a+d=b+c..........\left( 1 \right)$
And, $b+c=e+f.............\left( 2 \right)$
From equations (1) and (2), we can say that a+d = e+f.
So, (a + d, e + f) $\in $ R.
This implies that R is transitive.
So, R is reflexive, symmetric as well as transitive.
We know that a relation which is symmetric, reflexive as well as transitive is an equivalence relation.
Therefore, the assertion is correct and reason is the correct explanation.
Hence, option (a) is the correct answer.

Note: Students should note here that for a relation to be a particular type of relation, i.e. reflexive, symmetric or transitive, all its elements must satisfy the required conditions. If we are able to find even a single counter example, i.e. if any of the elements doesn’t satisfy the criteria, then we can’t proceed further.