# Show that the relation R on N × N, defined by(a, b) R (c, d) ⇔ a + d = b + cis an equivalent relation.

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, R be the relation in N ×N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in N ×N.

R is Reflexive if (a, b) R (a, b) for (a, b) in N ×N

Let (a,b) R (a,b)

a+b = b+a

which is true since addition is commutative on N.

R is reflexive.

R is Symmetric if (a,b) R (c,d) (c,d) R (a,b) for (a, b), (c, d) in N ×N

Let (a,b) R (c,d)

a+d = b+c

b+c = a+d

c+b = d+a [since addition is commutative on N]

(c,d) R (a,b)

R is symmetric.

R is Transitive if (a,b) R (c,d) and (c,d) R (e,f) (a,b) R (e,f) for (a, b), (c, d),(e,f) in N ×N

Let (a,b) R (c,d) and (c,d) R (e,f)

a+d = b+c and c+f = d+e

(a+d) – (d+e) = (b+c ) – (c+f)

a-e= b-f

a+f = b+e

(a,b) R (e,f)

R is transitive.

Hence, R is an equivalence relation.

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