Answer :
Given that, R be the relation in N ×N defined by (a, b) R (c, d) if ad(b + c) = bc(a + d) for (a, b), (c, d) in N ×N.
Reflexivity:
Let (a,b) R (a,b)
⇒ ab(a+b) =ba( a+b)
which is true since multiplication is commutative on N.
⇒ R is reflexive.
Symmetricity:
Let (a,b) R (c,d)
⇒ ad(b + c) = bc(a + d)
⇒ da(c + b) = cb(d + a)
[since addition and multiplication is commutative on N]
⇒ cb(d + a) = da(c + b)
⇒ (c,d) R (a,b)
∴ (a,b) R (c,d) ⇒ (c,d) R (a,b) ∀ (a,b), (c,d), (e,f) ∈ N ×N.
⇒ R is symmetric.
Transitivity:
Let (a,b) R (c,d)
⇒ ad(b + c) = bc(a + d)
also (c,d) R (e,f)
and cf(d + e) = de(c + f)
Adding (i) and (ii) , we get
⇒ af(b + e) = be(a + f)
⇒ (a,b) R (e,f)
Thus , (a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) ∀ (a,b), (c,d), (e,f) ∈ N ×N.
⇒ R is transitive.
Hence, R is an equivalence relation.
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