Q. 204.7( 3 Votes )

Let N denote the

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.


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

ab(a+b) =ba( a+b)

which is true since multiplication is commutative on N.

R is reflexive.


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.


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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