If N denote the set of all natural numbers and R be the relation on N N defined by (a, b) R (c, d), if ad(b + c) = bc(a + d), then R is
symmetric only
reflexive only
transitive only
an equivalence relation
D.
an equivalence relation
For (a, b), (c, d) N N
(a, b) R (c, d)
ad(b + c) = ba(a + b)
Reflexive: Since, ab(b + a) = ba(a + b), ∀ ab ∈ N
(a, b) R (a, b)
So, R is reflexive.
Symmetric: For (a, b), (c, d)
Let (a, b) R (c, d)
So, R is symmetric.
Transitive: For(a, b), (c, d), (e, f) ∈ N N
Let (a, b) R (c, d), (c, d) R (e, f)
ad(b + c) = bc(a + d) , cf(d + e) = de(c + f)
On multiplying Eq. (i) by ef and Eq. (ii) by ab, then we get
adbef + adcef + cfdab + cfeab
= bcaef + bcdef + decab + defab
So, R is transitive.
Hence R is an equivalence relation.
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