Question

Let 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). Show that R is an equivalence relation.

Answer 1

To prove a relation R is an equivalence relation, it will be sufficient to prove it as a reflexive, symmetric and transitive relation.

i) Reflexivity:
Let (a, b) be an arbitrary element of N × N.
Now,
a, b ∈ N

⇒ab(a+b)=ba(a+b)

⇒(a,b)R(a,b)

∴ (a, b)R(a, b) for all (a, b) ∈ N × N
Hence, R is reflexive.

ii) Symmetry:
Let (a, b), (c, d) be an arbitrary element of N × N such that (a, b)R(c, d).

∴ ad(b+c)=bc(a+d)

⇒cb(d+a)=da(c+b)

⇒(c,d)R(a,b)

∴ (a, b)R(c, d) ⇒ (c, d)R(a, b) for all (a, b), (c, d) ∈ N × N
Hence, R is symmetric.

iii) Transitivity:
Let (a, b), (c, d), (e, f) be an arbitrary element of N × N such that (a, b)R(c, d) and (c, d)R(e, f).

ad(b+c)=bc(a+d)

⇒adb+adc=abc+bcd     

⇒cd(a−b)=ab(c−d)      .....(1)

Also,cf(d+e)=de(c+f)

⇒cfd+cfe=dec+def       

⇒cd(f−e)=ef(d−c)  ....(2)

From (1) and (2), we have

`(a−b)/(f−e)=−(ab)/(ef)`

⇒aef−bef=−abf+aeb

⇒aef+abf=aeb+bef

⇒af(b+e)=be(a+f)

⇒(a, b)R(e, f)

∴(a, b)R(c, d) and (c, d)R(e, f) ⇒ (a, b)R(e, f) for all (a, b), (c, d), (e, f) ∈ N × N
Hence, R is transitive.
Thus, R being reflexive, symmetric and transitive, is an equivalence relation on N × N.