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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.

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#### Solution

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.

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