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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.
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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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