VITEEE 2013 Paper

For (a, b), (c, d) ∊ N × N
(a, b) R (c, d)
⇒ ad (b + c) = be (a + d)
Reflexive : ab (b +a) = ba (a+ b), y ab ∊ N (a, b) R (a, b)
So, R is reflexive,
Symmetric : ad (b + c) = bc (a + d)
⇒ bc (a + d) = ad (b + c)
⇒ cd (d + a) = da (c + b)
⇒ (c, d) R (a, b)
So, R is symmetric.
Transitive : For (a, b), (c, d), (e, f) ∊ N x N
Let (a, b) R (c, d), (c,d) R (e,f)
∴ ad (b + c) = bc(a + d), cf(d + e) = de (c +1)
⇒ adb + adc = bca + bcd ...(i)
and cfd + cfe = dec + def ...(ii)
On multiplying eq. (i) by ef and eq. (ii) by ab and then adding,
we have adbef + adcef + cfdab + cfeab = bcaef + bedef + decab + defab
⇒ adcf (b + e) = bcde (a + f)
⇒ af (b + e) = be (a + f)
⇒ (a, b) R (e,f)
So, R is transitive.
Hence R is an equivalence relation.