Q.
Let N denotes the set of natural numbers and r is a relation in N×N. Which of the following is not an equivalence relation in N
see full answer
Start JEE / NEET / Foundation preparation at rupees 99/day !!
21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya
a
b
c
2. (a, b) R (c, d) ⇔ad = bc
d
3. (a, b) R (c, d)⇔ad(b + c)= bc ( a+ d)
e
4. (a, b)R (c, d) ⇔bc(b + c) =ad(a + d)
answer is D.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Let (a, b),(c, d)∈N×N.The relation R defined by (a, b) R (c, d)⇔a + b = b + c is an equivalence reaction.The relation R defined by (a, b) R (c, d)⇔ad = bc is an equivalence relation.Let R = {[(a, b), (c, d)]:ad (b+ c) = bc (a + d), (a, b), (c, d) ∈N×N} ad (b + c) = bc (a +d)⇔b + cbc=a + dad⇔1c+1b+1d+1a⇔1a-1b=1c-1dLet (a, b) ∈N×NWe have, 1a-1b=1a-1b ∴ (a, b)R (c, d)R is reflexiveLet (a, b)R (c, d)∴ 1a-1b=1c-1d⇒ 1c-1d=1a-1b⇒ (c, d) R (a, b) R is Symmetric.Let (a, b) R (c, d) and (c, d) R (e, f)Let (a, b) R (c, d) and (c, d) R (e, f)∴ 1a-1b=1c-1dand ⇒ 1a-1b=1e-1f (a, b) R (e, f)⇒ R is transitive.∴R is equivalence relation.Let R={[(a, b), (c, d)] : bc (b + c) = ad ( a + d) , (a , b), (c ,d )bc N×N}Let (a, b)∈ N×NWe have, b a(b +a) = ab ( a +b) R is reflexive.Let (a, b) R (c, d)∴ bc (b +c) = ad (a +d)⇒ ad (a + d) = (b + c)⇒ da(d + a) = cb (c + b)⇒ (c, d) R (a, b)⇒ R symmetricLet (a, b) R ( c, d) and r (e, f).∴ bc (b +c) = ad (a +d) and de ( d+ e) = c f (c, f).These equations need no imply be (b + e) = a f (a + f).∴ (a, b) may not be R - related to ( c, f)∴R is not transitve.∴R is not an equivalence relation.
Watch 3-min video & get full concept clarity