Let R be the set of real numbers. Statement-1 : A = {(x, y) ∈ R × R : y – x is an integer} is an equivalence relation on R.
Let R be the set of real numbers.
Statement-1 : A = {(x, y) ∈ R × R : y – x is an integer} is an equivalence relation on R.
Statement-2 : B = {(x, y) ∈ R × R : x = ay for some rational number a} is an equivalence relation on R.
(a) Statement-1 is true, Statement-2 is false.
(b) Statement-1 is false, Statement-2 is true.
(c) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
1 Answers
(a) : y – x = integer and z – y = integer
⇒ z – x = integer
∴ (x, y) ∈ A and (y, z) ∈ A ⇒ (x, z)
⇒ Transitive
Also (x, x) ∈ A is true ⇒ Reflexive
As (x, y) ∈ A ⇒ (y, x) ⇒ Symmetric
Hence A is a equivalence relation but B is not.
(0, y) is in B but (y, 0) is not in B.