Let R be a relation on the set of all real numbers R R = {(a, b) ∈R

1.	Let R be a relation on the set of all real numbers R R = {(a, b) ∈R × R : a2 + b2 = 1}, then R is(a) Equivalence (b) Only transitive (c) Only symmetric (d) None of these
Answer» (c) Only symmetric • V a ∈ R, a² + a² ≠ 1 ⇒ (a, a) ∉R ⇒ R is not reflexive. For example, 0² + 0² = 0, 1² + 1² = \(\big(\frac{1}{2}\big)^2\) + \(\big(\frac{1}{2}\big)^2\) = \(\frac{1}{2}\) and so on. • V a, b ∈R, (a, b) ∈R ⇒ a² + b² = 1 ⇒ b² + a² = 1 ⇒ (b, a) ∈R ⇒ R is symmetric • V a, b, c ∈R, (a, b) ∈R and (b, c) ∈R ⇒ a² + b² = 1 and b² + c² = 1 which does not necessarily mean a²+ c² = 1 ⇒ (a, c) ∉R For example, let a = 0, b = 1, c = 0 0² + 1² = 1 and 1² + 0² = 1 But 0² + 0² ≠ 1.

Let R be a relation on the set of all real numbers R R = {(a, b) ∈R × R : a2 + b2 = 1}, then R is(a) Equivalence (b) Only transitive (c) Only symmetric (d) None of these

Answer»

(c) Only symmetric

• V a ∈ R, a² + a² ≠ 1 ⇒ (a, a) ∉R ⇒ R is not reflexive.

For example, 0² + 0² = 0, 1² + 1² = \(\big(\frac{1}{2}\big)^2\) + \(\big(\frac{1}{2}\big)^2\) = \(\frac{1}{2}\)

and so on.

• V a, b ∈R, (a, b) ∈R ⇒ a² + b² = 1 ⇒ b² + a² = 1

⇒ (b, a) ∈R

⇒ R is symmetric

• V a, b, c ∈R, (a, b) ∈R and (b, c) ∈R

⇒ a² + b² = 1 and b² + c² = 1 which does not necessarily mean

a²+ c² = 1 ⇒ (a, c) ∉R

For example, let a = 0, b = 1, c = 0

0² + 1² = 1 and 1² + 0² = 1

But 0² + 0² ≠ 1.

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