Question

Let R = {(x, y) : x, y ∈ N and x² – 4xy + 3y² = 0}, where N is the set of all natural numbers. Then the relation R is ______.

Options

• reflexive but neither symmetric nor transitive.

• symmetric and transitive.

• reflexive and symmetric.

• reflexive and transitive.

Answer 1

Let R = {(x, y) : x, y ∈ N and x² – 4xy + 3y² = 0}, where N is the set of all natural numbers. Then the relation R is reflexive and transitive.

Explanation:

R = {(x, y) : x, y ∈ N and x² – 4xy + 3y² = 0}

Now, x² – 4xy + 3y² = 0

⇒ (x – y) (x – 3y) = 0

∴ x = y or x = 3y

∴ R = {(1, 1), (3, 1), (2, 2), (6, 2), (3, 3), (9, 3),......}

Since (1, 1), (2, 2), (3, 3),...... are present in the relation, therefore R is reflexive.

Since (3, 1) is an element of R but (1, 3) is not the element of R, therefore R is not symmetric

Here (3, 1) ∈ R and (1, 1) ∈ R `\implies` (3, 1) ∈ R

(6, 2) ∈ R and (2, 2) ∈ R `\implies` (6, 2) ∈ R

For all such (a, b) ∈ R and (b, c) ∈ R `\implies` (a, c) ∈ R

Hence R is transitive.