Question

Let R₁ and R₂ be two relations defined as follows :

R₁ = {(a, b) ∈ R² : a² + b² ∈ Q} and

R₂ = {(a, b) ∈ R² : a² + b² ∉ Q}, where Q is the set of all rational numbers. Then ______

Options

• Neither R₁ nor R₂ is transitive.

• R₂ is transitive but R₁ is not transitive.

• R₁ is transitive but R₂ is not transitive.

• R₁ and R₂ are both transitive.

Answer 1

Let R₁ and R₂ be two relations defined as follows :

R₁ = {(a, b) ∈ R² : a² + b² ∈ Q} and

R₂ = {(a, b) ∈ R² : a² + b² ∉ Q}, where Q is the set of all rational numbers. Then neither R₁ nor R₂ is transitive.

Explanation:

For R₁ let a = `1 + sqrt(2)`, b = `1 - sqrt(2)`, c = `8^(1//4)`

aR₁b `\implies` a² + b² = `(1 + sqrt(2))^2 + (1 - sqrt(2))^2` = 6 ∈ Q

bR₁c `\implies` b² + c² = `(1 - sqrt(2))^2 + (8^(1//4))^2` = 3 ∈ Q

aR₁c `\implies` a² + c² = `(1 + sqrt(2))^2 + (8^(1//4))^2 = 3 + 4sqrt(2) ∉ Q`

∴ R₁ is not transitive.

For R₂ let a = `1 + sqrt(2)`, b = `sqrt(2)`, c = `1 - sqrt(2)`

aR₂b `\implies` a² + b² = `(1 + sqrt(2))^2 + (sqrt(2))^2 = 5 + 2sqrt(2) ∉ Q` 

bR₂c `\implies` b² + c² = `(sqrt(2))^2 + (1 - sqrt(2))^2 = 5 - 2sqrt(2) ∉ Q` 

aR₂c `\implies` a² + c² = `(1 + sqrt(2))^2 + (1 - sqrt(2))^2` = 6 ∈ Q

∴ R₂ is not transitive.