Let R? and R? be two relations defined as follows:
R? = {(a, b) ∈ R²: a² + b² ∈ Q} and
R? = {(a, b) ∈ R²: a² + b² ∉ Q}, where Q is the set of all rational numbers. Then:
Let R? and R? be two relations defined as follows:
R? = {(a, b) ∈ R²: a² + b² ∈ Q} and
R? = {(a, b) ∈ R²: a² + b² ∉ Q}, where Q is the set of all rational numbers. Then:
Option 1 -
Neither R₁ nor R₂ is transitive.
Option 2 -
R₁ is transitive but R₂ is not transitive
Option 3 -
R₁ and R₂ are both transitive
Option 4 -
R₂ is transitive but R₁ is not transitive
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1 Answer
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Correct Option - 1
Detailed Solution:For R? let a = 1+√2, b=1-√2, c = 8¹/?
aR? b => a² + b² = (1+√2)² + (1-√2)² = 6 ∈ Q
aR? c => b² + c² = (1-√2)² + (8¹/? )² = 3 ∈ Q
aR? c => a² + c² = (1+√2)² + (8¹/? )² = 3 + 4√2 ∉ Q
∴ R? is not transitive.
For R? let a = 1+√2, b=√2, c=1-√2
aR? B => a² + b² = (1+√2)² + (√2)² = 5+2√2 ∉ Q
bR? b => b² + c² = (√2)² + (1-√2)² = 5-2√2 ∉ Q
aR? c => a² + c² = (1+√2)² + (1-√2)² =...more
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