Let a relation R on N × ℕ be defined as: (x₁, y₁) R(x₂, y₂) if and only if x₁ ≤ x₂ or y₁ ≤ y₂. Consider the two statements:
(I) R is reflexive but not symmetric.
(II) R is transitive
Then which one of the following is true?

Solution

Maths sir · 4 steps

With an "or" condition, each link in a chain can use a different coordinate, so transitivity breaks.

1. Reflexive. For any pair, x₁ ≤ x₁ is always true. So (x₁, y₁) R (x₁, y₁) holds for every pair, and R is reflexive.
2. Not symmetric. (1, 1) R (2, 2) because 1 ≤ 2. But (2, 2) is not related to (1, 1), since 2 ≤ 1 fails for both coordinates. So statement (I) is correct.

   Pair checked	First coordinate	Second coordinate	Related?
   (3, 3) to (1, 4)	3 ≤ 1 fails	3 ≤ 4 holds	yes
   (1, 4) to (2, 2)	1 ≤ 2 holds	4 ≤ 2 fails	yes
   (3, 3) to (2, 2)	3 ≤ 2 fails	3 ≤ 2 fails	no

3. Not transitive. The first link uses the second coordinate and the second link uses the first coordinate. Neither inequality carries through, so (3, 3) is not related to (2, 2). Statement (II) is false.
4. So. R is reflexive but not symmetric, and it is not transitive. Only (I) is correct, which is option (d).

The idea to keep. When a relation is defined with "or", test transitivity with a chain where each step uses a different condition. An "or" condition usually fails to chain.

Answer from the NTA final answer key. Solution by PrepWiser.