Transitive relation

Step 1: If (a<b) and (b<c), then (a<c). Step 2: Therefore, the less-than relation is transitive. Step 3: The same rule works for finite as well as infinite sets.

Step 1: (1R2) and (2R3) are true. Step 2: Transitivity would require (1R3), but (1+1=3) is false. Step 3: A next-number relation is generally not transitive.

Step 1: In this set, the possible pairs are ((1,3)) and ((2,4)). Step 2: There is no second pair starting with (3) or (4) to form a chain. Step 3: If no chain triggers transitivity, the condition is not violated.

Step 1: (1+2=3), so ((1,3)) is present, and (3+2=5), so ((3,5)) is present. Step 2: Transitivity requires ((1,5)), but (1+2=5) is false. Step 3: The same rule can behave differently when the set changes.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: Required pairs formed with self-pairs are also already present. Step 3: A relation need not contain every self-pair to be transitive.

Step 1: From ((a,b)) and ((b,c)), we get ((a,c)). Step 2: Then from ((a,c)) and ((c,d)), we get ((a,d)). Step 3: Transitivity can be applied repeatedly in a chain.

Step 1: ((1,2)) and ((2,2)) require ((1,2)), which is present. Step 2: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 3: Check all active chains systematically.

Step 1: If (A \subseteq B) and (B \subseteq C), then (A \subseteq C). Step 2: Therefore, the subset relation is transitive. Step 3: In set-based questions, form the inclusion chain.

Step 1: Take (A={1}), (B={2}), and (C={1}). Step 2: (A \cap B=\varnothing) and (B \cap C=\varnothing), but (A \cap C={1}). Step 3: The disjointness relation is not necessarily transitive.

Step 1: If line (l) is parallel to line (m), and (m) is parallel to line (n), then (l) is parallel to (n). Step 2: Hence, the relation is transitive. Step 3: In geometry, focus on the chain of same direction.

Step 1: If (a) is brother of (b), and (b) is brother of (c), it does not always force (a) to be brother of (c) in every possible setting. Step 2: The relation may depend on family context, so general transitivity is not guaranteed. Step 3: Apply mathematical relation rules carefully to social examples.

Step 1: Check only the pairs that are actually present in the relation. Step 2: ((1,2)) and ((2,3)) are present, so ((1,3)) is required. Step 3: Do not start a transitivity check from pairs that are not in the relation.

Step 1: ((2,1)) and ((1,3)) are present. Step 2: They require ((2,3)), and it is already in the relation. Step 3: So among the options, no new required pair is represented; the need for ((2,3)) is already fulfilled.

Step 1: ((2,1)) and ((1,2)) require ((2,2)). Step 2: This pair is missing, so transitivity fails. Step 3: Whenever reverse pairs are present, check the needed self-pair.

Step 1: The relation has many two-way pairs among (1,2,3). Step 2: ((1,2)) and ((2,1)) require ((1,1)), and similarly ((2,2)) and ((3,3)) are needed. Step 3: When a cycle appears, quickly check for missing self-pairs.

Step 1: If ((a,b)) and ((b,c)) are in (R \cap S), they are in both (R) and (S). Step 2: Since both relations are transitive, ((a,c)) is in both. Step 3: Therefore, ((a,c)) is in (R \cap S), so the intersection is transitive.

Step 1: In a union, one pair may come from one relation and the second pair from another relation. Step 2: Then the required ((a,c)) may not be in the union. Step 3: Intersection of transitive relations is safe, but union is not always safe.

Step 1: The relation has only ((1,2)). Step 2: There is no pair starting with (2) to form a chain. Step 3: A single pair often does not violate transitivity unless a chain is formed.

Step 1: ((1,2)) and ((2,2)) require ((1,2)). Step 2: That pair is already present. Step 3: Check conclusions formed with self-pairs carefully.

Step 1: ((2,1)) and ((1,2)) are both in the relation. Step 2: They require ((2,2)). Step 3: Since ((2,2)) is missing, transitivity fails.

Step 1: A valid chain must use two starting pairs that are actually in the relation. Step 2: ((1,2)) and ((2,3)) are present and give ((1,3)). Step 3: Chains using absent pairs are not valid transitivity checks.

Step 1: If (a \equiv b \pmod{3}) and (b \equiv c \pmod{3}), then (a) and (c) have the same remainder. Step 2: So (a \equiv c \pmod{3}). Step 3: A relation based on the same remainder is transitive.

Step 1: (a+b) even means (a) and (b) have the same parity. Step 2: (b+c) even means (b) and (c) have the same parity, so (a) and (c) also have the same parity. Step 3: Hence (a+c) is even, and the relation is transitive.

Step 1: (1+2) is odd, so (1R2). Step 2: (2+3) is odd, so (2R3). But (1+3) is even, so (1R3) is false. Step 3: For an odd-sum relation, a counterexample is easy to find.

Step 1: ((1,3)) and ((3,3)) require ((1,3)), which is present. Step 2: ((1,3)) and ((3,4)) require ((1,4)), which is present. Step 3: If all required conclusions are present, the relation is transitive.