Transitive relation

Step 1: To prove non-transitivity, we must show the transitive rule fails. Step 2: That means two connected pairs are in (R), but the required third pair is not in (R). This is the correct situation. Step 3: Always write a counterexample using clear ordered pairs.

Step 1: ((1,2)) and ((2,1)) require ((1,1)), which is present. Step 2: ((2,1)) and ((1,2)) require ((2,2)), which is present. Reflexive chains also create no missing pair. So (R) is transitive. Step 3: The absence of pairs involving the third element is not automatically a problem.

Step 1: If (a) is an ancestor of (b), and (b) is an ancestor of (c), then (a) is also an ancestor of (c). Step 2: Therefore ((a,b)) and ((b,c)) imply ((a,c)). Hence the relation is transitive. Step 3: An ancestor relation continues through generations.

Step 1: The less-than relation moves forward in order. Step 2: Since (1<3) and (3<5), we get (1<5). Hence ((1,5)\in R). Step 3: Do not reverse the direction in inequality relations.

Step 1: ((2,3)) and ((3,2)) are both in (R). Step 2: They require ((2,2)), which is not in (R). Hence the relation is not transitive. Step 3: Opposite connected pairs often demand reflexive pairs.

Step 1: The transitive condition becomes active only when pairs like ((a,b)) and ((b,c)) exist. Step 2: If no such connected pairs exist, there is no chance for the condition to fail. So the relation is transitive. Step 3: First search for connected pairs when testing transitivity.

Step 1: After ((1,2)), there is no pair starting with (2). After ((3,4)), there is no pair starting with (4). Step 2: So no transitive requirement is created. Hence (R) is transitive. Step 3: Do not assume every small relation is non-transitive.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((2,2)) with ((2,3)) requires ((2,3)), and ((2,3)) with ((3,3)) again requires ((2,3)), which is present. Hence (R) is transitive. Step 3: Check only the pairs that are actually required.

Step 1: From ((1,2)) and ((2,3)), we get ((1,3)). Step 2: Then from ((1,3)) and ((3,4)), we get ((1,4)). So ((1,4)) is certain. Step 3: In a transitive relation, extend the chain step by step.

Step 1: ((1,2),(2,3)) require ((1,3)); ((2,3),(3,4)) require ((2,4)); and ((1,3),(3,4)) require ((1,4)). Step 2: All these pairs are present in (R). ((4,4)) does not create a missing pair. Hence (R) is transitive. Step 3: In a long list, mark the required pairs while checking.

Step 1: In a transitive relation, if ((a,b) \in R) and ((b,c) \in R), then ((a,c) \in R) must be present. Step 2: Here ((1,2)) and ((2,3)) are present, and ((1,3)) is also present. So the condition holds. Step 3: In exams, always check the missing direct pair after linking two pairs.

Step 1: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 2: The relation does not contain ((1,3)), so the condition fails. Step 3: One missing required pair is enough to make a relation non-transitive.

Step 1: In the identity relation, each ordered pair has the same first and second element. Step 2: If ((a,a)) and ((a,a)) are present, then ((a,a)) is already present. So transitivity holds. Step 3: Remember that the identity relation is both reflexive and transitive.

A relation is transitive if every occurrence of (a,b) and (b,c) implies (a,c). The empty relation contains no ordered pairs, so there is no pair of premises that can produce a counterexample. The implication is therefore true vacuously. This remains true even though A has three elements; the set need not be empty. Hence the empty relation is transitive, and option A is correct.

Step 1: If (a) divides (b) and (b) divides (c), then (a) divides (c). Step 2: Hence the divisibility relation is transitive. Step 3: In divisibility questions, connect the multiplication factors.

Step 1: In the less-than relation, order moves forward. Step 2: If (a<b) and (b<c), then (a<c). So the relation is transitive. Step 3: For inequality relations, remember the direction on the number line.

Step 1: If (a\le b) and (b\le c), then (a\le c) must hold. Step 2: Therefore the less-than-or-equal relation is transitive. Step 3: Both (<) and (\le) are transitive in the same order direction.

Step 1: From ((2,1)) and ((1,2)), transitivity requires ((2,2)). Step 2: The pair ((2,2)) is not in the relation, so it is not transitive. Step 3: Opposite pairs often create a required self-pair.

Step 1: The main check is from ((1,2)) and ((2,3)), which requires ((1,3)). Step 2: ((1,3)) is present, and ((3,3)) creates no missing required pair. So the relation is transitive. Step 3: Check all linked pairs systematically.

Step 1: In transitivity, ((a,b)) and ((b,c)) imply ((a,c)). Step 2: Here (a=4), (b=6), and (c=9), so ((4,9)) is required. Step 3: When the middle element matches, connect the first and last elements.