Transitive relation

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.

Step 1: ((1,2)) and ((2,4)) require ((1,4)). Step 2: ((1,4)) is present, and ((4,4)) does not create any missing pair. Therefore the relation is transitive. Step 3: Find each chain and match the required final pair.

Step 1: ((2,4)) and ((4,1)) require ((2,1)). Step 2: ((2,1)) is not in the given relation, so the relation is not transitive. Step 3: Do not stop after one chain; check all possible chains.

Step 1: The universal relation contains every possible ordered pair from (A). Step 2: So if ((a,b)) and ((b,c)) are present, ((a,c)) is also definitely present. Step 3: In a universal relation, no required pair can be missing.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: The given self-pairs do not create any other missing required pair. So ((1,3)) must be added. Step 3: To make a relation transitive, first locate the missing direct pair.

Step 1: If (a-b) is even and (b-c) is even, then their sum (a-c) is also even. Step 2: Hence (a) and (c) are related. Step 3: In parity-based relations, add the differences.

Step 1: Adding an odd difference and an odd difference gives an even difference. Step 2: If (a-b) is odd and (b-c) is odd, then (a-c) is even, so the relation fails. Step 3: Remember that odd plus odd is even.

Step 1: If (a=b) and (b=c), then (a=c) must hold. Step 2: Therefore the equality relation is transitive. Step 3: Equality is reflexive, symmetric, and transitive.

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 also present. Step 3: When opposite pairs create self-pair requirements, both must be checked.

Step 1: For transitivity, the relation should pass forward like a chain. Step 2: If (a<b) and (b<c), then (a<c), so less than is transitive. Step 3: Everyday relations are often not transitive, so test them mathematically.

Step 1: Symmetry deals with ((a,b)) implying ((b,a)). Step 2: Transitivity needs ((a,b)) and ((b,c)) to imply ((a,c)). The conditions are different. Step 3: Check each property by its own definition.

Step 1: In the definition of transitivity, the middle element is common. Step 2: From ((x,y)) and ((y,z)), the first and last elements form ((x,z)). Step 3: Do not reverse the order of ordered pairs.

A relation R is transitive when, for every a, b, c in A, the presence of (a,b) and (b,c) guarantees the presence of (a,c). Here, (1,2) and (2,3) require (1,3), and (1,3) is present. The pair (2,2) also creates checks such as (2,2) followed by (2,3), which requires (2,3), already included, and (2,2) followed by itself, which requires (2,2), also included. No other available pair creates a missing consequence. Therefore R satisfies the transitivity condition. Option B is incorrect because it overlooks the required pair that is already present. Options C and D describe different properties and do not answer the transitivity question.

Step 1: Both ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is missing. Step 3: Missing reflexive pairs do not always break transitivity; focus on required chains.