Transitive relation

The correct answer is C, (1,3). Transitivity says that if (a,b) and (b,c) are in R, then (a,c) must also be in R. Here the first pair is (1,2) and the second is (2,3). Their middle entries match, so the required consequence is (1,3). Adding this one pair makes the relation transitive: the new pair does not create a further missing chain, because there is no pair beginning with 3 and no pair ending in 1 that would force another result. Option A, (2,1), does not supply the consequence of the existing chain. Option B, (3,1), and option D, (3,2), likewise do not address the required pair. Do not reverse the chain; follow it from 1 to 2 to 3, then join the first and last elements.

Correct answer: A. A relation is transitive when, whenever (x,y) and (y,z) belong to it, (x,z) must also belong to it. It is symmetric when (x,y) implies (y,x). In option A, (1,2) is present but (2,1) is absent, so the relation is not symmetric. It is transitive because the only possible chains either involve a diagonal pair, which leaves the other pair unchanged, or use (1,2) followed by (2,2), which again gives (1,2), already present. Thus A satisfies both required conditions. Option B is symmetric because it contains both (1,2) and (2,1), but it is not transitive: these two pairs would require (1,1), which is absent. Option C is not transitive because (1,2) and (2,3) are present but (1,3) is missing; it is also not symmetric. Option D is not transitive: (2,1) and (1,2) would require (2,2), which is absent. The key test is to search for a two-step chain and then check the required shortcut pair.

Step 1: A transitive relation must pass from the first pair to the third pair. Step 2: If ((a,b)) and ((b,c)) are present, then ((a,c)) must also be present. Step 3: In exams, match the middle element and check the direct pair.

Step 1: Look at ((1,2)) and ((2,3)). Step 2: They require ((1,3)), which is already present. Step 3: Check only the connected pairs that can trigger transitivity.

A relation R is transitive when, for any a, b, and c, the presence of (a,b) and (b,c) guarantees the presence of (a,c). Here, (1,2) and (2,3) are both in R, so transitivity requires (1,3). That ordered pair is absent. The reverse pair (2,1) and self-pair (1,1) are not required by transitivity, so option A is correct.

Step 1: Check all possible connected pairs. Step 2: ((1,2)) and ((2,2)) require ((1,2)), which is present. Step 3: When all required pairs are already present, the relation is transitive.

Step 1: Combine ((1,2)) and ((2,3)). Step 2: Transitivity requires ((1,3)). Step 3: In such questions, find the compulsory missing pair only.

Step 1: If (a) divides (b), and (b) divides (c), then (a) divides (c). Step 2: So the condition of transitivity is satisfied. Step 3: For divisibility relations, remember the chain of division.

Step 1: If (a \le b) and (b \le c), then (a \le c). Step 2: This is exactly the condition for transitivity. Step 3: For inequality relations, join the inequality chain directly.

Step 1: Observe the order of smaller and larger numbers. Step 2: If (a) is less than (b), and (b) is less than (c), then (a) is less than (c). Step 3: Read inequality relations as a chain.

Step 1: Assume (a-b) is even and (b-c) is even. Step 2: Adding them gives (a-c) even. Step 3: In such questions, adding the differences is a useful exam method.

Step 1: Take (1R2) because (1-2) is odd, and (2R3) because (2-3) is odd. Step 2: But (1-3) is even, so (1R3) is false. Step 3: One small counterexample is enough to disprove transitivity.

Use the rule: whenever (a,b) and (b,c) belong to R, (a,c) must also belong to R. The only nontrivial chain is (1,2) followed by (2,4), and its required result (1,4) is already present. The pair (3,3) leads only to itself, which is present. No reverse pair such as (2,1) or (4,1) is demanded, so R is transitive and option A is correct.

Step 1: ((1,2)) and ((2,4)) trigger the transitivity check. Step 2: They require ((1,4)). Step 3: For a minimum addition question, add only the necessary pair.

Step 1: The transitivity condition is tested only when both ((a,b)) and ((b,c)) are present. Step 2: In the empty relation, no such pairs exist, so the condition is never violated. Step 3: Remember that the empty relation is transitive by vacuous truth.

Step 1: The universal relation contains all ordered pairs from (A). Step 2: Therefore, whenever ((a,b)) and ((b,c)) are present, ((a,c)) is also present. Step 3: A relation containing all pairs cannot fail transitivity.

Step 1: The identity relation contains only pairs of equal elements. Step 2: ((a,a)) followed by ((a,a)) requires ((a,a)), which is present. Step 3: The identity relation is always transitive.

Step 1: In ((4,7)) and ((7,9)), the middle element (7) matches. Step 2: Transitivity gives ((4,9)) directly. Step 3: Be careful not to reverse the order of an ordered pair.

Step 1: First, ((2,5)) and ((5,8)) give ((2,8)). Step 2: Then ((2,8)) and ((8,10)) give ((2,10)). Step 3: In a long chain, connect the first element to the last element.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((2,3)) and ((3,4)) require ((2,4)), and ((1,3)) and ((3,4)) require ((1,4)), both present. Step 3: If every required pair is present, mark the relation as transitive.

Step 1: ((1,3)) and ((3,4)) are in the relation. Step 2: Transitivity requires ((1,4)), but it is missing. Step 3: If even one required pair is absent, the relation is not transitive.

Transitivity concerns only pairs that can be joined through a common middle element. The chain (1,2),(2,1) requires (1,1), which is present; the reverse chain (2,1),(1,2) requires (2,2), also present. Chains involving 1,1 or 2,2 reproduce already listed pairs. Element 3 appears in no pair, so no pair involving 3 is required. Thus R is transitive, making A correct.

Step 1: ((1,2)) and ((2,1)) require ((1,1)). Step 2: ((2,1)) and ((1,2)) require ((2,2)). Step 3: When reverse pairs are present, self-pairs often become necessary for transitivity.

Step 1: If (a=b) and (b=c), then (a=c). Step 2: This satisfies the condition of transitivity. Step 3: Equality relation is always transitive.

Step 1: (1 \ne 2) and (2 \ne 1) are true. Step 2: But (1 \ne 1) is false, so ((1,1)) cannot be obtained. Step 3: For the not-equal relation, test a counterexample before marking it transitive.