Transitive relation

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.

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.