Transitive relation

Step 1: In a transitive relation, two connected pairs force a third pair. Step 2: If ((a,b)\in R) and ((b,c)\in R) always give ((a,c)\in R), the relation is transitive. Step 3: In exams, first spot the common middle element.

Transitivity requires that whenever (a,b) and (b,c) belong to R, the pair (a,c) must also belong to R. In this relation, the only nontrivial chain is (1,2) followed by (2,3), which requires (1,3). That required pair is explicitly present. Other possible chains do not create a missing requirement: there are no pairs beginning with 3, and the available links do not produce another unfulfilled chain. Therefore R is transitive, so option A is correct. Reverse pairs are relevant to symmetry, not transitivity.

The governing rule for transitivity is: if (a,b) ∈ R and (b,c) ∈ R, then (a,c) must be in R. Here, (1,2) and (2,3) are both present, so transitivity requires (1,3). However, (1,3) is not listed in R. This single missing consequence is enough to disprove transitivity. The absence of (1,1), (3,2), or (2,1) is irrelevant because those pairs are not required by the displayed linked chain. Therefore option A correctly identifies the failure.

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

Step 1: The universal relation contains every possible ordered pair from (A). Step 2: So whenever ((a,b)) and ((b,c)) are present, ((a,c)) is also present. Step 3: A relation with all possible pairs automatically satisfies transitivity.

Step 1: The identity relation contains only pairs of the form ((a,a)). Step 2: If ((a,a)) and ((a,a)) are used, the required pair ((a,a)) is already present. Step 3: Remember the identity relation as reflexive, symmetric, and transitive.

Step 1: From ((1,2)) and ((2,1)), we need ((1,1)), which is present. Step 2: From ((2,1)) and ((1,2)), we need ((2,2)), which is also present. Step 3: Check every possible chain before deciding.

Step 1: The pairs ((1,2)) and ((2,1)) require ((1,1)). Step 2: The pairs ((2,1)) and ((1,2)) require ((2,2)). Since both are missing, the relation is not transitive. Step 3: Reverse pairs often force self-pairs.

Step 1: If (a\le b) and (b\le c), then on the number line (a\le c). Step 2: So (aRc) is also true. Step 3: Order relations like (\le) and (<) are common examples of transitive relations.

Step 1: If (a<b) and (b<c), then (a<c) must hold. Step 2: Hence (aRc) is true. Step 3: In inequality questions, connect the order directly.

Step 1: If (A\subseteq B) and (B\subseteq C), every element of (A) is also in (C). Step 2: Therefore (A\subseteq C), so (ARC) is true. Step 3: The subset relation is a standard example of transitivity.

Step 1: If (a=b) and (b=c), all three numbers are equal. Step 2: Therefore (a=c), so transitivity is satisfied. Step 3: Treat equality as an always transitive relation.

Step 1: If (a\mid b), then (b=ak) for some integer (k). Step 2: If (b\mid c), then (c=bl), so (c=a(kl)). Hence (a\mid c). Step 3: For divisibility, connect the chain of multiples.

Step 1: The important chain is ((1,2)) and ((2,4)). Step 2: These require ((1,4)), which is present. Self-pairs do not create a problem here. Step 3: In a larger relation, first check chains of non-self pairs.

Step 1: The pairs ((1,2)) and ((2,4)) form a chain. Step 2: Transitivity directly requires ((1,4)). Step 3: In pair-addition questions, identify the missing required pair first.

Step 1: The common middle element is (5). Step 2: Transitivity connects the first element (3) to the last element (7), giving ((3,7)). Step 3: Join the first element of the first pair with the second element of the second pair.

Step 1: In ((a,b)) and ((b,a)), the middle element (b) matches. Step 2: So transitivity requires ((a,a)). Step 3: When reverse pairs appear, check the required self-pair.

Step 1: In equality, if the first equals the second and the second equals the third, then the first equals the third. Step 2: Therefore equality is transitive. Step 3: For real-life wording, translate it into the mathematical rule.

Step 1: The main chain is ((1,2)) and ((2,3)). Step 2: They require ((1,3)), which is present. The self-pairs do not break the rule. Step 3: Separate the required chain and decide from it.

Step 1: Transitivity needs the second element of one pair to match the first element of the next pair. Step 2: ((2,3)) and ((3,4)) require ((2,4)), which is present. Step 3: Do not draw conclusions from pairs that do not connect.

Step 1: First connect ((1,2)) and ((2,3)). Step 2: They require ((1,3)). Later, ((2,4)) and then ((1,4)) may also need checking, but among the options the required pair is ((1,3)). Step 3: In a long chain, build pairs step by step.

Step 1: ((1,2)) and ((2,2)) require ((1,2)), which is present. Step 2: ((1,2)) and ((2,3)) require ((1,3)), and ((2,2)) with ((2,3)) requires ((2,3)). Both are present. Step 3: Also check chains involving self-pairs.

Step 1: The pairs ((5,6)) and ((6,8)) are connected. Step 2: By transitivity, the first element (5) connects directly to the last element (8). Step 3: Transitivity does not give a reverse-direction pair.

Step 1: ((1,1)) and ((1,2)) require ((1,2)), which is present. Step 2: ((1,2)) and ((2,2)) again require ((1,2)), also present. Step 3: A small relation is not automatically non-transitive; check actual chains.

Step 1: The reverse pair ((b,a)) belongs to the idea of symmetry, not transitivity. Step 2: Transitivity checks the forward linked pair ((a,c)). Step 3: When a question asks for the false statement, avoid mixing relation properties.