Transitive relation

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: In hard questions, tick every formed chain separately.

Step 1: ((2,3)) and ((3,1)) require ((2,1)) for transitivity. Step 2: ((2,1)) is not present in the list, so the condition fails. Step 3: Do not check only the first visible chain; check later chains too.

Step 1: If (a-b) is divisible by (4) and (b-c) is also divisible by (4), then their sum (a-c) is divisible by (4). Step 2: Hence (aRc). Step 3: For divisibility-based relations, add the differences.

Step 1: (a+b) being even means (a) and (b) have the same parity. Step 2: If (a) and (b) have the same parity, and (b) and (c) also have the same parity, then (a) and (c) have the same parity. Step 3: In parity relations, focus on the nature of the elements.

Step 1: If (a+b) is odd, (a) and (b) have opposite parity. Step 2: If (b+c) is also odd, then (a) and (c) have the same parity, so (a+c) is even. Step 3: Two consecutive odd-sum conditions do not give the third odd-sum condition.

Step 1: ((2,1)) and ((1,3)) are in the relation. Step 2: They require ((2,3)), which is present; ((1,2)) and ((2,1)) require ((1,1)), and ((2,1)) and ((1,2)) require ((2,2)), both present. Step 3: After all checks, the relation is transitive, so saying no is wrong.

Step 1: In transitivity, we look for any missing required pair. Step 2: If for every ((a,b)) and ((b,c)), ((a,c)) is present, the relation is transitive. Step 3: Do not be confused by many pairs; follow the rule.

Option B is correct. The only composable pair in B is (1,2) followed by (2,3), and the required pair (1,3) is missing, so B is not transitive; therefore this option cannot be correct. Option A is transitive but also reflexive. Option C is transitive on {1,2} but fails reflexivity on A because (3,3) is absent. Option D is not transitive because (2,1) and (1,2) would require (2,2). Hence option C is the required relation.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((2,3)) and ((3,1)) require ((2,1)), and ((3,1)) and ((1,2)) require ((3,2)). Step 3: In cyclic relations, new pairs may create further pairs.

Step 1: If (a) and (b) have the same remainder on division by (5), and (b) and (c) also have the same remainder, then (a) and (c) have the same remainder. Step 2: Hence (a \equiv c \pmod{5}). Step 3: In congruence questions, treat equal remainders as a chain.

Step 1: ((1,4)) and ((4,3)) require ((1,3)). Step 2: ((1,3)) is not in the relation, so transitivity fails. Step 3: Use already present linked pairs like ((1,4)) in further checks.

Step 1: ((2,5)) and ((5,8)) imply ((2,8)). Step 2: Then ((2,8)) and ((8,11)) imply ((2,11)). Step 3: In a long chain, apply transitivity step by step.

Step 1: If ((a,b)) and ((b,c)) are in (R\cap S), they are in both (R) and (S). Step 2: Since both are transitive, ((a,c)) is in both, hence in (R\cap S). Step 3: In intersection, both relation conditions work together.

Step 1: Chains from different relations may combine inside the union. Step 2: It is possible that ((a,b)) comes from one relation and ((b,c)) from another, but ((a,c)) is not in the union. Step 3: The union must be checked separately.

Step 1: (R\cup S={(1,2),(2,3)}). Step 2: ((1,2)) and ((2,3)) require ((1,3)), which is missing. Step 3: This example shows that union is not always transitive.

Step 1: ((a,b)\in R^{-1}) means ((b,a)\in R). Step 2: If ((a,b)) and ((b,c)) are in (R^{-1}), then ((b,a)) and ((c,b)) are in (R); transitivity gives ((c,a)\in R), hence ((a,c)\in R^{-1}). Step 3: Inverse reverses order, but transitivity remains.

Step 1: The given (R) acts like an increasing chain from (1) to (2) to (3) and includes the required direct pairs. Step 2: The inverse of a transitive relation is also transitive. Step 3: While taking inverse, every pair reverses, but the chain condition remains valid.

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. Step 2: Self-pairs like ((2,2)) and ((4,4)) only demand already present pairs again. Step 3: Check self-pairs too, but they often do not create new missing pairs.

Step 1: ((2,4)) and ((4,2)) require ((2,2)). Step 2: ((2,2)) is not in the relation, so it is not transitive. Step 3: Opposite-direction pairs often force a self-pair.

Step 1: If (a) is a factor of (b) and (b) is a factor of (c), then (a) is a factor of (c). Step 2: Therefore the relation is transitive. Step 3: Factor and divisibility language may differ, but the logic is the same.