Transitive relation

Answer: A. Transitivity means that whenever (a,b) and (b,c) are in R, the pair (a,c) must also be in R. The relation contains (1,2) and (2,4), so it must contain (1,4), but (1,4) is absent. Hence this is the failed requirement. Option B is wrong because transitivity from (1,3) and (3,4) would require (1,4), not (3,1). Option C uses (3,2), which is not in the stated relation; moreover, its proposed conclusion is already present but the premise is unavailable. Option D is unrelated: (4,4) alone does not require (4,1). Memory cue: match the second coordinate of the first pair with the first coordinate of the second pair, then connect the two outer elements.

The correct answer is B. Transitivity requires that whenever (a,b) and (b,c) belong to the relation, (a,c) must also belong to it. In a union, the first pair may come from R and the second from S, but neither relation individually is required to contain the composite pair. For example, on {1,2,3}, let R={(1,2)} and S={(2,3)}. Each relation is transitive: there is no complete two-step chain inside either one. However, R∪S={(1,2),(2,3)} does not contain (1,3), so it is not transitive. A is disproved by this example. C is false because the union can contain pairs. D is unrelated and is not guaranteed; a transitive relation may contain reverse pairs. Thus only B is always valid.

The correct answer is B, (1,4). To disprove transitivity, find two pairs (a,b) and (b,c) that are present while (a,c) is absent. Here (1,2) belongs to R and (2,4) also belongs to R. Therefore transitivity would require (1,4), but (1,4) is missing. This is a direct counterexample. A is not forced by any useful chain through the listed pairs, and C is not forced by a chain beginning at 3. D is actually not the required composite pair for the forward chain 1→2→4; although (4,2) is present, it does not replace (1,4). The relation is symmetric because reverse pairs are included, but symmetry alone does not guarantee transitivity. Always follow the arrows in order: a→b and b→c demand a→c.

Direct answer: A. To prove that a relation T is transitive, we must show that whenever (a,b) and (b,c) belong to T, the pair (a,c) also belongs to T. Let T=R∩S. If the first two pairs are in R∩S, each is in both R and S. Since R is transitive, (a,c)∈R; since S is transitive, (a,c)∈S. Therefore (a,c) lies in R∩S, exactly as option A states. Option B is the definition of symmetry, not transitivity. Option C expresses reflexivity and is not required. Option D says the intersection is universal, which is much stronger and need not hold. A useful proof cue is “intersection means satisfy both”: apply transitivity in both relations, then intersect the conclusions.

Direct answer: A. A relation is transitive when two linked pairs (a,b) and (b,c) force (a,c). Each singleton relation in option A is itself transitive: R has only (1,2), and S has only (2,3), so neither relation contains two consecutive pairs requiring a further pair. However, their union contains (1,2) and (2,3), while it does not contain (1,3). Therefore the union fails transitivity. Option B gives {(1,1),(2,2)}, an identity-type relation, which is transitive. Option C gives the empty relation; transitivity is vacuously true because there are no starting pairs. Option D gives A×A, since adding the empty relation changes nothing; the universal relation is transitive. Common warning: transitivity need not pass from two relations to their union, because the first link may come from one relation and the second from the other.

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.

Direct answer: C. A relation is reflexive on A only when (1,1), (2,2), and (3,3) are all present. Option C omits (3,3), so it is not reflexive. To check transitivity, the pairs (1,2) and (2,1) require (1,1), which is present; (1,2) followed by (2,2) again gives (1,2); (2,1) followed by (1,2) gives (2,2), present; and the remaining compositions stay within the listed pairs. Thus C is transitive. Option A contains every diagonal pair and is therefore reflexive as well as transitive. Option B has (1,2) and (2,3) but lacks (1,3), so it is not transitive. Option D has (2,1) and (1,2) but lacks (2,2), so it is not transitive. Check reflexivity first, then test every composable two-step chain.

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.