Transitive relation

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is absent. Step 3: The presence of another diagonal pair does not remove this missing requirement.

Step 1: ((3,2)) is in the relation because (3=2+1), and ((2,1)) is also in it. Step 2: Transitivity would require ((3,1)), but (3\ne1+1). Step 3: Relations based on a fixed difference of one are generally not transitive.

Step 1: If (a\ge b) and (b\ge c), then (a\ge c). Step 2: Therefore two connected pairs force a third pair that also belongs to the relation. Step 3: Both (\ge) and (\le) are transitive order relations.

Step 1: If (A\subseteq B) and (B\subseteq C), then every element of (A) is also in (C). Step 2: Hence (A\subseteq C). Step 3: The subset relation is a standard example of a transitive relation.

Step 1: ((1,3)) and ((3,4)) are in the relation. Step 2: They require ((1,4)), but it is not in the list. Step 3: Check not only the first chain, but also chains formed by existing pairs.

Step 1: The common middle element in the two pairs is (4). Step 2: By the transitive rule, ((1,6)) must be present. Step 3: Form the answer using the first element of the first pair and the second element of the second pair.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((1,2)) and ((2,2)) require ((1,2)), which is already present. Step 3: When all possible chains give pairs already in the relation, the relation is transitive.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: They require ((1,3)), but it is not present. Step 3: The presence of ((2,2)) does not fix the missing transitive pair.

Step 1: The transitive rule applies directly to ((a,b)) and ((b,c)). Step 2: These two pairs require ((a,c)). Step 3: Do not reverse the order; take the first element of the first pair and the second element of the second pair.

Step 1: The main chain ((1,2)) and ((2,3)) requires ((1,3)), which is present. Step 2: Chains involving diagonal pairs return pairs that are already present. Step 3: Do not get confused by diagonal pairs; just match the required pairs.

Step 1: Symmetry and transitivity are different properties. Step 2: For example, ({(1,2),(2,1)}) is symmetric, but it is not transitive because ((1,1)) and ((2,2)) are missing. Step 3: Always check each property separately.

Step 1: Reflexivity requires all ((a,a)) pairs. Step 2: Transitivity requires ((a,c)) whenever ((a,b)) and ((b,c)) are present. These are different conditions. Step 3: Assuming transitivity from reflexivity is a common exam mistake.

Step 1: If (a^2=b^2) and (b^2=c^2), then equality gives (a^2=c^2). Step 2: Hence ((a,c)) also belongs to the relation. Step 3: A chain of equalities is an easy way to prove transitivity.

Step 1: If ((a,b)) is in the relation, then both (a) and (b) are even. Step 2: If ((b,c)) is also in the relation, then (c) is even too, so (a) and (c) are both even. Step 3: Conditions based on belonging to the same group often give transitivity.

The key concept is parity. The sum a+b is even exactly when a and b have the same parity: both are even or both are odd. Suppose (a,b) ∈ R and (b,c) ∈ R. Then a and b have the same parity, and b and c also have the same parity. Consequently, a and c have the same parity, so a+c is even and (a,c) ∈ R. This proves transitivity, making option A correct. In fact, the relation is also reflexive and symmetric, but those facts are not needed for this question.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((1,3)) and ((3,3)) require ((1,3)), which is also present. Step 3: If every formed requirement is found in the list, the relation is transitive.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is not in the list. Step 3: The presence of ((3,3)) does not repair this missing pair.

Step 1: The union of two transitive relations can create a new chain. Step 2: For example, (R={(1,2)}) and (S={(2,3)}) are both transitive, but their union lacks ((1,3)). Step 3: Always check transitivity of a union separately.

Step 1: If ((a,b)) and ((b,c)) are both in (R\cap S), then they are in both (R) and (S). Step 2: Since both are transitive, ((a,c)) is in both, hence in the intersection. Step 3: The intersection of transitive relations remains transitive.

Step 1: ((1,2)) and ((2,4)) also form a chain. Step 2: They require ((1,4)), which is present in the relation. Step 3: One starting pair can form several chains, so check all of them.