Transitive relation

Step 1: ((1,2)) and ((3,4)) do not form a chain because (2\ne 3). Step 2: Since there is no connected pair, the transitivity condition is not violated. Step 3: Do not connect every two pairs; only check matching middle elements.

Step 1: In transitivity, take the first element of the first pair and the second element of the second pair. Step 2: Thus ((x,y)) and ((y,z)) give ((x,z)\in R). Step 3: In symbolic questions, avoid changing the order.

Step 1: ((a,b)) and ((b,c)) require ((a,c)), which is present. Step 2: ((a,c)) with ((c,c)) requires ((a,c)), and ((b,c)) with ((c,c)) requires ((b,c)). These are present. Step 3: The same rule applies to letter-based pairs.

Step 1: ((a,b)) and ((b,c)) require ((a,c)). Step 2: ((a,c)) is not in the relation, so transitivity fails. Step 3: A cyclic-looking relation is not transitive unless all required pairs are present.

Step 1: Transitivity gives ((1,3)) from ((1,2)) and ((2,3)). Step 2: It does not give the reverse pair ((3,1)). Step 3: Keep transitive and symmetric properties separate.

Step 1: In the first option, ((1,2)) and ((2,3)) are present. Step 2: The required pair ((1,3)) is also present, so the condition is satisfied. Step 3: In MCQs, first find the option containing both a chain and its result.

Step 1: In ((2,2)) and ((2,5)), the middle element (2) matches. Step 2: Transitivity requires ((2,5)), which is already present. Step 3: With a self-pair, the required pair may be the same pair again.

Step 1: ((4,6)) and ((6,6)) are connected. Step 2: The first element (4) and last element (6) give ((4,6)), already present. Step 3: Do not assume an unnecessary new pair just because a self-pair appears.

Step 1: (aRb) means (a) is related to (b). Step 2: Together with (bRc), the chain reaches from (a) to (c), so (aRc) is correct. Step 3: Think of symbolic notation as ordered pairs ((a,b)), ((b,c)), and ((a,c)).

Step 1: Testing transitivity begins with two connected pairs. Step 2: So first check whether a chain like ((a,b)) and ((b,c)) exists. Step 3: After finding a valid chain, check whether ((a,c)) is present.

Step 1: A transitive relation must pass from the first pair to the third pair. Step 2: If ((a,b)) and ((b,c)) are present, then ((a,c)) must also be present. Step 3: In exams, match the middle element and check the direct pair.

Step 1: Look at ((1,2)) and ((2,3)). Step 2: They require ((1,3)), which is already present. Step 3: Check only the connected pairs that can trigger transitivity.

A relation R is transitive when, for any a, b, and c, the presence of (a,b) and (b,c) guarantees the presence of (a,c). Here, (1,2) and (2,3) are both in R, so transitivity requires (1,3). That ordered pair is absent. The reverse pair (2,1) and self-pair (1,1) are not required by transitivity, so option A is correct.

Step 1: Check all possible connected pairs. Step 2: ((1,2)) and ((2,2)) require ((1,2)), which is present. Step 3: When all required pairs are already present, the relation is transitive.

Step 1: Combine ((1,2)) and ((2,3)). Step 2: Transitivity requires ((1,3)). Step 3: In such questions, find the compulsory missing pair only.

Step 1: If (a) divides (b), and (b) divides (c), then (a) divides (c). Step 2: So the condition of transitivity is satisfied. Step 3: For divisibility relations, remember the chain of division.

Step 1: If (a \le b) and (b \le c), then (a \le c). Step 2: This is exactly the condition for transitivity. Step 3: For inequality relations, join the inequality chain directly.

Step 1: Observe the order of smaller and larger numbers. Step 2: If (a) is less than (b), and (b) is less than (c), then (a) is less than (c). Step 3: Read inequality relations as a chain.

Step 1: Assume (a-b) is even and (b-c) is even. Step 2: Adding them gives (a-c) even. Step 3: In such questions, adding the differences is a useful exam method.

Step 1: Take (1R2) because (1-2) is odd, and (2R3) because (2-3) is odd. Step 2: But (1-3) is even, so (1R3) is false. Step 3: One small counterexample is enough to disprove transitivity.