Transitive relation

Step 1: From ((3,4)) and ((4,4)), ((3,4)) is required. Step 2: From ((3,3)) and ((3,4)), ((3,4)) is also required and present. Step 3: Include cases involving self-pairs while checking.

Step 1: The second element of the first pair and the first element of the second pair are both (7). Step 2: By transitivity, ((5,9)) must be present. Step 3: Do not reverse the order in an ordered pair.

Step 1: The main condition of transitivity requires ((a,c)\in R). Step 2: The question states that ((a,c)\notin R), so the condition fails. Step 3: One counterexample is enough to show non-transitivity.

Step 1: The main chain is ((1,2)) and ((2,3)), which requires ((1,3)). Step 2: ((1,3)) is present, and ((3,3)) creates no missing required pair. Step 3: An extra self-pair does not make a relation non-transitive.

Step 1: In an identity relation, each element is related only to itself. Step 2: Any chain formed in it keeps the needed self-pair inside the relation. Step 3: Remember the identity relation as a standard example of a transitive relation.

Step 1: From ((1,2)) and ((2,3)), ((1,3)) is required. Step 2: ((1,3)) is present, and the requirements involving ((2,2)) are also satisfied. Step 3: Checking each possible chain is the safest method.

Step 1: In ((1,3)) and ((3,2)), the middle element is (3). Step 2: Transitivity requires ((1,2)), but it is not in the relation. Step 3: As soon as a chain appears, write the required outer pair.

Step 1: Transitivity is based on a chain of two pairs. Step 2: A chain is formed when the second element of one pair equals the first element of another pair. Step 3: Identifying the correct chain matters more than counting pairs.

Step 1: Transitivity requires a chain of two pairs. Step 2: A single pair ((2,5)) does not form a pair chain of the required type. Step 3: If no violating chain exists, the relation can be considered transitive.

Step 1: From ((1,2)) and ((2,1)), ((1,1)) is required. Step 2: From ((2,1)) and ((1,2)), ((2,2)) is required; both are present. Step 3: With reverse pairs, always check the self-pairs.

Step 1: Combining ((2,1)) and ((1,2)) requires ((2,2)). Step 2: ((2,2)) is not present in the relation, so transitivity fails. Step 3: One self-pair may not be enough when reverse pairs exist.

Step 1: The key chain is from ((1,2)) to ((2,4)). Step 2: This requires ((1,4)), which is present. Step 3: Wrong options often show reverse pairs, so be careful.

Step 1: The middle element (b) connects the two pairs. Step 2: By transitivity, the pair from (a) to (d) must be in (R). Step 3: Apply the same rule to letters as you do to numbers.

Step 1: From ((1,1)) and ((1,2)), ((1,2)) is required and present. Step 2: From ((1,2)) and ((2,3)), ((1,3)) is required and present. Step 3: List all possible chains and check them.

Step 1: Combining ((1,3)) and ((3,4)) requires ((1,4)). Step 2: ((1,4)) is missing, so the relation is not transitive. Step 3: In a longer list, also check chains formed by later pairs.

Step 1: To show non-transitivity, find a chain where the required pair is missing. Step 2: In the first option, ((1,2)) and ((2,3)) are present, but ((1,3)) is not. Step 3: Recognising a counterexample gives quick marks.

Step 1: The transitivity condition becomes relevant only when both ((a,b)) and ((b,c)) are present. Step 2: If no such chain exists, the condition is not violated. Step 3: Do not quickly call a relation non-transitive when no chain exists.

Step 1: From ((2,3)) and ((3,5)), ((2,5)) is required. Step 2: ((2,5)) is given, and ((5,5)) creates no missing pair. Step 3: Check the main chain first, then inspect the remaining pairs.

Step 1: From ((2,5)) and ((5,6)), ((2,6)) is required. Step 2: This pair is not present in the given relation, so it must be added. Step 3: Also consider chains formed by already implied paths.

Step 1: From ((1,2)) and ((2,2)), ((1,2)) is required. Step 2: ((1,2)) is present in the relation, so the condition is satisfied. Step 3: A self-pair can require the same pair again.