Transitive relation

Check every composable pair. From (1,2) and (2,2), the required pair is (1,2), already present. From (2,2) and (2,3), the required pair is (2,3), also present. From (1,2) and (2,3), the required pair is (1,3), present as well. There are no pairs beginning with 3, so no further condition arises. Therefore R is transitive and option A is correct.

Step 1: In a transitive relation, ((a,b)) and ((b,c)) imply ((a,c)). Step 2: Here (a=4), (b=6), (c=9), so ((4,9)\in R) is certain. Step 3: Reverse pairs do not follow from transitivity alone.

Step 1: From ((1,2)) and ((2,1)), ((1,1)) is required and present. Step 2: From ((2,1)) and ((1,2)), ((2,2)) is required and present. Other reflexive pairs satisfy their own chains. Hence (R) is transitive. Step 3: When reverse pairs exist, check whether the needed reflexive pairs are present.

Step 1: The second element of ((2,3)) matches the first element of ((3,4)). Step 2: By transitivity, combine the first and last elements to get ((2,4)). Step 3: Keep the order of ordered pairs unchanged.

Step 1: The definition of transitivity is exactly in this form. Step 2: From ((x,y)) and ((y,z)), ((x,z)) must belong to (R). Reverse pairs do not follow. Step 3: In symbolic questions, read (x,y,z) just like (a,b,c).

Step 1: The main chains are ((1,2),(2,3)), ((2,3),(3,4)), and ((1,3),(3,4)). Step 2: They require ((1,3)), ((2,4)), and ((1,4)), all of which are present. Hence (R) is transitive. Step 3: Transitivity does not require every reflexive or reverse pair.

Step 1: Same remainder means the difference of two numbers is divisible by (5). Step 2: If (a-b) and (b-c) are divisible by (5), then (a-c) is also divisible by (5). Hence the relation is transitive. Step 3: For congruence relations, think in terms of differences.

Step 1: ((1,2)) and ((2,3)) are in (R). Step 2: They require ((1,3)), which is missing. Reflexive pairs do not create a problem here. So ((1,3)) must be added. Step 3: First check pairs that create a new reach from start to end.

Step 1: Transitivity is identified using two connected ordered pairs. Step 2: If ((a,b)) and ((b,c)) are present, ((a,c)) must also be present. This is the correct definition. Step 3: Keep it separate from symmetric and reflexive definitions.

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. Step 2: Chains involving ((2,2)) give existing pairs again, and ((4,4)) does not create a missing pair. Hence (R) is transitive. Step 3: Check all connected pairs systematically.

Step 1: If (a) is the father of (b), and (b) is the father of (c), then (a) may be the grandfather of (c), not the father. Step 2: So ((a,c)) need not belong to the same relation. Hence it is not transitive. Step 3: In family relation questions, read the exact relation name carefully.

Step 1: If (a>b) and (b>c), then (a>c). Step 2: Therefore ((a,b)) and ((b,c)) imply ((a,c)). This satisfies transitivity. Step 3: Usual order relations are generally transitive.

Step 1: ((2,3)) and ((3,1)) are in (R). Step 2: Transitivity requires ((2,1)), but it is missing. Hence (R) is not transitive. Step 3: One broken chain is enough to prove non-transitivity.

Step 1: Adding ((1,1)) makes chains such as ((1,1)) with ((1,2)) requiring ((1,2)), and ((1,1)) with ((1,3)) requiring ((1,3)). Step 2: These pairs are already present, so no new missing requirement appears. Step 3: When adding a pair, check all chains it creates.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: Chains with ((1,1)) and ((2,2)) require already existing pairs. Therefore (R) is transitive. Step 3: Do not be confused by missing reflexive pairs that are not required by a chain.

Step 1: If (a\ge b) and (b\ge c), then (a\ge c). Step 2: Thus the transitive condition is satisfied, so the relation is transitive. Step 3: Understand (\ge) and (\le) as order relations.

Step 1: The main chain is ((1,3)) and ((3,4)). Step 2: It requires ((1,4)), which is present. ((2,2)) only requires itself with itself. Hence (R) is transitive. Step 3: For each pair, look for another pair that starts from its second element.

Step 1: To prove non-transitivity, we must show the transitive rule fails. Step 2: That means two connected pairs are in (R), but the required third pair is not in (R). This is the correct situation. Step 3: Always write a counterexample using clear ordered pairs.

Step 1: ((1,2)) and ((2,1)) require ((1,1)), which is present. Step 2: ((2,1)) and ((1,2)) require ((2,2)), which is present. Reflexive chains also create no missing pair. So (R) is transitive. Step 3: The absence of pairs involving the third element is not automatically a problem.

Step 1: If (a) is an ancestor of (b), and (b) is an ancestor of (c), then (a) is also an ancestor of (c). Step 2: Therefore ((a,b)) and ((b,c)) imply ((a,c)). Hence the relation is transitive. Step 3: An ancestor relation continues through generations.

Step 1: The less-than relation moves forward in order. Step 2: Since (1<3) and (3<5), we get (1<5). Hence ((1,5)\in R). Step 3: Do not reverse the direction in inequality relations.

Step 1: ((2,3)) and ((3,2)) are both in (R). Step 2: They require ((2,2)), which is not in (R). Hence the relation is not transitive. Step 3: Opposite connected pairs often demand reflexive pairs.

Step 1: The transitive condition becomes active only when pairs like ((a,b)) and ((b,c)) exist. Step 2: If no such connected pairs exist, there is no chance for the condition to fail. So the relation is transitive. Step 3: First search for connected pairs when testing transitivity.

Step 1: After ((1,2)), there is no pair starting with (2). After ((3,4)), there is no pair starting with (4). Step 2: So no transitive requirement is created. Hence (R) is transitive. Step 3: Do not assume every small relation is non-transitive.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((2,2)) with ((2,3)) requires ((2,3)), and ((2,3)) with ((3,3)) again requires ((2,3)), which is present. Hence (R) is transitive. Step 3: Check only the pairs that are actually required.