Transitive relation

Step 1: ((1,2)) and ((2,4)) require ((1,4)). Step 2: ((1,4)) is present, and ((3,3)) creates no missing new pair. Step 3: Do not invent conditions from pairs that do not form a chain.

Step 1: Connecting ((2,1)) and ((1,2)) requires ((2,2)). Step 2: This pair is not in (R), so transitivity fails. Step 3: For reverse chains, always check the forced self-pairs.

Step 1: Transitivity requires ((a,c)) whenever ((a,b)) and ((b,c)) are present. Step 2: Here ((a,c)) is absent, so the condition fails. Step 3: One broken chain is enough to prove a relation is not transitive.

Step 1: (a\equiv b \pmod{2}) means (a) and (b) have the same parity. Step 2: If (b) and (c) also have the same parity, then (a) and (c) have the same parity. Step 3: Congruence-type relations are usually transitive.

Step 1: If (a-b) is even, then (a) and (b) have the same parity. Step 2: If (b-c) is also even, then (a) and (c) have the same parity, so (a-c) is even. Step 3: Use parity to decide quickly in such questions.

Step 1: If (a\ge b) and (b\ge c), then by order (a\ge c). Step 2: So (aRc) is true and the relation is transitive. Step 3: Both (\ge) and (\le) are transitive order relations.

Step 1: Take the example (1\ne 2) and (2\ne 1). Step 2: Transitivity would require (1\ne 1), which is false. So the relation is not transitive. Step 3: The not-equal relation looks simple but is not transitive.

Step 1: If the first student has the same age as the second, and the second has the same age as the third, then the first and third have the same age. Step 2: So the relation is transitive. Step 3: For same-property relations, connect equality through the middle object.

Step 1: If (a) is brother of (b), and (b) is brother of (c), it does not always force (a) to be brother of (c). Step 2: Family relations depend on roles and gender, so transitivity is not guaranteed. Step 3: For real-life relations, test with a counterexample.

Step 1: If (l\parallel m) and (m\parallel n) in the same plane, then (l\parallel n). Step 2: So the parallel relation is transitive. Step 3: For geometry relations, a small diagram helps you see the chain.

Step 1: If (l\perp m) and (m\perp n), then (l) and (n) may be parallel. Step 2: So (l\perp n) is not necessary, and transitivity fails. Step 3: Do not confuse perpendicularity with parallelism.

Step 1: First connect ((p,q)) and ((q,r)). Step 2: By transitivity, ((p,r)) must be in (R). Further pairs like ((q,s)) and ((p,s)) may also follow, but ((p,r)) is listed. Step 3: In a long chain, begin with two adjacent pairs.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: The pair ((2,2)) only re-requires ((1,2)) and ((2,3)), which are already present. ((3,1)) is not forced. Step 3: Transitivity does not automatically create reverse direction.

Step 1: To prove non-transitivity, one counterexample is enough. Step 2: If ((a,b)) and ((b,c)) are present but ((a,c)) is absent, transitivity fails. Step 3: In proof questions, one valid broken chain is sufficient.

Step 1: The transitivity condition needs two connected pairs. Step 2: With only ((2,3)), there is no pair of the form ((3,c)), so no condition is violated. Step 3: For one-pair relations, first check whether any chain exists at all.

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.