Transitive relation

Step 1: If (\gcd(a,b)=a), then (a) divides (b). Step 2: Divisibility is transitive because (a\mid b) and (b\mid c) imply (a\mid c). Step 3: Convert greatest common divisor conditions into divisibility before testing.

Step 1: ((1,3)) and ((3,4)) require ((1,4)). Step 2: ((1,2)) and ((2,4)) also require ((1,4)), and it is missing. Step 3: If the same missing pair is forced by multiple chains, it is definitely needed.

Step 1: Check that (5\le3+2) and (3\le1+2) are both true. Step 2: But (5\le1+2) is false, so ((5,1)) is not in the relation. Step 3: For modified inequalities, do not decide only by seeing (\le); look for a counterexample.

Step 1: If (a\ge b) and (b\ge c), an ordered chain is formed. Step 2: This chain gives (a\ge c), so ((a,c)\in R). Step 3: The same chain rule applies to decreasing inequalities.

Step 1: If (a) and (b) are in the same class, and (b) and (c) are also in the same class, then (a) and (c) are in that same class. Step 2: So ((a,c)) belongs to the relation, and transitivity holds. Step 3: Same-group relations often form equivalence relations.

Step 1: (1+2) is odd, so ((1,2)\in R). Also (2+3) is odd, so ((2,3)\in R). Step 2: Transitivity would require ((1,3)\in R), but (1+3) is even. Step 3: In parity-based relations, one clear counterexample can decide the result.

Step 1: If (a) and (b) have the same remainder, and (b) and (c) have the same remainder, then (a) and (c) also have the same remainder. Step 2: Hence ((a,c)\in R). Step 3: Same-remainder relations are usually transitive.

Step 1: ((1,2)) and ((2,1)) require ((1,1)), which is present. ((2,1)) and ((1,2)) require ((2,2)), also present. Step 2: After ((3,4)), no pair starts with (4), so no new requirement arises. Step 3: In listed relations, check only connectable chains.

Step 1: Combining ((2,3)) and ((3,1)) requires ((2,1)). Step 2: This pair is missing, so transitivity fails. Step 3: Identifying the first forced missing pair is very useful in transitive closure questions.

Step 1: Suppose (a-b) and (b-c) are integers. Step 2: Then (a-c=(a-b)+(b-c)), and the sum of two integers is an integer. Step 3: For difference-based relations, adding the two differences is the simplest method.

Step 1: (b/a) being an integer means (a) divides (b). Step 2: If (a\mid b) and (b\mid c), then (a\mid c), so the relation is transitive. Step 3: Convert quotient conditions into divisibility first.

Step 1: (1+3=4), so (1R3), and (3+1=4), so (3R1). Step 2: Transitivity would require (1R1), but (1+1=2) is not divisible by (4). Step 3: Do not assume sum-based divisibility is always transitive.

Step 1: ((1,2)\in R) and ((2,3)\in R), because the order is correct and each sum is odd. Step 2: But ((1,3)\notin R), because (1+3) is even. Step 3: For two-condition relations, verify both conditions on the final pair too.

Step 1: (R\subseteq S) only tells us that all pairs of (R) are in (S). Step 2: New pairs in (S) may create a chain whose required final pair is missing. Step 3: Subset information alone does not guarantee transitivity.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: Self-pairs such as ((1,1)), ((2,2)), and ((3,3)) do not create a missing requirement because the needed pairs are already listed. Step 3: Check ordered relations like small paths.

Step 1: Combining ((2,4)) and ((4,2)) requires ((2,2)). Step 2: This pair is not in the relation, so transitivity is not complete. Step 3: When reverse pairs appear, check the self-pairs forced by them.

Step 1: The condition (a=b) or (a<b) is the same as (a\le b). Step 2: If (a\le b) and (b\le c), then (a\le c), so transitivity holds. Step 3: Sometimes recognizing the simplified condition is the key step.

Step 1: If (a-b) and (b-c) are even, then (a-c=(a-b)+(b-c)) is even. Step 2: Also, (a\le b) and (b\le c) imply (a\le c). Step 3: For two conditions, build a separate chain for each condition.

Step 1: ((1,3)) and ((3,4)) require ((1,4)), which is present. Step 2: Chains involving ((3,3)) require pairs already present, and ((4,4)) also creates no gap. Step 3: Do not panic over self-pairs; they often only repeat existing requirements.

Step 1: If (a^3\le b^3) and (b^3\le c^3), the inequality chain gives (a^3\le c^3). Step 2: This is exactly the condition for (aRc), so the relation is transitive. Step 3: For power-based conditions, first check the inequality chain on that power.

Step 1: (|1-2|\le1), so ((1,2)\in R), and (|2-3|\le1), so ((2,3)\in R). Step 2: But (|1-3|=2), so ((1,3)\notin R). Step 3: In bounded-distance relations, two small steps may exceed the limit.

Step 1: The identity relation contains only pairs like ((a,a)). Step 2: If ((a,a)) and ((a,a)) are combined, the required ((a,a)) is already present. Step 3: Remember the identity relation as a basic example of transitivity.

Step 1: ((3,2)\in R) and ((2,3)\in R) can be chained. Step 2: Transitivity requires ((3,3)\in R), but it is not listed. Step 3: Wrong options often claim a pair is missing even though it is actually present.

Step 1: (1R2) because (2=2\cdot1), and (2R4) because (4=2\cdot2). Step 2: Transitivity would require (1R4), but (4\ne2\cdot1). Step 3: For multiplicative next-term relations, always test a two-step chain.

Step 1: If (a) and (b) have the same remainder, and (b) and (c) have the same remainder, then all three have the same remainder. Step 2: So (a) and (c) also have the same remainder, meaning ((a,c)\in R). Step 3: Such relations can be checked quickly by forming remainder classes.