Transitive relation

Step 1: ((3,1)) and ((1,2)) require ((3,2)), which is present. Step 2: Requirements formed with ((1,1)) and ((2,2)) are already in the relation. Step 3: In a downward chain, check the final direct pair.

Step 1: ((3,1)) and ((1,2)) require ((3,2)). Step 2: This pair is not in the relation, so it must be added. Step 3: When an outside starting point enters a chain, find the direct pair.

Step 1: In transitivity, not every missing pair matters. Step 2: A missing pair matters only if it is required by a chain like ((a,c)) and ((c,b)). Step 3: Do not conclude immediately from a missing pair; prove that it is required.

Step 1: ((2,3)) and ((3,5)) require ((2,5)). Step 2: ((3,5)) and ((5,2)) require ((3,2)), and ((5,2)) with ((2,3)) requires ((5,3)). Step 3: In a cycle, first-layer pairs may later create self-pairs too.

Step 1: In transitivity, after ((a,b)), we need a pair starting with (b). Step 2: Grouping by the second component helps find all chains quickly. Step 3: For long lists, this method reduces mistakes.

Step 1: For transitivity, if ((a,b)\in R) and ((b,c)\in R), then ((a,c)\in R) must hold. Step 2: Here (a\le b) and (b\le c) directly imply (a\le c). Step 3: In exams, check inequality relations by forming a chain.

Step 1: Check only pairs where the second element of one pair becomes the first element of another. Step 2: ((1,2)) and ((2,3)) require ((1,3)), which is present. ((1,2)) and ((2,2)) require ((1,2)), also present. Step 3: Do not treat missing irrelevant pairs as failure of transitivity.

Step 1: If (a\mid b) and (b\mid c), write (b=ak) and (c=bl). Step 2: Then (c=a(kl)), so (a\mid c). Step 3: For divisibility relations, use a multiplication chain to test transitivity.

Step 1: Combining ((2,1)) and ((1,2)) requires ((2,2)). Step 2: Combining ((1,2)) and ((2,1)) requires ((1,1)), which is already present. Step 3: Add only the pairs forced by actual chains.

Step 1: Suppose (a-b) and (b-c) are rational. Step 2: Then (a-c=(a-b)+(b-c)), and the sum of two rational numbers is rational. Step 3: In such questions, add the two differences to form the required difference.

The condition aRb means that a and b have the same parity: both are even or both are odd. If aRb and bRc, then a and b have the same parity, and b and c also have the same parity. Consequently, a and c have the same parity, so a+c is even and aRc. The argument works for all integers, including negative values, so A is correct.

Step 1: One counterexample is enough to disprove transitivity. Step 2: ((1,2)\in R) and ((2,3)\in R), but ((1,3)\notin R) because (|1-3|=2). Step 3: In distance-based relations, two small steps may become a larger step.

Step 1: Transitivity is checked only when both ((a,b)) and ((b,c)) are in the relation. Step 2: In the empty relation, no such pair exists, so no violation occurs. Step 3: The empty relation is commonly treated as transitive by vacuous truth.

Step 1: A universal relation contains every possible ordered pair. Step 2: So if ((a,b)) and ((b,c)) are present, ((a,c)) is also certainly present. Step 3: Treat (A\times A) as a safe example for transitivity.

Step 1: If (aRb), then (b=a+1). If (bRc), then (c=b+1=a+2). Step 2: For (aRc), we need (c=a+1), but here (c=a+2). Step 3: Successor-type relations usually fail transitivity after two steps.

Step 1: A strict inequality forms an ordered chain. Step 2: (a<b) and (b<c) together give (a<c), so ((a,c)\in R). Step 3: The chain method works well for both (<) and (\le).

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. Step 2: ((2,4)) and ((4,4)) require ((2,4)), and ((1,4)) with ((4,4)) requires ((1,4)); both are present. Step 3: Not every missing pair is required for transitivity.

Step 1: In transitive closure, we add only the pairs needed to remove transitivity gaps. Step 2: Since ((1,2)) and ((2,3)) are present, ((1,3)) is required. Step 3: Do not add arbitrary pairs; add only those forced by chains.

Step 1: Suppose (a-b) and (b-c) are both divisible by (3). Step 2: Then (a-c=(a-b)+(b-c)), so (a-c) is also divisible by (3). Step 3: For remainder-based relations, add the differences.

Step 1: Since (a+b=5), both ((1,4)) and ((4,1)) are in the relation. Step 2: Transitivity would require ((1,1)), but (1+1\ne5), so it is absent. Step 3: Symmetry does not guarantee transitivity.

Step 1: If ((a,b)) and ((b,c)) are in (R\cap S), then they are in both (R) and (S). Step 2: Since (R) and (S) are transitive, ((a,c)) is in both, hence in (R\cap S). Step 3: For intersection questions, apply both relation conditions together.

Step 1: In a union, pairs may come from different relations. Step 2: For example, (R={(1,2)}) and (S={(2,3)}) are individually transitive, but (R\cup S) lacks ((1,3)). Step 3: Always test chains separately for union.

Step 1: From (a<b) and (b<c), we get (a<c). Step 2: If (b-a) and (c-b) are both even, then (c-a=(c-b)+(b-a)) is even. Step 3: For relations with two conditions, verify both conditions separately.

Step 1: Possible chains from ((1,2)) connect with ((2,4)) or ((2,2)). Step 2: These require ((1,4)) and ((1,2)), both present. ((2,4)) with ((4,4)) requires ((2,4)), also present. Step 3: In list-based questions, check all connectable chains systematically.

Step 1: Suppose (a^2=b^2) and (b^2=c^2). Step 2: Then (a^2=c^2), so (aRc). Step 3: In equality-based chains, eliminate the middle equal quantity.