Transitive relation

Step 1: If (a^2\le b^2) and (b^2\le c^2), then by the usual order (a^2\le c^2). Step 2: Hence (aRc) holds. Step 3: When a relation compares a derived value, apply order to that value.

Step 1: (2+4=6), so ((2,4)) is in the relation, and (4+2=6), so ((4,2)) is also in the relation. Step 2: Transitivity requires ((2,2)), but (2+2=4), so it is not in the relation. Step 3: In a counterexample, the middle element must match.

Step 1: If (a\le b) and (b\le c), then (a\le c). Step 2: 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 3: In combined conditions, each condition must pass forward separately.

Step 1: Both ((3,1)) and ((1,2)) are in the relation. Step 2: Transitivity requires ((3,2)), but it is missing. Step 3: Choose the failure where the first two pairs are true and the required third pair is absent.

Step 1: (4\le 3+1), so (4R3) is true. Also (3\le 2+1), so (3R2) is true. Step 2: But (4\le 2+1) is false, so (4R2) does not hold. Step 3: In relaxed inequalities, two small relaxations may break the direct rule.

Step 1: If ((a,b)) is in the relation, then both (a) and (b) are less than (5). If ((b,c)) is also in the relation, then both (b) and (c) are less than (5). Step 2: Hence (a) and (c) are both less than (5), so ((a,c)) is in the relation. Step 3: For property-based relations, check whether the first and third elements keep the property.

Step 1: For real numbers, (a^2+b^2=0) occurs only when (a=0) and (b=0). Step 2: The relation contains only a pair like ((0,0)), which does not break transitivity. Step 3: When the relation list is very small, write possible pairs directly.

Step 1: For positive numbers, (a\mid b) and (b\mid a) imply (a=b). Step 2: So this relation behaves like equality, and equality is transitive. Step 3: Understand two-way divisibility through equality.

Step 1: Since (1\ne 2), ((1,2)) is in the relation, and since (2\ne 1), ((2,1)) is also in the relation. Step 2: Transitivity requires ((1,1)), but (1\ne 1) is false. Step 3: The not-equal relation is not transitive; remember it through a counterexample.

Step 1: If ((a,b)) and ((b,c)) are in (R\cap S), then they are in both (R) and (S). Step 2: Since both (R) and (S) are transitive, ((a,c)) belongs to both, hence to (R\cap S). Step 3: The intersection of two transitive relations is transitive, but the union is not always transitive.

Step 1: In transitivity, check only pairs where the second element of the first pair matches the first element of the next pair. Step 2: Here ((1,2)) and ((2,5)) require ((1,5)), which is present. After ((3,4)), no pair starts with (4). Step 3: Do not treat every missing self-pair as an error; check only required chains.

Step 1: In this relation, ((1,3)) and ((3,5)) form the main chain. Step 2: This chain requires ((1,5)), which is already present. ((2,4)) creates no further requirement. Step 3: In transitivity, find required pairs only from actual chains.

Step 1: ((1,4)) and ((4,5)) are in the relation, so ((1,5)) is required. Step 2: ((1,2)) and ((2,5)) also require ((1,5)), but it is missing. Step 3: When the same pair is demanded by two chains, it is important for transitive closure.

Step 1: If (a-b) and (b-c) are both divisible by (9), their sum (a-c) is also divisible by (9). Step 2: Hence (aRc) is also true. Step 3: For divisibility-based relations, adding differences gives a simple proof of transitivity.

Step 1: From (a+2\le b+2), subtracting (2) from both sides gives (a\le b). Step 2: Since (a\le b) and (b\le c) imply (a\le c), the relation is transitive. Step 3: Adding or subtracting the same number does not change the order.

Step 1: (1R3) is true because (3=3\cdot 1), and (3R9) is true because (9=3\cdot 3). Step 2: Transitivity would require (1R9), but (9\ne 3\cdot 1). Step 3: Applying a rule twice does not necessarily make the direct rule true.

Step 1: If (a\mid b) and (b\mid c), then (b=ak) and (c=bl). Step 2: Then (c=a(kl)), so (a\mid c). Step 3: Whether the relation list is small or large, this divisibility rule proves transitivity.

Step 1: ((2,1)) and ((1,2)) are in the relation. Step 2: Transitivity requires ((2,2)), but it is missing. Step 3: Never skip self-pairs forced by reverse ordered pairs.

Step 1: If (a\le b) and (b\le c), then (a\le c). Step 2: If (b-a) and (c-b) are both divisible by (2), then (c-a=(c-b)+(b-a)) is also divisible by (2). Step 3: In combined conditions, check order and divisibility separately.

Step 1: The relation compares (a^4), (b^4), and (c^4). Step 2: If (a^4\le b^4) and (b^4\le c^4), then by the usual order (a^4\le c^4). Step 3: Whatever the power value is, apply the order rule to the compared values.