Transitive relation

Step 1: In the usual order, (a\le b) and (b\le c) imply (a\le c). Step 2: Hence ((a,c)) also belongs to the relation. Step 3: In order relations, keep the same direction and do not assume reverse order.

Step 1: If (a>b) and (b>c), then by the usual order (a>c). Step 2: Therefore ((a,c)) belongs to the relation. Step 3: Both (>) and (<) are transitive when the direction is kept consistent.

Step 1: (3-2=1) is odd, and (2-1=1) is also odd. Step 2: Transitivity requires (3R1), but (3-1=2) is even, so (3R1) does not hold. Step 3: When choosing a counterexample, both starting relations must be true.

Step 1: Same remainder means both numbers lie in the same parity class. 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: For remainder relations, think in classes.

Step 1: From ((1,2)) and ((2,3)), ((1,3)) is required, and it is present. Step 2: From ((2,2)) and ((2,3)), ((2,3)) itself is required and present; ((3,3)) creates no missing pair. Step 3: Also check chains involving self-pairs carefully.

Step 1: (a<b) or (a=b) means (a\le b). Step 2: If (a\le b) and (b\le c), then (a\le c), so ((a,c)) is in the relation. Step 3: Converting the statement into a simpler form helps in relation questions.

Step 1: (1+(-1)=0), so (1R(-1)), and ((-1)+1=0), so ((-1)R1). Step 2: Transitivity requires (1R1), but (1+1=2), so it is not in the relation. Step 3: When reverse or opposite pairs appear, check the required self-pair.

Step 1: ((1,3)) and ((3,4)) are in the relation. Step 2: Therefore ((1,4)) is required, but it is not in the relation. Step 3: Even when self-pairs are present, a later chain may still have a missing pair.

Step 1: (4=2^2), so (2R4). Also (16=4^2), so (4R16). Step 2: Transitivity would require (2R16), but (16\ne 2^2). Step 3: Applying a rule twice does not necessarily mean the same rule applies directly.

Step 1: The relation means (a) and (b) have the same parity. Step 2: If (a) has the same parity as (b), and (b) has the same parity as (c), then (a) and (c) also have the same parity. Step 3: In classification-based relations, use the idea of the same class.

Step 1: (\gcd(2,3)=1), and (\gcd(3,4)=1). Step 2: But (\gcd(2,4)=2), so ((2,4)) is not in the relation. Step 3: Coprimality may be symmetric, but it is not necessarily transitive.

Step 1: (3R6) is true because (6) is divisible by (3), and (6R12) is true because (12) is divisible by (6). Step 2: Transitivity requires (3R12), and (12) is divisible by (3). Step 3: Choose a chain whose first two relations are both true.

Step 1: (a-b>0) means (a>b). Step 2: If (a>b) and (b>c), then (a>c), so (a-c>0). Step 3: First convert the condition into a simple inequality.

Step 1: If (a<b) and (b<c), then (a<c). Step 2: If (b-a) and (c-b) are both divisible by (3), then (c-a) is also divisible by (3). Step 3: In a two-condition relation, check order and divisibility separately.

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. Step 2: ((2,4)) and ((4,4)) require ((2,4)), which is present; ((1,4)) and ((4,4)) require ((1,4)), also present. Step 3: Do not skip chains involving self-pairs.

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.