Transitive relation

The relation contains every pair needed from the displayed increasing chains. Specifically, (1,2) and (2,3) give (1,3); (2,3) and (3,4) give (2,4); and (1,3) and (3,4) give (1,4). Other composable pairs either repeat these requirements or have no matching middle element. Therefore every transitivity implication is satisfied. The relation need not be reflexive or symmetric to be transitive, so option A is correct.

Step 1: ((1,4)) and ((4,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is not in the relation. Step 3: If one missing pair is forced by many chains, identifying it is very useful.

Step 1: If (a-b) and (b-c) are both divisible by (7), then their sum ((a-b)+(b-c)=a-c) is also divisible by (7). Step 2: Hence (aRc) holds. Step 3: For divisibility relations, adding differences gives the fastest check.

Step 1: (5<4+2), so (5R4) is true, and (4<3+2), so (4R3) is true. Step 2: Transitivity would require (5R3), but (5<3+2) is false. Step 3: For inequalities with a fixed added number, a counterexample is a safe test.

Step 1: ((1,2)) and ((2,3)) are in the relation because each has difference (1). Step 2: Transitivity requires ((1,3)), but (3-1=2), so it is not in the relation. Step 3: A next-number relation is generally not transitive.

Step 1: If (a\mid b), then (b=ak), and if (b\mid c), then (c=bl). Step 2: Then (c=(ak)l=a(kl)), so (a\mid c). Step 3: In divisibility questions, writing the multiplication form reduces mistakes.

Step 1: ((3,4)) and ((4,3)) form a chain. Step 2: Transitivity requires ((3,3)), but it is missing. Step 3: Reverse pairs create the need for self-pairs, so always check them.

Step 1: The comparison is between (a^3), (b^3), and (c^3) using the usual order. Step 2: If (a^3\le b^3) and (b^3\le c^3), then (a^3\le c^3), so (aRc). Step 3: In power-based relations, first identify the quantity being compared.

Step 1: (a\equiv b \pmod{3}) means (a-b) is divisible by (3). Step 2: If (a-b) and (b-c) are both divisible by (3), then (a-c) is also divisible by (3). Step 3: In congruence relations, transitivity follows from adding differences.

Step 1: ((1,2)) and ((2,3)) are present in the relation. Step 2: Transitivity requires ((1,3)), but it is absent. Step 3: Even if all self-pairs are present, a missing required direct pair makes the relation non-transitive.

Step 1: From ((a,b)) and ((b,c)), we get ((a,c)). Step 2: Now from ((a,c)) and ((c,d)), we get ((a,d)). Step 3: In a long chain, apply transitivity in order and do not reverse direction.

Step 1: (a+b) being even means (a) and (b) have the same parity. Step 2: If (b+c) is even, then (b) and (c) have the same parity, so (a) and (c) have the same parity. Hence (a+c) is even. Step 3: For such questions, think using parity.

Step 1: ((1,2)) is in the relation because (1+2=3) is odd. ((2,3)) is also in the relation because (2+3=5) is odd. Step 2: Transitivity requires ((1,3)), but (1+3=4) is even. Step 3: One valid counterexample disproves transitivity.

Step 1: The relation compares absolute values of numbers. Step 2: If (|a|\ge |b|) and (|b|\ge |c|), then by the usual order (|a|\ge |c|), so (aRc). Step 3: Treat absolute value as the compared quantity and apply the order rule.

Step 1: ((1,3)) and ((3,4)) are in the relation. Step 2: Transitivity requires ((1,4)), but it is missing. Step 3: Complete an existing long reach by adding the direct pair.

Step 1: If (l\parallel m) and (m\parallel n), then the lines have the same direction. Step 2: Hence (l\parallel n), so (lRn) holds. Step 3: In geometry relations, drawing a figure helps identify the common direction.

Step 1: If (a) and (b) have the same date of birth, and (b) and (c) have the same date of birth, then (a) and (c) also have the same date of birth. Step 2: Hence (aRc) holds. Step 3: Relations based on equality are directly proved transitive.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: They require ((1,3)), which is missing. Adding it fixes this one failure, though another failure may also need checking. Step 3: When the question asks for one failure, focus on the given chain.

Step 1: (a=b) or (a<b) together mean (a\le b). Step 2: If (a\le b) and (b\le c), then (a\le c), so the relation is transitive. Step 3: First convert a combined statement into a simpler symbol.

Step 1: The relation uses the (<) order only among even numbers. Step 2: If (a<b) and (b<c), then (a<c); if all are even, ((a,c)) also belongs to the relation. Step 3: Even with an extra condition, check the basic order rule separately.

Step 1: (|1-2|=1), so ((1,2)) is in the relation. Also (|2-3|=1), so ((2,3)) is in the relation. Step 2: (|1-3|=2), so ((1,3)) is not in the relation. Step 3: Bounded distance conditions often break transitivity.

Step 1: (a\equiv b \pmod{4}) means (a-b) is divisible by (4). Step 2: Similarly, if (b-c) is divisible by (4), then (a-c=(a-b)+(b-c)) is also divisible by (4). Step 3: Adding differences is the key idea in congruence.

Step 1: From ((1,3)) and ((3,1)), ((1,1)) is required, and it is present. From ((3,1)) and ((1,3)), ((3,3)) is required, and it is also present. Step 2: After ((2,4)), there is no pair starting with (4), so no new requirement arises. Step 3: Every missing self-pair is not a transitivity failure.

Step 1: If (a^2=b^2) and (b^2=c^2), the two equalities give (a^2=c^2). Step 2: Hence (aRc) holds. Step 3: In equality-based relations, connect the same quantity through equality.

Step 1: ((1,3)) is in the relation because (1\le 3) and (3-1=2). ((3,5)) is also in the relation. Step 2: For ((1,5)), (5-1=4), which exceeds the limit, so it is not in the relation. Step 3: Two small jumps can form a larger jump, so check bounded conditions carefully.