Transitive relation

Step 1: (2=2\cdot 1), so (1R2). Also (4=2\cdot 2), so (2R4). Step 2: If the relation were transitive, (1R4) would hold, but (4\ne 2\cdot 1). Step 3: In multiplication-based relations, two steps may change the rule, so check the direct pair.

Step 1: If (b) is a multiple of (a), then (b=ak). If (c) is a multiple of (b), then (c=bl). Step 2: Then (c=(ak)l=a(kl)), so (c) is a multiple of (a). Step 3: For multiple and divisibility relations, writing the multiplication form is safest.

Step 1: ((1,2)) and ((2,3)) form a valid chain. Step 2: Transitivity requires ((1,3)), but it is not in the relation. Step 3: Existing self-pairs are not the problem; the real problem is the missing required direct pair.

Step 1: From ((p,q)) and ((q,r)), we get ((p,r)). Step 2: Now from ((p,r)) and ((r,s)), we get ((p,s)). Step 3: In a long chain, apply transitivity step by step and do not assume reverse pairs.

Step 1: The relation contains all required ordered pairs in the increasing direction. Step 2: ((1,2)) and ((2,3)) require ((1,3)), ((2,3)) and ((3,4)) require ((2,4)), and ((1,3)) and ((3,4)) require ((1,4)); all are present. Step 3: In a larger relation, check the main chains like a table.

Step 1: The comparison is between absolute values, not directly between the numbers. Step 2: If (|a|\le |b|) and (|b|\le |c|), then by the usual order (|a|\le |c|). Step 3: For absolute value relations, the same order rule applies to the compared quantities.

Step 1: (3<2.5+1), so (3R2.5) is true. Also (2.5<2+1), so (2.5R2) is true. Step 2: Transitivity would require (3R2), but (3<2+1) is false. Step 3: When a fixed number is added in an inequality relation, using a counterexample is safer.

Step 1: ((1,2)) is in the relation because (1\le 2) and (2-1=1). Also ((2,3)) is in the relation. Step 2: Transitivity requires ((1,3)), but (3-1=2), so it is not in the relation. Step 3: Relations with a closeness condition are often not transitive.

Step 1: In divisibility, (a\mid b) means (b=ak). Step 2: If (b\mid c), then (c=bl), so (c=a(kl)), which means (a\mid c). Step 3: In proof-based questions, writing the definition in algebraic form is best.

Step 1: The given pairs include both ((1,2)) and ((2,3)). Step 2: Transitivity requires ((1,3)), but it is not given. Step 3: Even with all self-pairs present, a missing chain pair can make the relation non-transitive.

Step 1: (a \equiv b \pmod{5}) means (a-b) is divisible by (5). Step 2: If (b-c) is also divisible by (5), then (a-c=(a-b)+(b-c)) is divisible by (5). Step 3: For congruence relations, remember the sum of differences.

Step 1: (a\equiv b \pmod{2}) 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) have the same parity. Step 3: Thinking in parity classes makes such questions quick.

Step 1: A correct transitivity check must start with two pairs that actually belong to the relation. Step 2: ((1,2)) and ((2,4)) are present, and the required ((1,4)) is also present. Step 3: Do not build an argument from pairs that are not given in the relation.

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

Step 1: In the same plane, if two distinct lines are parallel to the same line, they are parallel to each other. Step 2: Thus (l\parallel m) and (m\parallel n) imply (l\parallel n). Step 3: In geometry-based relations, drawing a figure helps you understand common direction.

Step 1: The relation uses the less-than order only among odd numbers. Step 2: If (a<b) and (b<c), then (a<c), and all three remain odd. Hence ((a,c)) is in the relation. Step 3: Even with an extra condition, apply the basic order rule carefully.

Step 1: (1+2=3) is odd, so ((1,2)) is in the relation. Also (2+3=5) is odd, so ((2,3)) is in the relation. Step 2: For ((1,3)), (1+3=4), which is even, so ((1,3)) is not in the relation. Step 3: In a counterexample, the first two pairs must be true and the required third pair must fail.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), and ((2,3)) and ((3,4)) require ((2,4)). Step 2: ((1,3)) and ((3,4)) also require ((1,4)). All these pairs are present. Step 3: A relation without self-pairs can still be transitive.

Step 1: If (a<b) and (b<c), then (a<c). Step 2: If (b-a) and (c-b) are both even, then (c-a=(c-b)+(b-a)) is also even. Hence ((a,c)) is in the relation. Step 3: In relations with combined conditions, check each condition separately.

Step 1: From ((1,2)) and ((2,1)), ((1,1)) is required and present. From ((2,1)) and ((1,2)), ((2,2)) is required and also present. Step 2: Other required pairs involving self-pairs are already given. Step 3: Making a small table helps avoid missing such pairs.