Transitive relation

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.

Step 1: (a\equiv b \pmod{4}) means (a-b) is divisible by (4). Step 2: If (a-b) and (b-c) are both divisible by (4), then (a-c) is also divisible by (4). Step 3: In congruence questions, remember the sum of differences.

Step 1: ((2,5)) is in the relation because (2+5=7), and ((5,2)) is also in the relation because (5+2=7). Step 2: Transitivity requires ((2,2)), but (2+2=4), so it is not in the relation. Step 3: In constant-sum relations, reverse pairs often give a quick counterexample.

Step 1: From ((x,y)) and ((y,z)), we get ((x,z)). Step 2: Then ((x,z)) and ((z,u)) give ((x,u)), and ((x,u)) with ((u,v)) gives ((x,v)). Step 3: In a long chain, apply transitivity step by step.

Step 1: (|1-3|=2), so ((1,3)) is in the relation. Also (|3-5|=2), so ((3,5)) is in the relation. Step 2: Transitivity requires ((1,5)), but (|1-5|=4). Step 3: A fixed-distance relation need not be transitive.

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: For relaxed absolute value inequalities, a counterexample is a good test.

Step 1: Same remainder means the numbers lie in the same remainder 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, thinking in classes is simple.

Step 1: ((1,4)) and ((4,5)) are in the relation. Step 2: Transitivity requires ((1,5)), but it is missing. ((2,4)) and ((4,5)) also suggest checking ((2,5)), but among the options the key required pair is ((1,5)). Step 3: When a last link is added to a chain, new direct pairs are formed.

Step 1: For real numbers, (a^2=b^2) and (a^3=b^3) together imply (a=b). Step 2: Equality is transitive because (a=b) and (b=c) imply (a=c). Step 3: When multiple equality conditions appear, see which simpler relation they form.

Step 1: ((1,3)) is in the relation because (1\ne 3) and (1+3) is even. ((3,1)) is also in the relation. Step 2: Transitivity requires ((1,1)), but (1\ne 1) is false. Step 3: Adding an inequality condition to a same-class relation can break transitivity.

Step 1: If (a\mid b) and (b\mid c), then (a\mid c). Step 2: Also, in this forward chain of positive divisibility with (a\ne b) and (b\ne c), we still have (a\ne c). Hence ((a,c)) belongs to the relation. Step 3: In divisibility, check both direction and the inequality condition.

Step 1: If (P) and (Q) have the same (y)-coordinate, and (Q) and (S) have the same (y)-coordinate, then (P) and (S) also have the same (y)-coordinate. Step 2: Hence (P R S) is true. Step 3: In a relation based on the same property, check that the property passes through the chain.

Step 1: (9R5) is true because (9-5=4<5), and (5R1) is true because (5-1=4<5). Step 2: Transitivity would require (9R1), but (9-1=8), which is not less than (5). Step 3: In bounded inequalities, two small differences can combine into a larger one.

Step 1: ((1,2)) and ((2,1)) are in the relation. Step 2: Transitivity requires ((1,1)), but it is missing. Step 3: In cyclic pairs, self-pairs often become necessary.

Step 1: The relation takes the less-than order only among odd numbers. Step 2: If (a<b) and (b<c), then (a<c), and since all three are odd, ((a,c)) is also in the relation. Step 3: Even with an extra condition, do not forget the basic order rule.

Step 1: ((1,4)) is in the relation because (1\le 4) and (4-1=3). ((4,5)) is also in the relation. Step 2: For ((1,5)), (5-1=4), which is greater than the bound (3). So ((1,5)) is not in the relation. Step 3: In bounded relations, two valid steps can form an invalid longer step.