Transitive relation

Step 1: For transitivity, if ((a,b) \in R) and ((b,c) \in R), then ((a,c) \in R) must also be present. Step 2: Here, ((1,2)) and ((2,3)) imply ((1,3)), which is present. Other required cases also hold. Step 3: In exams, check only pairs where the second element of one pair matches the first element of another.

Step 1: ((1,2) \in R) and ((2,3) \in R) are present. Step 2: Transitivity requires ((1,3) \in R), but it is missing. Step 3: For such questions, identify a two-step path and add the direct ordered pair.

Step 1: If (a-b) and (b-c) are both divisible by (3), then ((a-b)+(b-c)=a-c) is also divisible by (3). Step 2: Hence (aRc) is true. Step 3: For divisibility relations, add the differences to test transitivity quickly.

Step 1: Transitivity means two connected relations must imply a direct relation between the first and third elements. Step 2: For real numbers, (a<b) and (b<c) definitely imply (a<c). Step 3: In inequality questions, remember the direction on the number line.

Step 1: If (a\le b) and (b\le c), then by the order property (a\le c). Step 2: So (aRc) is also true. Step 3: Both (\le) and (<) are transitive in the usual order.

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: For divides relations, use the multiplication form.

Step 1: Transitivity is checked only when both ((a,b)) and ((b,c)) are present. Step 2: Here, ((1,2)) and ((2,4)) require ((1,4)), which is present. No other necessary chain appears. Step 3: A transitive relation need not contain all self-pairs.

Step 1: From ((1,2)) and ((2,1)), transitivity requires ((1,1)). Step 2: Similarly, ((2,1)) and ((1,2)) require ((2,2)). Both are missing, so the relation is not transitive. Step 3: Reverse pairs can create the need for self-pairs, so do not ignore them.

Step 1: To break transitivity, two pairs like ((a,b)) and ((b,c)) must exist while ((a,c)) is absent. Step 2: In the empty relation, no ordered pair exists, so no counterexample can occur. Step 3: Remember that the empty relation is considered transitive.

Step 1: If ((a,b) \in A\times A) and ((b,c) \in A\times A), then (a,c \in A). Step 2: Hence ((a,c)) is also in (A\times A). Step 3: In a universal relation, no required ordered pair is missing.

Step 1: If (a<b) and (b<c), then by order property (a<c). Step 2: Therefore ((a,c)) also belongs to (R). Step 3: When the increasing direction continues, the relation is transitive.

Step 1: Since (|1-2|=1), ((1,2) \in R). Since (|2-3|=1), ((2,3) \in R). Step 2: Transitivity would require ((1,3)), but (|1-3|=2), so it is not in (R). Step 3: Being consecutive does not guarantee transitivity.

Step 1: (1\mid 2) and (2\mid 4) are both true. Step 2: Then (1\mid 4) is also true, so ((1,4)) belongs to the relation. Step 3: In a check, both starting pairs must actually be true; guessing is not enough.

Step 1: From ((1,2)) and ((2,1)), ((1,1)) is required, and it is present. Step 2: From ((2,1)) and ((1,2)), ((2,2)) is required, and it is also present. Self-pairs create no missing requirement. Step 3: For small relations, check each possible chain systematically.

Step 1: From ((1,3)) and ((3,4)), ((1,4)) is required. Step 2: From ((1,2)) and ((2,4)), ((1,4)) is also required, but it is missing. Step 3: In a chain, finding the missing direct pair is the key to transitivity.

Step 1: (a+b) being even means (a) and (b) have the same parity. Step 2: (b+c) being even means (b) and (c) have the same parity, so (a) and (c) also have the same parity. Hence (a+c) is even. Step 3: For parity relations, think in terms of same or different parity.

Step 1: (1+2=3) is odd, so (1R2). Also, (2+3=5) is odd, so (2R3). Step 2: But (1+3=4) is even, so (1R3) does not hold. Step 3: One valid counterexample is enough to disprove transitivity.

Step 1: Both ((1,2)) and ((2,3)) belong to the relation. Step 2: Transitivity requires ((1,3)), but it is absent. Step 3: Having self-pairs alone does not guarantee transitivity.

Step 1: If (a\ge b) and (b\ge c), then by order property (a\ge c). Step 2: Therefore ((a,c)) also belongs to the relation. Step 3: The usual order (\ge) is transitive just like (\le).

Step 1: ((1,2)) and ((2,3)) form a chain. Step 2: Transitivity requires ((1,3)), which is not in the relation. Step 3: Missing self-pairs do not always break transitivity; the key issue is the missing required direct pair.

Step 1: Transitivity gives ((a,c)) from ((a,b)) and ((b,c)). Step 2: Here (a=5), (b=7), and (c=9), so ((5,9) \in R). Step 3: Do not reverse the order; transitivity is not symmetry.

Step 1: ((3,1)) and ((1,2)) are in the relation. Step 2: Transitivity requires ((3,2)), but it is missing. Step 3: In cyclic-looking pairs, carefully check the direct pair formed by every two consecutive steps.

Step 1: If (a^2=b^2) and (b^2=c^2), then (a^2=c^2). Step 2: Hence (aRc) is true. Step 3: In equality-based relations, equality of the same quantity passes forward.

Step 1: If (P) and (Q) have the same (x)-coordinate, and (Q) and (S) have the same (x)-coordinate, then (P) and (S) also have the same (x)-coordinate. Step 2: Hence (P R S) is true. Step 3: Relations based on the same property are often transitive, but always justify it.

Step 1: ((1,2)) is in the relation because (1+1=2). ((2,3)) is also in the relation because (2+1=3). Step 2: Transitivity requires ((1,3)), but (1+1=3) is false. Step 3: A next-number relation is generally not transitive.