Transitive relation

Step 1: (|1-2|\le1), so ((1,2)\in R), and (|2-3|\le1), so ((2,3)\in R). Step 2: But (|1-3|=2), so ((1,3)\notin R). Step 3: In bounded-distance relations, two small steps may exceed the limit.

Step 1: The identity relation contains only pairs like ((a,a)). Step 2: If ((a,a)) and ((a,a)) are combined, the required ((a,a)) is already present. Step 3: Remember the identity relation as a basic example of transitivity.

Step 1: ((3,2)\in R) and ((2,3)\in R) can be chained. Step 2: Transitivity requires ((3,3)\in R), but it is not listed. Step 3: Wrong options often claim a pair is missing even though it is actually present.

Step 1: (1R2) because (2=2\cdot1), and (2R4) because (4=2\cdot2). Step 2: Transitivity would require (1R4), but (4\ne2\cdot1). Step 3: For multiplicative next-term relations, always test a two-step chain.

Step 1: If (a) and (b) have the same remainder, and (b) and (c) have the same remainder, then all three have the same remainder. Step 2: So (a) and (c) also have the same remainder, meaning ((a,c)\in R). Step 3: Such relations can be checked quickly by forming remainder classes.

Step 1: ((1,2)) with ((2,3)) requires ((1,3)), and ((2,3)) with ((3,4)) requires ((2,4)). Both are present. Step 2: ((1,3)) with ((3,4)) requires ((1,4)), also present. Step 3: Missing self-pairs matter only when forced by a chain.

Step 1: (2<1.5+1) is true, so (2R1.5). Also (1.5<1+1) is true, so (1.5R1). Step 2: But (2<1+1) is false, so (2R1) does not hold. Step 3: For shifted inequalities, a numerical counterexample is very effective.

Step 1: If ((a,b)\in R^{-1}) and ((b,c)\in R^{-1}), then ((b,a)\in R) and ((c,b)\in R). Step 2: Since (R) is transitive, ((c,a)\in R), so ((a,c)\in R^{-1}). Step 3: In the inverse relation, order reverses, but transitivity is preserved.

Step 1: ((1,3)\in R) because (1\le3) and (3-1=2). Also ((3,5)\in R). Step 2: Transitivity requires ((1,5)\in R), but (5-1=4), which is greater than (2). Step 3: With upper distance limits, two valid steps may combine into an invalid one.

Step 1: The case (a=b) gives (a-b=0), and (0) is divisible by (5), so the relation is a same-remainder type condition. Step 2: If (a-b) and (b-c) are divisible by (5), then (a-c) is also divisible by (5). Step 3: Try to merge equality conditions with difference conditions when possible.

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.