Transitive relation

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.

Direct answer: B. The union of two transitive relations need not be transitive. Take R={(1,2)} and S={(2,3)}. Each relation is transitive because neither contains a pair of consecutive links within itself. But R∪S contains (1,2) and (2,3) without containing (1,3), so the union is not transitive. Option A is always true: if two pairs are in R∩S, they are in both R and S, so both transitivity conditions produce (a,c), which remains in the intersection. Option C is true because when R=S, R∩S=R, already transitive. Option D is true vacuously: the empty relation has no two pairs that could violate transitivity. Common confusion: intersections preserve transitivity, but unions can combine a first link from one relation with a second link from the other.