Transitive relation

Step 1: (\gcd(2,3)=1), and (\gcd(3,4)=1). Step 2: But (\gcd(2,4)=2), so ((2,4)) is not in the relation. Step 3: Being coprime does not automatically pass forward.

Step 1: ((1,3)) and ((3,5)) are in the relation. Step 2: Transitivity requires ((1,5)), but it is absent. Step 3: Even if all reflexive pairs are present, missing chain pairs must be checked separately.

Step 1: (a\equiv b \pmod{6}) means (a-b) is divisible by (6). Step 2: If (b-c) is also divisible by (6), then (a-c=(a-b)+(b-c)) is divisible by (6). Step 3: For congruence relations, prove transitivity by adding differences.

Step 1: ((1,2)) is in the relation because (1<2) and (2-1) is odd. ((2,3)) is also in the relation. Step 2: Transitivity requires ((1,3)), but (3-1=2) is even. Step 3: Two odd-difference links can combine into an even difference.

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) is also even. Hence ((a,c)) is in the relation. Step 3: An increasing chain with even differences remains transitive.

Step 1: (3R2) is true because (3=2+1), and (2R1) is true because (2=1+1). Step 2: Transitivity requires (3R1), but neither (3=1) nor (3=1+1) is true. Step 3: Adding one-step allowance to equality can fail after two steps.

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. ((1,4)) and ((4,2)) require ((1,2)), which is present. Step 2: ((2,4)) and ((4,2)) require ((2,2)), and ((4,2)) and ((2,4)) require ((4,4)); both are present. Step 3: When reverse links appear, check the self-pairs they force.

Step 1: The given rule is the same as (a\le b). Step 2: If (a\le b) and (b\le c), then (a\le c), so ((a,c)) is in the relation. Step 3: First convert a word-based relation into mathematical form.

Step 1: ((-1)R2) is true because (-1+2=1\ge 0). Also (2R(-1)) is true because (2+(-1)=1\ge 0). Step 2: Transitivity would require ((-1)R(-1)), but (-1+(-1)=-2), which is not at least (0). Step 3: Choose a chain where the required third pair truly fails.

Step 1: If ((a,b)) is in the relation, then both (a) and (b) are multiples of (3). If ((b,c)) is also in the relation, then (c) is also a multiple of (3). Step 2: Hence (a) and (c) are both multiples of (3), so ((a,c)) is in the relation. Step 3: For shared-property group relations, check the first and third elements.

Step 1: ((3,1)) and ((1,3)) are in the relation. Step 2: Transitivity requires ((3,3)), but it is missing. Step 3: In cyclic relations, each element may require its self-pair.

Option B is correct because the union of two transitive relations need not be transitive. For example, let R={(1,2)} and S={(2,3)}; each relation is transitive because it has no composable pair, but R∪S contains (1,2) and (2,3) without containing (1,3). The intersection is always transitive: any pair present in it belongs to both relations, so transitivity in both supplies the composite pair. Options C and D are also true.

Step 1: ((1,2)) and ((2,2)) require ((1,2)), which is present. Step 2: ((2,2)) with itself requires ((2,2)), which is also present. Step 3: In a small list, a self-pair may only require the same pair again.

Step 1: (0.8R0) is true because (0.8^2+0^2=0.64\le 1). Also (0R0.8) is true. Step 2: Transitivity requires (0.8R0.8), but (0.8^2+0.8^2=1.28>1). Step 3: For region or bound-based relations, counterexamples are very useful.

Step 1: If (a\le b) and (b\le c), then (a\le c). Step 2: If (a,b,c) are all divisible by (3), then ((a,c)) also satisfies the rule. Step 3: Check whether both order and property pass forward.

Step 1: (1+2=3) is odd, and (2+3=5) is also odd. Step 2: But (1+3=4) 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 be false.

Step 1: This relation is equivalent to (a^2=b^2). Step 2: If (a^2=b^2) and (b^2=c^2), then (a^2=c^2), so (aRc). Step 3: A relation of equal or opposite sign is easier to understand through equality of squares.

Step 1: ((2,3)) and ((3,2)) are in the relation. Step 2: Transitivity requires ((2,2)), but it is missing. Step 3: Check self-pairs forced by reverse pairs in each separate group.

Step 1: All pairs among even numbers are in the relation, and every number is related to itself. Step 2: If both links are within the even group, the first and third are also even; if an equality link is involved, the required pair remains valid. Step 3: In a mixed relation, think through both types of cases separately.

Step 1: (R\cap S) contains pairs where (a<b) and both numbers are odd. Step 2: If (a<b) and (b<c), then (a<c); and if all three are in the odd group, ((a,c)) also belongs to the relation. Step 3: In an intersection, check that both conditions pass forward together.