Transitive relation

Step 1: If (a-b>2) and (b-c>2), adding gives (a-c>4). Step 2: Since (a-c>4) implies (a-c>2), transitivity holds. Step 3: Adding inequalities can give a stronger conclusion.

Step 1: Take (a=3), (b=1.5), and (c=0). Step 2: (0<3-1.5<2) and (0<1.5-0<2), but (3-0=3), which is not less than (2). Step 3: In bounded inequalities, two small gaps can cross the bound.

Step 1: ((1,2)) and ((2,3)) imply ((1,3)). Step 2: Then ((1,3)) and ((3,1)) imply ((1,1)). Step 3: Cyclic chains produce self-pairs.

Step 1: The three pairs form a cycle. Step 2: Repeated transitivity gives ((1,3),(2,1),(3,2)) and then self-pairs like ((1,1),(2,2),(3,3)). Step 3: The transitive closure of a full cycle often connects the whole small set.

Step 1: ((1,3)) and ((3,1)) require ((1,1)), which is present. Step 2: ((3,1)) and ((1,3)) require ((3,3)), also present. Step 3: Reverse pairs do not break transitivity if both required self-pairs are present.

Step 1: ((1,3)) and ((3,1)) require ((1,1)), and ((3,1)) with ((1,3)) requires ((3,3)). Step 2: Both are present; ((2,4)) does not form any further linked chain. Step 3: An isolated pair does not break transitivity by itself.

Step 1: ((3,1)) and ((1,3)) require ((3,3)). Step 2: ((3,3)) is not in the relation, so transitivity fails. Step 3: Reverse pairs must be checked in both orders.

Step 1: ((a,b)) and ((b,c)) imply ((a,c)). Step 2: Then ((a,c)) and ((c,a)) imply ((a,a)); rotating the cycle gives ((b,b)) and ((c,c)) too. Step 3: Cyclic relations force self-pairs under transitivity.

Step 1: From (a\le 2b) and (b\le 2c), we only get (a\le 4c). Step 2: Transitivity requires (a\le 2c), which need not hold. Take (a=4), (b=2), (c=1); (4\le4), (2\le2), but (4\le2) is false. Step 3: In multiplier-based relations, the bound may change.

Step 1: If (a=kb) and (b=kc), then (a=k^2c). Step 2: Transitivity requires (a=kc), so generally (k^2=k); for positive (k), this gives (k=1). Step 3: In fixed-multiplier relations, apply the multiplier twice.

Step 1: Self-pairs in (\Delta) only carry a pair forward as itself. Step 2: Chains already inside (R) are complete because (R) is transitive. Adding (\Delta) creates no missing direct pair. Step 3: Adding the identity relation preserves transitivity.

Step 1: Identity pairs only fill self-pair gaps like ((a,a)). Step 2: If the missing pair is like ((1,3)), (\Delta) will not add it. For example, ({(1,2),(2,3)}) still lacks ((1,3)). Step 3: Not every missing pair is a self-pair.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), and ((2,3)) with ((3,4)) requires ((2,4)); both are present. Step 2: ((1,3)) and ((3,4)) require ((1,4)), also present. ((4,4)) creates no new missing pair. Step 3: Find every jump in the long chain.

Step 1: ((1,3)) and ((3,4)) require ((1,4)). Step 2: This pair is missing, so transitivity fails. Step 3: Use an already present shorter jump to check the longer jump.

Step 1: The relation (a-b) rational is transitive. Step 2: The inverse of a transitive relation is also transitive. Here (b-a=-(a-b)) remains rational. Step 3: In inverse, the sign of the difference changes, not the property.

Step 1: (R\cap S) contains only those self-pairs that are also in (R). Step 2: Any relation containing only self-pairs is transitive, because it only requires the same pair again. Step 3: Identify the nature of pairs left after intersection.

Step 1: (R\cup S) contains both ((1,2)) and ((2,3)). Step 2: Transitivity requires ((1,3)), but it is not in the union. Step 3: Union of two relation parts can create a new missing pair.

Step 1: (R\cap S={(2,2)}). Step 2: ((2,2)) with ((2,2)) requires ((2,2)), which is present. Step 3: A relation with one self-pair is transitive.

Step 1: Over real numbers, (a^2+b^2=0) happens only when (a=0) and (b=0). Step 2: So the relation has only ((0,0)), which is transitive. Step 3: First identify the actual pairs, then check the property.

Step 1: (ab>0) means (a) and (b) have the same sign and both are non-zero. Step 2: If (a) and (b) have the same sign, and (b) and (c) have the same sign, then (a) and (c) also have the same sign. Step 3: In product-sign relations, track the sign chain.

Step 1: (ab<0) means (a) and (b) have opposite signs. Step 2: If (a) and (b) have opposite signs and (b) and (c) also have opposite signs, then (a) and (c) have the same sign. Step 3: The example (1,-1,1) breaks transitivity.

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. Step 2: ((2,4)) and ((4,8)) require ((2,8)), and ((1,4)) with ((4,8)) requires ((1,8)); both are present. Step 3: In a chain-like relation, check all direct jumps.

Step 1: ((1,4)) and ((4,8)) require ((1,8)). Step 2: This pair is missing from the list, so it is the main gap. Step 3: Use already obtained long pairs to form the next long pair.

Step 1: From ((x,y)) and ((y,y)), transitivity requires ((x,y)) itself. Step 2: This pair is already present, so no new missing pair is created. Step 3: A self-pair does not always create a new pair.

Step 1: Combining ((x,x)) and ((x,y)), the middle element is (x). Step 2: Transitivity requires ((x,y)), which is already present. Step 3: Do not assume a reverse pair just because a self-pair is present.