Transitive relation

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.

Step 1: ((3,1)) and ((1,2)) require ((3,2)), which is present. Step 2: Requirements formed with ((1,1)) and ((2,2)) are already in the relation. Step 3: In a downward chain, check the final direct pair.

Step 1: ((3,1)) and ((1,2)) require ((3,2)). Step 2: This pair is not in the relation, so it must be added. Step 3: When an outside starting point enters a chain, find the direct pair.

Step 1: In transitivity, not every missing pair matters. Step 2: A missing pair matters only if it is required by a chain like ((a,c)) and ((c,b)). Step 3: Do not conclude immediately from a missing pair; prove that it is required.

Step 1: ((2,3)) and ((3,5)) require ((2,5)). Step 2: ((3,5)) and ((5,2)) require ((3,2)), and ((5,2)) with ((2,3)) requires ((5,3)). Step 3: In a cycle, first-layer pairs may later create self-pairs too.

Step 1: In transitivity, after ((a,b)), we need a pair starting with (b). Step 2: Grouping by the second component helps find all chains quickly. Step 3: For long lists, this method reduces mistakes.