Transitive relation

Step 1: A transitive relation is checked using two connected ordered pairs. Step 2: If one pair goes from (a) to (b) and the next goes from (b) to (c), then the pair from (a) to (c) must also be present. Step 3: In exams, check transitivity only when the middle element is common.

Step 1: Here ((1,2)) and ((2,3)) are connected pairs. Step 2: They require ((1,3)), and it is already present in the relation. Step 3: For transitivity checks, find the pair that is forced and then look for it in the list.

Step 1: Since ((1,2)) and ((2,3)) are in the relation, transitivity requires ((1,3)). Step 2: The pair ((1,3)) is not present. Step 3: One missing required pair is enough to show that a relation is not transitive.

Step 1: If (a\le b) and (b\le c), then (a\le c) on the number line. Step 2: So ((a,b)) and ((b,c)) imply ((a,c)). Step 3: Order relations such as (\le) are usually transitive.

Step 1: If (a<b) and (b<c), then (a<c) must be true. Step 2: Hence ((a,b)) and ((b,c)) imply ((a,c)). Step 3: A chain from smaller to larger is a simple way to understand transitivity.

Step 1: This relation contains only diagonal pairs of the form ((a,a)). Step 2: If ((a,a)) and ((a,a)) are used, the required pair is again ((a,a)), which is present. Step 3: A relation containing only diagonal pairs is transitive.

Step 1: The transitive condition is checked only when both ((a,b)) and ((b,c)) are present. Step 2: The empty relation has no pair, so no example violates the condition. Step 3: Remember that the empty relation is transitive by vacuous truth.

Step 1: (A\times A) contains all possible ordered pairs from (A). Step 2: If ((a,b)) and ((b,c)) are present, then ((a,c)) is also surely present. Step 3: A relation with all possible pairs automatically satisfies transitivity.

Step 1: ((1,2)) and ((2,4)) form a chain. Step 2: They require ((1,4)), and it is present in the relation. Step 3: In a short list, check all possible chains carefully.

Step 1: ((1,2)) and ((2,4)) are connected because the middle element (2) is common. Step 2: Transitivity requires ((1,4)). Step 3: The required pair uses the first element of the first pair and the second element of the second pair.

Step 1: If (a=b) and (b=c), then by equality (a=c). Step 2: Thus ((a,b)) and ((b,c)) imply ((a,c)). Step 3: The equality relation is a very direct example of a transitive relation.

Step 1: From ((1,2)) and ((2,1)), the pair ((1,1)) is required. Step 2: From ((2,1)) and ((1,2)), the pair ((2,2)) is also required. Both are absent. Step 3: Reverse pairs can create a need for diagonal pairs in transitivity.

Step 1: To check transitivity, we need two pairs with a common middle element. Step 2: Here there is only one pair, so no chain exists to violate the condition. Step 3: A relation with a single pair can be transitive when no required next pair is formed.

Step 1: If (a\mid b) and (b\mid c), then (a) divides (c). Step 2: Hence ((a,b)) and ((b,c)) imply ((a,c)). Step 3: Divisibility is an important example of a transitive relation.

Step 1: (a \equiv b \pmod{3}) and (b \equiv c \pmod{3}) mean the remainders are linked through the same middle number. Step 2: Therefore (a \equiv c \pmod{3}). Step 3: Same-remainder relations are transitive.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((2,3)) and ((3,4)) require ((2,4)), and ((1,3)) and ((3,4)) require ((1,4)); both are present. Step 3: A relation is transitive when all required chain pairs are present.

Step 1: In ((1,2)) and ((2,3)), the middle element (2) is common. Step 2: So ((1,3)) becomes necessary. Step 3: In a longer chain, first check requirements made by adjacent pairs.

Step 1: The transitive rule gives ((a,c)) from ((a,b)) and ((b,c)). Step 2: Here (a=2), (b=5), and (c=7), so ((2,7)) is required. Step 3: Identify the common middle element and form the final pair.

Step 1: Using ((3,3)) with itself requires ((3,3)) again. Step 2: That pair is already present, so no new different pair is forced. Step 3: Diagonal pairs alone do not create new elements without extra information.

Step 1: ((1,1)) and ((1,2)) require ((1,2)), which is present. Step 2: ((1,2)) and ((2,2)) also require ((1,2)), which is present. Step 3: If all required pairs are already present, the relation is transitive.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is absent. Step 3: The presence of another diagonal pair does not remove this missing requirement.

Step 1: ((3,2)) is in the relation because (3=2+1), and ((2,1)) is also in it. Step 2: Transitivity would require ((3,1)), but (3\ne1+1). Step 3: Relations based on a fixed difference of one are generally not transitive.

Step 1: If (a\ge b) and (b\ge c), then (a\ge c). Step 2: Therefore two connected pairs force a third pair that also belongs to the relation. Step 3: Both (\ge) and (\le) are transitive order relations.

Step 1: If (A\subseteq B) and (B\subseteq C), then every element of (A) is also in (C). Step 2: Hence (A\subseteq C). Step 3: The subset relation is a standard example of a transitive relation.

Step 1: ((1,3)) and ((3,4)) are in the relation. Step 2: They require ((1,4)), but it is not in the list. Step 3: Check not only the first chain, but also chains formed by existing pairs.