Transitive relation

Step 1: If (a \ge b) and (b \ge c), then (a \ge c). Step 2: The other options can fail transitivity by small counterexamples. Step 3: For order relations, form a chain and choose the answer.

Step 1: ((1,3)) and ((3,1)) are in the relation. Step 2: Transitivity requires ((1,1)). Step 3: When an element returns to itself through a chain, check the required self-pair.

Step 1: A relation is transitive if ((a,b) \in R) and ((b,c) \in R) imply ((a,c) \in R). Step 2: Here ((1,2)) and ((2,3)) require ((1,3)), which is present. Other required cases also do not fail. Step 3: In exams, first check pairs with matching middle elements.

Step 1: To test transitivity, look for ((a,b)) and ((b,c)), then check ((a,c)). Step 2: Since ((1,2)) and ((2,3)) are in (R), ((1,3)) must also be in (R). It is missing, so (R) is not transitive. Step 3: A single missing required pair is enough to disprove transitivity.

Step 1: If (a) divides (b) and (b) divides (c), then (a) divides (c). Step 2: Thus ((a,b)) and ((b,c)) imply ((a,c)), so the relation is transitive. Step 3: For divisibility questions, write the numbers in multiplication form.

Step 1: If (a-b) is even and (b-c) is even, then their sum ((a-b)+(b-c)=a-c) is also even. Step 2: Hence ((a,c)) belongs to (R), so the relation is transitive. Step 3: In such questions, add the given differences.

Step 1: If (a<b) and (b<c), then by order property (a<c). Step 2: Therefore ((a,b)) and ((b,c)) imply ((a,c)), so (R) is transitive. Step 3: For inequality relations, remember the direction on the number line.

Step 1: If (a\le b) and (b\le c), then (a\le c) must hold. Step 2: This is exactly the condition for transitivity, so the relation is transitive. Step 3: Both (\le) and (<) are transitive in usual order.

Step 1: In ((1,2)) and ((2,4)), the middle element (2) matches. Step 2: Transitivity requires ((1,4)), and it is present in (R). Step 3: Quickly identify pairs with a common middle element.

Step 1: Transitivity does not require every reflexive pair; it only requires the pairs forced by chains. Step 2: ((a,b)) and ((b,c)) require ((a,c)), which is present. ((a,b)) and ((b,b)) again require ((a,b)). Step 3: Do not confuse transitivity with reflexivity.

Step 1: To disprove transitivity, we need ((a,b)) and ((b,c)) present but ((a,c)) absent. Step 2: In the empty relation, no starting pair exists, so the condition is never violated. Hence it is transitive. Step 3: Do not assume an empty relation is non-transitive.

Step 1: A universal relation contains every possible ordered pair from (A). Step 2: Therefore, if ((a,b)) and ((b,c)) are present, ((a,c)) is also certainly present. Hence it is transitive. Step 3: In (A\times A), no required pair is missing.

Step 1: From ((1,3)) and ((3,4)), ((1,4)) is required. Step 2: From ((1,2)) and ((2,4)), ((1,4)) is also required. Since it is missing, it must be added. Step 3: In longer chains, check the end-to-end pair.

Step 1: If (a-b) and (b-c) are both divisible by (3), then their sum (a-c) is also divisible by (3). Step 2: Thus ((a,b)) and ((b,c)) imply ((a,c)), so the relation is transitive. Step 3: Use the addition property of divisibility.

Step 1: From ((1,2)) and ((2,1)), transitivity requires ((1,1)). Step 2: Similarly, from ((2,1)) and ((1,2)), ((2,2)) is required. Both are missing, so (R) is not transitive. Step 3: Reverse pairs can force reflexive pairs in transitivity checks.

Step 1: In the transitive rule, the second element of the first pair matches the first element of the second pair. Step 2: From ((2,5)) and ((5,7)), the required pair is ((2,7)). Step 3: Do not reverse the order of ordered pairs.

Step 1: This behaves like the equality relation, containing only pairs of the form ((a,a)). Step 2: From ((a,a)) and ((a,a)), the required pair is again ((a,a)), which is present. So the relation is transitive. Step 3: A relation with only identity pairs is transitive.

Step 1: If (a=b) and (b=c), then (a=c). Step 2: Thus ((a,b)) and ((b,c)) imply ((a,c)), so equality is transitive. Step 3: Equality is reflexive, symmetric, and transitive.

Step 1: One counterexample is enough to show a relation is not transitive. Step 2: (1\ne2) and (2\ne1), but (1\ne1) is false. So ((1,2)) and ((2,1)) are present, but ((1,1)) is not. Step 3: Do not assume the relation (a\ne b) is transitive.

Step 1: Both ((1,3)) and ((3,1)) are in (R). Step 2: They require ((1,1)), but ((1,1)) is not present. Hence (R) is not transitive. Step 3: In cyclic pairs, check the required reflexive pairs carefully.

Step 1: If ((a,b)) and ((b,c)) are in (R \cap S), then they are in both (R) and (S). Step 2: Since both are transitive, ((a,c)) belongs to both, so it belongs to (R \cap S). Step 3: For intersection, check membership in both relations.

Step 1: In a union, the two pairs may come from different relations. Step 2: For example, (R={(1,2)}) and (S={(2,3)}) are individually transitive, but (R \cup S) has ((1,2)) and ((2,3)) without ((1,3)). Step 3: Do not treat union like intersection for transitivity.

Step 1: ((1,2)) and ((2,3)) are in (R). Step 2: Transitivity requires ((1,3)), but it is absent. This is a clear reason (R) is not transitive. Step 3: Start checking from the shortest visible chain.

Step 1: If person (a) has the same age as (b), and (b) has the same age as (c), then (a) has the same age as (c). Step 2: Hence ((a,c)) also belongs to the relation, so it is transitive. Step 3: Relations based on the same property are often transitive.

Step 1: In friendship, if (a) is a friend of (b), and (b) is a friend of (c), it is not necessary that (a) is a friend of (c). Step 2: So the transitive condition does not generally hold. Step 3: For real-life relations, test the rule using examples.