Transitive relation

Step 1: ((1,2)) and ((2,4)) require ((1,4)). Step 2: ((1,4)) is present, and ((4,4)) does not create any missing pair. Therefore the relation is transitive. Step 3: Find each chain and match the required final pair.

Step 1: ((2,4)) and ((4,1)) require ((2,1)). Step 2: ((2,1)) is not in the given relation, so the relation is not transitive. Step 3: Do not stop after one chain; check all possible chains.

Step 1: The universal relation contains every possible ordered pair from (A). Step 2: So if ((a,b)) and ((b,c)) are present, ((a,c)) is also definitely present. Step 3: In a universal relation, no required pair can be missing.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: The given self-pairs do not create any other missing required pair. So ((1,3)) must be added. Step 3: To make a relation transitive, first locate the missing direct pair.

Step 1: If (a-b) is even and (b-c) is even, then their sum (a-c) is also even. Step 2: Hence (a) and (c) are related. Step 3: In parity-based relations, add the differences.

Step 1: Adding an odd difference and an odd difference gives an even difference. Step 2: If (a-b) is odd and (b-c) is odd, then (a-c) is even, so the relation fails. Step 3: Remember that odd plus odd is even.

Step 1: If (a=b) and (b=c), then (a=c) must hold. Step 2: Therefore the equality relation is transitive. Step 3: Equality is reflexive, symmetric, and transitive.

Step 1: ((1,2)) and ((2,1)) require ((1,1)), which is present. Step 2: ((2,1)) and ((1,2)) require ((2,2)), which is also present. Step 3: When opposite pairs create self-pair requirements, both must be checked.

Step 1: For transitivity, the relation should pass forward like a chain. Step 2: If (a<b) and (b<c), then (a<c), so less than is transitive. Step 3: Everyday relations are often not transitive, so test them mathematically.

Step 1: Symmetry deals with ((a,b)) implying ((b,a)). Step 2: Transitivity needs ((a,b)) and ((b,c)) to imply ((a,c)). The conditions are different. Step 3: Check each property by its own definition.

Step 1: In the definition of transitivity, the middle element is common. Step 2: From ((x,y)) and ((y,z)), the first and last elements form ((x,z)). Step 3: Do not reverse the order of ordered pairs.

A relation R is transitive when, for every a, b, c in A, the presence of (a,b) and (b,c) guarantees the presence of (a,c). Here, (1,2) and (2,3) require (1,3), and (1,3) is present. The pair (2,2) also creates checks such as (2,2) followed by (2,3), which requires (2,3), already included, and (2,2) followed by itself, which requires (2,2), also included. No other available pair creates a missing consequence. Therefore R satisfies the transitivity condition. Option B is incorrect because it overlooks the required pair that is already present. Options C and D describe different properties and do not answer the transitivity question.

Step 1: Both ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is missing. Step 3: Missing reflexive pairs do not always break transitivity; focus on required chains.

Step 1: If (a) is a multiple of (b), and (b) is a multiple of (c), then (a) is a multiple of (c). Step 2: The multiple relation passes along the chain. Step 3: Divisibility and multiple relations use similar transitivity logic.

Step 1: First, ((2,5)) and ((5,7)) imply ((2,7)). Step 2: Then ((2,7)) and ((7,9)) imply ((2,9)). Step 3: In longer chains, apply transitivity repeatedly.

Step 1: (1) is different from (2), and (2) is different from (1), but (1) is not different from (1). Step 2: So the different-from relation is not transitive. Step 3: A counterexample is the fastest way to disprove transitivity.

Step 1: The basic form of transitivity is making a third pair from two linked pairs. Step 2: ((a,b)) and ((b,c)) require ((a,c)). Step 3: Keep this separate from symmetry and reflexivity.

Step 1: With ((1,2)) and ((2,2)), transitivity requires ((1,2)), which is present. Step 2: With ((1,1)) and ((1,2)), it again requires ((1,2)), which is present. So no required pair is missing. Step 3: Extra reflexive pairs do not break transitivity.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: This pair is not given, so it is the first clear requirement. Step 3: More pairs may be needed later, but the first chain gives ((1,3)).

Step 1: If (a) and (b) have the same remainder, and (b) and (c) have the same remainder, then (a) and (c) also have the same remainder. Step 2: Hence (a \equiv c \pmod{3}). Step 3: Congruence relations are generally transitive.