Reflexive relation

Step 1: In a reflexive relation, only self-pairs are compulsory. Step 2: Since (4\in A), ((4,4)) must be present. Step 3: Other pairs of the form ((a,b)) are not compulsory.

Step 1: (A\times A) has (3^2=9) pairs. Step 2: The (3) diagonal pairs are compulsory, so (6) pairs are optional. Step 3: Total reflexive relations are (2^6=64).

Step 1: Check the self-pairs of (x,y,z). Step 2: ((x,x),(y,y),(z,z)) are all present. Step 3: ((x,y)) is extra and does not affect reflexivity.

Step 1: A reflexive relation needs self-pairs for (x,y,z). Step 2: ((x,x)) and ((z,z)) are present, but ((y,y)) is missing. Step 3: Add the self-pair of the missing element.

Step 1: The second relation is non-empty because it has pairs. Step 2: But it does not contain ((1,1),(2,2),(3,3)), so it is not reflexive. Step 3: Keep the difference between non-empty and reflexive clear.

Step 1: In an arrow diagram, an arrow from a point to itself represents ((a,a)). Step 2: Since every point has such an arrow, every element is related to itself. Step 3: A loop at every point is the sign of reflexivity in a diagram.

Step 1: A reflexive relation needs a self-arrow at every element. Step 2: Element (3) has no such arrow, so ((3,3)) is missing. Step 3: If one point lacks a loop, reflexivity fails.

To test reflexivity, we must check whether (a, a) belongs to R for every a in A. Substituting b = a into the defining condition gives a² = a², which is an identity and is true for a = 1, 2, and 3. Therefore (1, 1), (2, 2), and (3, 3) all belong to R. The relation may contain additional pairs as well, but those are irrelevant to reflexivity. Hence R is reflexive and option A is correct; options C and D wrongly limit the condition to selected elements.

Step 1: For reflexivity, the condition becomes (a^2<a^2). Step 2: A number cannot have its square less than itself squared. Step 3: Relations with strict inequality are generally not reflexive.

Step 1: For any (a), (a\equiv a \pmod{2}) is true. Step 2: Hence every ((a,a)) is included in the relation. Step 3: In same-remainder relations, always test the self-case first.

Step 1: For a self-pair, (a-a=0). Step 2: (0\equiv 0 \pmod{3}) is true. Step 3: When the difference becomes zero, such congruence relations are reflexive.

Step 1: For reflexivity, check (a+a). Step 2: (a+a=2a) is always divisible by (2). Step 3: In congruence relations, substitute the self-value to reach the conclusion quickly.

Step 1: For a self-pair, (a+a=2a). Step 2: (2a\equiv 0 \pmod{2}), not (1). Step 3: Thus no self-pair satisfies this condition.

Step 1: (I_A) contains all pairs ((a,a)). Step 2: A reflexive relation (R) must also contain all these pairs. Step 3: Therefore, the identity relation is a subset of (R).

Since R is reflexive on A, every self-pair (a, a) with a ∈ A belongs to R. Since S is also reflexive, the same self-pair belongs to S. Therefore every required self-pair belongs to both relations and consequently belongs to their intersection R ∩ S. This proves that R ∩ S is reflexive. The intersection need not be empty or equal to A, because relations consist of ordered pairs, not individual elements. Thus option A is correct.

Step 1: Both (R) and (S) contain all self-pairs. Step 2: Their union keeps the pairs from both relations. Step 3: Hence (R\cup S) also contains all ((a,a)) pairs and is reflexive.

Step 1: A reflexive (R) contains every ((a,a)). Step 2: The inverse of ((a,a)) is again ((a,a)). Step 3: Therefore, the inverse relation also contains all self-pairs.

Step 1: Both (R) and (S) contain ((1,1),(2,2),(3,3)). Step 2: These three pairs remain in the intersection. Step 3: Hence (R\cap S) is reflexive, even if extra pairs differ.

Step 1: (R) has all self-pairs, so (R) is reflexive. Step 2: On taking inverse, ((1,1),(2,2),(3,3)) remain unchanged. Step 3: Therefore, (R^{-1}) is also reflexive.

Step 1: The reflexive condition applies to every (a\in A). Step 2: In the empty set, there is no element, so no condition is violated. Step 3: The empty relation on the empty set is usually considered reflexive.

Step 1: A non-empty set has at least one element. Step 2: Reflexivity needs that element's self-pair, but the empty relation has no pair. Step 3: Understand the difference between an empty relation and an empty set.

The Cartesian product A × A contains every ordered pair from A, including the three self-pairs (1, 1), (2, 2), and (3, 3). The relation R is obtained by deleting (2, 2), so although (1, 1) and (3, 3) remain, the required self-pair for 2 is missing. Reflexivity requires all self-pairs, not merely most of them. Therefore R is not reflexive. It is neither the identity relation nor empty, so option B is correct.

A × A initially contains every ordered pair of A, including all diagonal pairs (1, 1), (2, 2), and (3, 3). The two removed pairs, (1, 2) and (2, 3), are off-diagonal pairs; neither is of the form (a, a). Therefore all required self-pairs remain in R, so R is reflexive. Removing some non-self-pairs does not destroy reflexivity. The relation is not empty and is much larger than the identity relation, making option A correct.

Step 1: The identity relation has one self-pair for each element. Step 2: Since (|A|=6), there are (6) such pairs. Step 3: A reflexive relation contains the whole identity relation.

Step 1: Reflexivity requires ((1,1),(2,2),(3,3)). Step 2: All three are present in the relation. Step 3: Other pairs may affect other properties, but reflexivity remains as long as all self-pairs are present.