Reflexive relation

Step 1: A relation is reflexive when every (a \in A) has ((a,a)) in the relation. Step 2: ((1,1)) and ((2,2)) are present but ((3,3)) is missing. Step 3: In exams, check diagonal pairs first.

Step 1: With (4) elements, there are (4^2=16) ordered pairs. Step 2: The (4) diagonal pairs are compulsory, so (16-4=12) pairs are optional. Step 3: Use (2^{n^2-n}) for quick counting.

Step 1: There are (5^2=25) total pairs and (5) diagonal pairs are compulsory. Step 2: There are (25-5=20) non-diagonal pairs, and exactly (3) must be chosen. Step 3: Separate compulsory and optional pairs before counting.

Step 1: In a reflexive relation, each element of the set must be related to itself. Step 2: For (A), all four diagonal pairs are required. Step 3: Extra pairs do not matter, but all diagonal pairs must be present.

Step 1: To check reflexivity from a matrix, look at the main diagonal. Step 2: The diagonal entries are (1,0,1), so the middle entry must be changed to (1). Step 3: Only the missing diagonal entry needs correction.

Step 1: Every ((a,a)) is present in both (R) and (S). Step 2: A pair common to both remains in (R \cap S), so all diagonal pairs remain. Step 3: The intersection of reflexive relations is reflexive.

Step 1: Since (R) is reflexive, every ((a,a) \in R). Step 2: The inverse of ((a,a)) is again ((a,a)). Step 3: Diagonal pairs remain unchanged in the inverse relation.

Step 1: (\Delta_A={(a,a):a \in A}) contains all diagonal pairs. Step 2: (R \cup \Delta_A) keeps the old pairs and adds any missing diagonal pairs. Step 3: This gives the smallest reflexive closure.

Step 1: To test reflexivity, check ((a,a)). Step 2: For every (a), (a+a=2a) is even, so ((a,a) \in R). Step 3: In such questions, put (b=a) first.

Step 1: For reflexivity, every ((a,a)) must belong to the relation. Step 2: (a-a=0), and (0) is divisible by any non-zero integer. Step 3: In divisibility-type relations, the diagonal difference is always (0).

Step 1: Put (a=b) for diagonal pairs. Step 2: Then (a+b=2a), which is always even, not odd. Step 3: If required diagonal pairs are missing, the relation is not reflexive.

Step 1: Reflexivity checks whether each element is related to itself. Step 2: For every number, (a \le a) is true, so ((a,a)) is included. Step 3: Be careful about the difference between (<) and (\le).

Step 1: Reflexivity requires ((a,a)). Step 2: The condition (a<a) is false for every (a), so no diagonal pair appears. Step 3: Strict inequality relations are usually not reflexive.

Step 1: Each element of (P(A)) is itself a set. Step 2: For every set (X), (X \subseteq X) is true. Step 3: Remember the difference between subset and proper subset.

Step 1: A proper subset does not allow equality. Step 2: No set is a proper subset of itself. Step 3: Distinguish (\subseteq) from (\subset) carefully.

Step 1: For reflexivity, check each ((a,a)). Step 2: Every non-zero number divides itself, so ((a,a)) is included. Step 3: Missing non-diagonal pairs do not affect reflexivity.

Step 1: Reflexivity requires ((1,1),(2,2),(3,3)). Step 2: (\gcd(1,1)=1), but (\gcd(2,2)=2) and (\gcd(3,3)=3). Step 3: If even one diagonal pair is missing, the relation is not reflexive.

Step 1: On the diagonal, put (b=a). Step 2: The condition becomes (a^2=a^2), which is always true. Step 3: Reflexivity only needs each element to be related to itself.

Step 1: Reflexivity requires every diagonal pair. Step 2: ((0,0)) satisfies the condition, but ((-1,-1)) and ((1,1)) do not. Step 3: Having only some diagonal pairs is not enough.

Step 1: Check the condition for each ((a,a)). Step 2: (|a-a|=0), and (0\le 2) is true. Step 3: Far non-diagonal pairs do not affect reflexivity.