Reflexive relation

Step 1: To check reflexivity, put ((a,a)) into the relation condition. Step 2: (a+a=2a) is even for every integer, so every self-pair belongs to the relation. Step 3: In condition-based questions, test self-pairs first.

Step 1: For a self-pair, (a-b=a-a=0). Step 2: (0) is not odd, so no self-pair satisfies the condition. Step 3: If self-pairs are missing, the relation is not reflexive.

Step 1: Reflexivity requires ((a,a)) for every (a\in A). Step 2: Every given positive number divides itself, so ((a,a)\in R). Step 3: In divisibility relations, first check self-division.

Step 1: For a self-pair in (a\le b), we get (a\le a). Step 2: Every number is equal to itself, so (a\le a) is true. Step 3: Relations based on (\le) are easily checked for reflexivity.

Step 1: Reflexivity requires ((a,a)) to be in the relation. Step 2: Putting (b=a) in (a<b) gives (a<a), which is never true. Step 3: Strict inequality relations are usually not reflexive.

For a set with n elements, a relation is a subset of the n² ordered pairs. Reflexivity requires all n diagonal pairs (a,a), while the remaining n² − n pairs may be chosen or omitted independently. For n = 4, there are 16 total pairs and 4 compulsory pairs, leaving 12 optional pairs. Therefore the number is 2^12, so option A is correct.

Step 1: For (A), the required self-pairs are ((1,1),(2,2),(3,3)). Step 2: The first two are present, but ((3,3)) is missing. Step 3: The minimum fix is to add the missing self-pair.

Step 1: (R\cap S) contains only pairs common to both relations. Step 2: The common self-pairs are ((a,a)) and ((c,c)), but ((b,b)) is missing. Step 3: Missing one self-pair makes the relation not reflexive.

Step 1: Both (R) and (S) contain ((a,a)) for every (a\in A). Step 2: A pair common to both relations belongs to (R\cap S). Step 3: Hence all self-pairs remain in the intersection.

Step 1: (R) already contains every self-pair. Step 2: Taking a union does not remove pairs of (R). Step 3: Therefore (R\cup S) is also reflexive.

Step 1: In an inverse relation, ((a,a)) remains ((a,a)). Step 2: Since (R) has all self-pairs, (R^{-1}) also has them. Step 3: The inverse of a reflexive relation is reflexive.

Step 1: Reflexive closure is obtained by adding missing self-pairs. Step 2: Here ((3,3)) is missing, so only one pair is added. Step 3: The total number of pairs becomes (4+1=5).

Step 1: With four elements, all four self-pairs are required. Step 2: ((2,2)) is absent, so a required pair is missing. Step 3: In reflexivity, no self-pair may be missing.

Step 1: For reflexivity, put (b=a) in the condition. Step 2: This gives (a^2=a^2), which is true for every element. Step 3: In equality-based conditions, check the self-case first.

Step 1: For every (a\in A), (a^2=a^2) is true. Step 2: So ((-1,-1),(0,0),(1,1)) all belong to (R). Step 3: Extra pairs such as ((-1,1)) do not affect reflexivity.

Step 1: For reflexivity, compare any element with itself. Step 2: Since (a-a=0), (a\equiv a \pmod{2}) is true for every (a). Step 3: In congruence relations, remember that an element is congruent to itself.

A relation R on A is reflexive only when (a, a) belongs to R for every a in A. Here, the required condition becomes a + a = 4, or 2a = 4. It is true for a = 2, so (2, 2) is present, but it is false for a = 1 and a = 3; therefore (1, 1) and (3, 3) are absent. Since even one required self-pair is missing, R is not reflexive. Thus option A is correct; options B and C merely identify true individual facts, while D is irrelevant.

Step 1: For a self-pair, (a+b=2a). Step 2: The smallest element in (A) is (1), so (2a\ge 2) is true for every element. Step 3: In inequality conditions, start with the smallest element.

Step 1: Reflexivity needs self-pairs for (0,1,2). Step 2: For ((0,0)), (0+0=0), which is not greater than (0). Step 3: One failed self-pair makes the relation not reflexive.

Step 1: (R) contains the self-pair of every element. Step 2: It contains only those self-pairs, so it is the identity relation. Step 3: The identity relation is always reflexive.