Reflexive relation

Step 1: All diagonal pairs are already present in (R). Step 2: Taking a union does not remove pairs from (R); it only adds pairs from (S). Step 3: A union with a reflexive relation remains reflexive.

Step 1: For every (a \in A), ((a,a) \in R) and ((a,a) \in S). Step 2: In the composition, choose the middle element as (a), so ((a,a) \in S \circ R). Step 3: The composition of reflexive relations is reflexive.

Step 1: Every element of (B) is also in (A). Step 2: Since (R) is reflexive, ((b,b) \in R) for every (b \in B), and also ((b,b) \in B \times B). Step 3: Thus the restricted relation is reflexive on (B).

Step 1: Since (S\subseteq R), some pairs of (R) may be removed. Step 2: If any required ((a,a)) is removed, (S) is not reflexive. Step 3: Reflexivity is not automatically preserved in subrelations.

Step 1: Reflexivity requires all diagonal pairs. Step 2: The smallest relation keeps only those compulsory pairs and no extra pair. Step 3: Therefore the identity relation is the smallest reflexive relation.

Step 1: With (3) elements, there are (9) pairs, and (3) diagonal pairs are compulsory. Step 2: Among the (6) non-diagonal pairs, ((1,2)) and ((2,3)) are also fixed. Step 3: The remaining (4) pairs are optional, so the count is (2^4=16).

Step 1: Reflexivity fixes the (4) diagonal pairs. Step 2: Symmetry makes each unordered off-diagonal pair either included both ways or excluded both ways. Step 3: There are ({4 \choose 2}=6) such pairs, so the count is (2^6=64).

Step 1: Reflexivity fixes the (3) diagonal pairs. Step 2: For each unordered pair of distinct elements, there are three choices: no direction, one direction, or the opposite direction. Step 3: Since there are ({3 \choose 2}=3) such pairs, the count is (3^3=27).

Step 1: For reflexivity in a matrix, all main diagonal entries must be (1). Step 2: Only the fourth matrix has diagonal entries (1,1,1). Step 3: Other entries do not affect the decision of reflexivity.

Step 1: Reflexivity depends only on the main diagonal. Step 2: Here the main diagonal is (1,1,1). Step 3: Symmetry is a separate property and is not required for reflexivity.

Step 1: Reflexivity requires every main diagonal entry to be (1). Step 2: If one diagonal entry is (0), one ((a,a)) is missing. Step 3: Having many off-diagonal pairs does not make a relation reflexive.

Step 1: An equivalence relation must be reflexive, symmetric, and transitive. Step 2: Therefore every equivalence relation is necessarily reflexive. Step 3: But a merely reflexive relation need not be an equivalence relation.

Step 1: For reflexivity, put (x=y). Step 2: For every (x), (f(x)=f(x)) is true, so ((x,x) \in R). Step 3: The function need not be one-one for this property.

Step 1: For self-relation, put (y=x). Step 2: The condition becomes (f(x)<f(x)), which is never true. Step 3: Relations defined by strict inequality are not reflexive.

Step 1: For reflexivity, put (x=y). Step 2: (f(x)\le f(x)) is true for every (x). Step 3: The function need not be increasing or decreasing for this.

Step 1: On the diagonal, put (y=x). Step 2: (x-x=0), and (0) is an integer. Step 3: Even on real numbers, the self-difference is always (0).

Step 1: For reflexivity, put (y=x). Step 2: The condition becomes (x<x+1), which is true for every real (x). Step 3: Substitute carefully instead of assuming it becomes (x<x).

Step 1: For self-relation, put (y=x). Step 2: Then (x-y=x-x=0), which is not greater than (0). Step 3: A positive-difference condition prevents reflexivity.

Step 1: Compare each element with itself in congruence. Step 2: (a-a=0), and (0) is divisible by (3), so (a \equiv a \pmod{3}). Step 3: Congruence relations are generally reflexive.

Step 1: Check the diagonal pairs ((0,0),(1,1),(2,2)). Step 2: (0+0=0) is divisible by (3), but (1+1=2) and (2+2=4) are not. Step 3: Therefore ((1,1)) and ((2,2)) must be added.

Step 1: Each separate relation may miss some diagonal pairs. Step 2: In the union, pairs from both relations are combined, so missing diagonal pairs may be supplied by the other relation. Step 3: For union questions, check the combined diagonal list.

Step 1: Each (R_i) is reflexive, so every ((a,a)) belongs to every (R_i). Step 2: A pair present in all relations remains in their intersection. Step 3: This works for both finite and infinite families.

Step 1: A four-element set has (4^2=16) total pairs. Step 2: To be non-reflexive, at least one diagonal pair must be missing. Step 3: The remaining (15) pairs can be present, and the relation will still not be reflexive.

Step 1: A reflexive relation must contain one diagonal pair for each element. Step 2: For (n) elements, there are (n) such diagonal pairs. Step 3: In the minimum case, only these (n) pairs are included.

Step 1: (\Delta_A) contains all diagonal pairs ((a,a)). Step 2: (R) is reflexive exactly when all these diagonal pairs are included in (R). Step 3: Therefore (\Delta_A \subseteq R) is the concise equivalent condition for reflexivity.