Reflexive relation

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.

Step 1: A relation is reflexive if ((a,a)) is present for every (a \in A). Step 2: Here (a+a=2a), which is always even, so every required diagonal pair is in (R). Step 3: In exams, check diagonal pairs first.

Step 1: A reflexive relation must contain ((a,a)) for every (a). Step 2: Since (a-a=0), and (0) is not odd, no diagonal pair belongs to (R). Step 3: For conditions involving (a-b), put (a=b) first.

Step 1: Reflexivity needs ((2,2),(3,3),(5,5),(7,7)). Step 2: Every number divides itself, so all these diagonal pairs already belong to the relation. Step 3: When minimum additions are asked, first check the existing diagonal pairs.

Step 1: In (a<b), (a=a) can never satisfy the condition. Step 2: So no diagonal pair is present, and all four diagonal pairs must be added. Step 3: To make a relation reflexive, only missing diagonal pairs are compulsory.

Step 1: The required diagonal pairs are ((1,1),(2,2),(3,3)). Step 2: The relation already has ((1,1)) and ((2,2)), but ((3,3)) is missing. Step 3: Do not get distracted by other pairs; check only ((a,a)) pairs.

Step 1: For reflexivity, put (b=a). Step 2: The condition becomes (a+a\geq 2a), or (2a\geq 2a), which is always true. Step 3: In inequality relations, carefully test the equality case.

Step 1: For ((a,a)), (|a-a|=0). Step 2: Since (0<1) is true, every diagonal pair is in the relation. Step 3: In modulus-based relations, equal elements usually make the expression (0).

Step 1: (|a-b|\leq 0) is possible only when (|a-b|=0). Step 2: This means (a=b), so all and only diagonal pairs occur. Step 3: A modulus condition with (\leq 0) is essentially an equality condition.

Step 1: For reflexivity, test ((a,a)). Step 2: The condition becomes (a^2=a^2), true for every (a \in A). Step 3: If substituting equal entries makes both sides identical, the relation is reflexive.

Step 1: Reflexivity does not require only (a=b) pairs. Step 2: Each self-pair ((a,a)) satisfies (a^2=a^2). Step 3: Extra pairs may exist, but no diagonal pair may be missing.