Reflexive relation

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.

Step 1: For ((a,a)), the condition becomes (a+a>2a). Step 2: This is (2a>2a), which is never true. Step 3: In strict inequalities, diagonal pairs often fail.

Step 1: Reflexivity needs every element to be related to itself. Step 2: The symbol (\leq) allows equality, so (a\leq a) is true. Step 3: The difference between (\leq) and (<) is very important in exams.

Step 1: In (a\neq b), no pair ((a,a)) is included. Step 2: Since (A) has 4 elements, four diagonal pairs must be added. Step 3: If a relation contains unequal pairs, add all equal self-pairs for reflexivity.

Step 1: To test reflexivity, put (b=a). Step 2: (a+a=2a), which is divisible by (2) for every integer (a). Step 3: Both even and odd numbers give an even sum with themselves.

Step 1: Reflexivity requires all ((a,a)) pairs. Step 2: For ((1,1)), (1+1=2), which is not divisible by (3). Step 3: If even one diagonal pair is missing, the relation is not reflexive.

Step 1: Every element of (A) is divisible by (3). Step 2: For ((a,a)), (a+a=2a), and if (a) is divisible by (3), then (2a) is also divisible by (3). Step 3: Always connect the nature of the set with the condition.

Step 1: Every number has the same remainder as itself. Step 2: Therefore (a \equiv a \pmod{2}) is true for every (a), so all diagonal pairs are present. Step 3: Congruence-based equality relations are usually reflexive.

Step 1: For reflexivity, put (b=a). Step 2: The condition becomes (a \equiv a+1 \pmod{2}), which means (0 \equiv 1 \pmod{2}), false. Step 3: Such shifted congruence conditions often fail for self-pairs.

Step 1: (a \equiv b \pmod{3}) means (a-b) is divisible by (3). Step 2: For ((a,a)), (a-a=0), and (0) is divisible by every positive integer. Step 3: In congruence relations, self-comparison is always true.

Step 1: Check the condition on all diagonal pairs. Step 2: For ((3,3)), (6\leq 5) is false, and for ((4,4)), (8\leq 5) is false. Step 3: Some diagonal pairs are not enough; all must be present.

Step 1: The diagonal pairs are ((1,1),(2,2),(3,3)). Step 2: Their sums are (2,4,6), all less than or equal to (6). Step 3: In bounded sum conditions, check the largest element with itself.

Step 1: Reflexivity requires even the smallest element to be related to itself. Step 2: For ((1,1)), (1+1=2), which is less than (4). Step 3: One failed diagonal pair makes the relation non-reflexive.

Step 1: For ((a,a)), the sum is (2a). Step 2: The smallest element is (2), and (2+2=4), so all larger elements also satisfy the condition. Step 3: In such questions, the smallest diagonal pair can decide reflexivity.

Step 1: Put ((a,a)), so (ab=a^2). Step 2: The condition becomes (a^2\geq a^2), true for every (a). Step 3: If equality is allowed in an inequality, diagonal pairs often satisfy it.

Step 1: For reflexivity, put (b=a). Step 2: The condition becomes (a^2>a^2), which is never true. Step 3: A strict (>) comparison with itself fails.