Reflexive relation

Step 1: A reflexive relation needs every ((a,a)). Step 2: On the diagonal, (a+b=2a), which is divisible by (6) only when (a) is divisible by (3). In (A), only (3) and (6) work. Step 3: (8) diagonal pairs are needed and (2) are present, so (6) must be added.

Step 1: On the diagonal, (a^2+b^2=2a^2). Step 2: For divisibility by (10), (2a^2) must include a factor (5), so (a) must be divisible by (5). Step 3: In the given set, only (a=5) works, so there is one diagonal pair.

Step 1: On the diagonal, put (b=a), giving (a^2\equiv a \pmod{5}). Step 2: This means (a(a-1)\equiv 0 \pmod{5}), so (a\equiv0) or (a\equiv1 \pmod{5}). Step 3: In the set, (0,1,5) work, so there are (3) diagonal pairs.

Step 1: On the diagonal, put (b=a), giving (a\equiv3a \pmod{4}). Step 2: This gives (2a\equiv0 \pmod{4}), so (a) must be even. In (A), (2,4,6) work. Step 3: Since (7) diagonal pairs are needed and (3) are present, (4) must be added.

Step 1: On the diagonal, put (b=a), giving (a^3-a). Step 2: (a^3-a=a(a-1)(a+1)), a product of three consecutive integers, so it is even. Step 3: Factoring the diagonal expression is a useful exam method.

Step 1: On the diagonal, put (b=a), giving (a^2+a=3a). Step 2: This becomes (a^2-2a=0), i.e. (a(a-2)=0). Step 3: (a=0) and (a=2) work, so there are (2) diagonal pairs.

Step 1: A reflexive relation on this set needs (5) diagonal pairs. Step 2: The condition gives diagonal pairs only for (a=0) and (a=2). Step 3: Hence (5-2=3) diagonal pairs must be added.

Step 1: On the diagonal, (\gcd(a,a)=a). Step 2: This is even exactly when (a) is even. In (A), (2,4,6) are even. Step 3: Therefore (3) diagonal pairs belong to the relation.

Step 1: A total of (6) diagonal pairs are required. Step 2: ((2,2),(4,4),(6,6)) are already present because their greatest common divisor is even. Step 3: ((1,1),(3,3),(5,5)) must be added, so the answer is (3).

Step 1: On the diagonal, (\operatorname{lcm}(a,a)=a). Step 2: Every element of the given set is less than or equal to (8). Step 3: Hence all (6) diagonal pairs are in the relation.

Step 1: On the diagonal, (\operatorname{lcm}(a,a)=a). Step 2: The condition (a<5) holds only for (2,3,4). Step 3: Since (6) diagonal pairs are needed and (3) are present, (3) must be added.

Step 1: For reflexivity, put (b=a). Step 2: The condition becomes (a\mid a^2), which is true for every positive integer (a). Step 3: In divisibility relations, checking the square of the same number is a quick method.

Step 1: On the diagonal, the condition becomes (a+1\mid a). Step 2: Since (a+1) is positive and greater than (a), it cannot divide (a). Step 3: No diagonal pair is present, so the count is (0).

Step 1: Put (b=a) on the diagonal. Step 2: Then (a+1\mid a+1), and any non-zero number divides itself. Step 3: Therefore every ((a,a)) belongs to the relation.

Step 1: On the diagonal, (a^2+b^2=2a^2). Step 2: (2a^2\leq9) gives (a^2\leq4.5), so (a=-2,-1,0,1,2) work. Step 3: ((-3,-3)) and ((3,3)) are missing, so (2) pairs must be added.

Step 1: On the diagonal, (ab=a^2) and (a+b=2a). Step 2: The condition is (a^2\geq2a), i.e. (a(a-2)\geq0). Step 3: In the set, (a=-2,-1,0,2) work, so there are (4) diagonal pairs.

Step 1: A total of (5) diagonal pairs are required. Step 2: The diagonal condition (a(a-2)\geq0) is true for (-2,-1,0,2) and false for (1). Step 3: Only ((1,1)) must be added.

Step 1: On the diagonal, (a-b=0). Step 2: Therefore ((a-b)^2=0), and (0<1) is true. Step 3: Every diagonal pair satisfies the condition, so the relation is reflexive.

Step 1: On the diagonal, (a-b=0). Step 2: Then ((a-b)^2=0), which is not greater than (0). Step 3: No diagonal pair is present, so all four diagonal pairs must be added.

Step 1: On the diagonal, (|a-a|=0). Step 2: The condition becomes (0\leq a-1), which is true for every (a\geq1) in (A). Step 3: Hence all (5) diagonal pairs are in the relation.

Step 1: On the diagonal, (|a-a|=0). Step 2: The condition is (0\leq a-1), which fails for (a=0) and holds for (a=1,2,3,4). Step 3: Only ((0,0)) must be added.

Step 1: On the diagonal, (a=b) is always true. Step 2: In an or condition, the first part already includes every ((a,a)). Step 3: So the second part does not need separate checking for reflexivity.

Step 1: In diagonal pairs, (a=b). Step 2: The condition requires (a\neq b), so no diagonal pair can be present. Step 3: Since all diagonal pairs are absent, the relation is not reflexive.

Step 1: The relation contains only diagonal pairs for even elements. Step 2: ((2,2),(4,4),(6,6)) are present, but ((1,1),(3,3),(5,5)) are missing. Step 3: Therefore (3) diagonal pairs must be added.

Step 1: A reflexive relation needs all four diagonal pairs. Step 2: Among the removed pairs, ((1,1)) and ((4,4)) are diagonal, while ((2,3)) is not. Step 3: Therefore only two diagonal pairs must be added.