Reflexive relation

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.

Step 1: On a five-element set, (5) diagonal pairs are compulsory. Step 2: To have exactly (12) pairs, choose (7) non-diagonal pairs. Step 3: There are (25-5=20) non-diagonal pairs, so the number is (\binom{20}{7}).

Step 1: For (6) elements, (A\times A) has (36) pairs. Step 2: The (6) diagonal pairs are compulsory, leaving (36-6=30) optional pairs. Step 3: Hence the number of reflexive relations is (2^{30}).

Step 1: A reflexive relation must contain the (n) diagonal pairs. Step 2: There are (n^2-n) non-diagonal pairs, and one fixed non-diagonal pair is also compulsory. Step 3: The remaining (n^2-n-1) pairs are optional, giving (2^{n^2-n-1}).

Step 1: On a four-element set, (12) non-diagonal pairs are optional. Step 2: For ((1,2)) and ((2,1)), they must not both be included, so there are (3) valid choices. Step 3: The remaining (10) non-diagonal pairs are optional, giving (3\cdot2^{10}).