Reflexive relation

Step 1: Since (R) is reflexive, all self-pairs are in (R). Step 2: Because (R\subseteq S), all those self-pairs are also in (S). Step 3: Therefore (S) is definitely reflexive.

Step 1: Reflexivity requires every ((a,a)). Step 2: On the diagonal, (a+b=2a), which is divisible by (4) only when (a) is even, so ((2,2),(4,4),(6,6)) are already present. Step 3: The missing diagonal pairs are ((1,1),(3,3),(5,5)), so (3) pairs are needed.

Step 1: For a diagonal pair, (a^2+b^2=2a^2). Step 2: Divisibility by (5) requires (a^2) to be divisible by (5), which happens only for (a=5) in the set. Step 3: Hence only ((5,5)) is a diagonal pair in (R).

Step 1: To test reflexivity, check ((a,a)). Step 2: The condition becomes (a^2\equiv a^2 \pmod{4}), which is true for every element. Step 3: Equality or same-remainder conditions usually include all diagonal pairs.

Step 1: On the diagonal, put (b=a), giving (a\equiv 2a \pmod{3}). Step 2: This means (a\equiv 0 \pmod{3}), so only (a=3,6) already work. Step 3: Since (6) diagonal pairs are required, (6-2=4) must be added.

Step 1: On the diagonal, put (b=a), giving (a^2-a=a(a-1)). Step 2: The product of two consecutive integers is always even. Step 3: Therefore every ((a,a)) belongs to the relation, so it is reflexive.

Step 1: On the diagonal, put (b=a), so (a^2+a=2a+1). Step 2: Rearranging gives (a^2-a-1=0), but checking the given finite set shows the condition holds for (a=1) and (a=2). Step 3: Thus the diagonal pairs are ((1,1)) and ((2,2)).

Step 1: Put (b=a) to test reflexivity. Step 2: Both sides become (a^2-a), so the equality is true for every (a). Step 3: If both sides become the same expression on the diagonal, all diagonal pairs are included.

Step 1: On the diagonal, (\gcd(a,a)=a). Step 2: The condition is true for (a>1), but for (a=1), (\gcd(1,1)=1). Step 3: Hence only ((1,1)) must be added.

Step 1: On the diagonal, (\operatorname{lcm}(a,a)=a). Step 2: Every element of (A) is less than or equal to (12). Step 3: Therefore every diagonal pair satisfies the condition, so (R) is reflexive.

Step 1: On the diagonal, (\operatorname{lcm}(a,a)=a). Step 2: The condition is true when (a<6), so (a=2,3,4,5) work. Step 3: (a=6) fails, so there are (4) diagonal pairs.

Step 1: Check ((a,a)) for reflexivity. Step 2: Every (a) divides itself, so both (a\mid a) and (a\mid a) are true. Step 3: In an and condition, both parts hold on the diagonal.

Step 1: On the diagonal, the condition becomes (a\mid a+1). Step 2: Since (a) and (a+1) differ by (1), this holds only for (a=1). Step 3: Hence only ((1,1)) is a diagonal pair in the relation.

Step 1: On the diagonal, put (b=a), giving (a+1\mid a+1). Step 2: Any non-zero number divides itself. Step 3: Therefore all diagonal pairs are included, so the relation is reflexive.

Step 1: On the diagonal, (a^2+b^2=2a^2). Step 2: (2a^2\leq4) gives (a^2\leq2), so (a=-1,0,1) work. Step 3: The missing diagonal pairs are ((-2,-2)) and ((2,2)), so (2) pairs are needed.

Step 1: For reflexivity, put (b=a). Step 2: Then (ab=a^2), so the condition becomes (a^2\leq a^2), which is true. Step 3: Equality is included in (\leq), so every diagonal pair works.

Step 1: On the diagonal, put (b=a), giving (ab=a^2). Step 2: The condition becomes (a^2<a^2), which is false for every (a). Step 3: Strict inequalities fail when both sides become equal on the diagonal.

Step 1: On the diagonal, (a-b=0). Step 2: The condition becomes (0^2\leq0), i.e. (0\leq0), which is true. Step 3: Therefore all four diagonal pairs are in the relation.

Step 1: The condition gives (|a-b|\leq0). Step 2: Absolute value is non-negative, so this is true exactly when (a=b). Step 3: All ((a,a)) pairs are present, so the relation is reflexive.

Step 1: On the diagonal, (|a-a|=0). Step 2: (0) is not positive, so no diagonal pair belongs to the relation. Step 3: A reflexive relation needs all diagonal pairs, so this relation is not reflexive.