Reflexive relation

Step 1: On the diagonal, (X=Y) is true. Step 2: The or condition includes every ((X,X)) through the equality part. Step 3: Therefore the relation is reflexive.

Step 1: On the diagonal, (X\triangle X=\varnothing). Step 2: The condition (X\triangle X=X) holds only when (X=\varnothing). Step 3: Hence only ((\varnothing,\varnothing)) is a diagonal pair.

Step 1: Reflexivity also requires ((1,1)). Step 2: The number (1) has no factor greater than (1). Step 3: Missing even one diagonal pair makes the relation non-reflexive.

Step 1: Every (a\in A) is divisible by some power of (2). Step 2: In ((a,a)), both entries are the same, so they share that same power. Step 3: Therefore all diagonal pairs are present.

Step 1: For reflexivity, put (a=b). Step 2: Then the ratio is (\frac{a}{a}=1=2^0), an integral power of (2). Step 3: Hence every ((a,a)) belongs to the relation.

Step 1: On the diagonal, the condition becomes (2a\leq a^2). Step 2: Since (a>0), this becomes (2\leq a). Step 3: (a=2,3,4,5) work, so there are (4) diagonal pairs.

Step 1: On the diagonal, the condition is (2a\leq a^2). Step 2: This fails for (a=1) and holds for (a=2,3,4,5). Step 3: Therefore only ((1,1)) must be added.

Step 1: On the diagonal, (a^3-a^3=0). Step 2: This is true for every (a\in A). Step 3: Therefore all diagonal pairs are in the relation, so (R) is reflexive.

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

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

Step 1: For reflexivity, check ((a,a)). Step 2: Then (a-a=0), and (0) is divisible by (6). Step 3: In difference-divisibility relations, the diagonal difference is always zero.

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).