Reflexive relation

Step 1: Five diagonal pairs are required. Step 2: ((1,1)) and ((2,2)) are already present because their sums are positive. Step 3: The remaining three diagonal pairs must be added.

Step 1: On the diagonal, (\min(a,a)=a). Step 2: This is true for every (a\in A), so all ((a,a)) are in the relation. Step 3: For minimum and maximum conditions, substitute equal elements first.

Step 1: For ((a,a)), (\max(a,a)=a). Step 2: Since (b=a) on the diagonal, (\max(a,a)=b) is true. Step 3: Reflexivity only needs the same-element cases to hold.

Step 1: On the diagonal, (\gcd(a,a)=a). Step 2: Hence ((a,a)\in R) for every (a\in A). Step 3: For greatest common divisor relations, remember the value for equal numbers.

Step 1: On the diagonal, (\gcd(a,a)=a). Step 2: This equals (1) only for (a=1), so only ((1,1)) is present. Step 3: The other four diagonal pairs must be added.

Step 1: For reflexivity, check ((a,a)). Step 2: The least common multiple of a number with itself is the number itself, so (\operatorname{lcm}(a,a)=a). Step 3: Evaluating the expression at equal numbers is the simplest method.

Step 1: Reflexivity requires (X\subseteq X) for every (X\in A). Step 2: Every set is a subset of itself. Step 3: The subset relation is reflexive on any collection of sets.

Step 1: (X\subset Y) means proper subset. Step 2: No set is a proper subset of itself, so (X\subset X) is false. Step 3: The difference between (\subseteq) and (\subset) is very important in exams.

Step 1: For ((X,X)), the condition becomes (X\cap X=X). Step 2: The intersection of a set with itself is the same set. Step 3: In set-operation relations, substitute the same set on the diagonal.

Step 1: (A=\mathcal{P}(S)) has (4) elements. Step 2: On the diagonal, (X\cup X=X), which equals (S) only when (X=S). Step 3: One diagonal pair is already present, so (3) must be added.

Step 1: On the diagonal, (X\triangle X=\varnothing). Step 2: The symmetric difference of a set with itself is the empty set. Step 3: For symmetric-difference relations, checking equal sets quickly proves reflexivity.

Step 1: Reflexivity requires ((1,1)) as well. Step 2: The number (1) has no prime factor, so ((1,1)\notin R). Step 3: If even one diagonal pair is missing, the relation is not reflexive.

Step 1: Every element of (A) is greater than (1), so each has at least one prime factor. Step 2: A number shares its own prime factor with itself. Step 3: Therefore, every ((a,a)) belongs to the relation.

Step 1: For reflexivity, test ((a,a)). Step 2: The condition (a=b) is true for every diagonal pair. Step 3: In an or condition, one true part is enough to include the pair.

Step 1: On the diagonal, (a=b) is true. Step 2: Also, (a+b=2a), which is even, so both conditions hold for every (a). Step 3: In an and condition, verify both parts.

Step 1: A reflexive relation must contain ((1,1),(2,2),(3,3),(4,4)). Step 2: In (P), ((2,2)) is already counted among these four, while ((1,3)) and ((3,1)) are two extra fixed pairs. Step 3: So 6 pairs are fixed, (16-6=10) pairs are free, and the count is (2^{10}).

Step 1: On 6 elements, 6 self-pairs are compulsory. Step 2: To have 9 total pairs, choose 3 pairs from the (36-6=30) non-self pairs. Step 3: Hence the number is (\binom{30}{3}).

Step 1: On 4 elements, the total number of reflexive relations is (2^{16-4}=2^{12}). Step 2: The identity relation is exactly one of them. Step 3: Only the identity relation is removed, so the count is (2^{12}-1).

Step 1: Four self-pairs are fixed by reflexivity. Step 2: At most 6 total pairs means choosing 0, 1, or 2 pairs from the 12 non-self pairs. Step 3: The count is (\binom{12}{0}+\binom{12}{1}+\binom{12}{2}=1+12+66=79).

Step 1: The 3 self-pairs are compulsory. Step 2: Choose exactly 1 of the given 3 non-self pairs in (\binom{3}{1}=3) ways. Step 3: The remaining 3 non-self pairs are free, so the count is (3\times2^3=24).