Reflexive relation

Step 1: A reflexive relation needs every ((a,a)) for (a\in A). Step 2: Since (a+a=2a) is always even, every diagonal pair is already present. Step 3: The exam trick is to check diagonal pairs first, not the whole relation.

Step 1: Reflexivity requires ((a,a)) for every (a\in A). Step 2: For every (a), (a-a=0), and (0) is divisible by any non-zero integer. Step 3: In exams, test the diagonal condition before listing all pairs.

Step 1: Check ((a,a)) for every (a\in A). Step 2: Since (a^2=a^2) for every element, all diagonal pairs belong to (R). Step 3: Equality-based conditions usually contain all diagonal pairs.

Step 1: A reflexive relation must contain ((1,1),(2,2),(3,3),(4,4)). Step 2: Since (a<a) is never true, no diagonal pair is present. Step 3: Therefore, all (4) diagonal pairs must be added.

Step 1: For ((a,a)), the condition becomes (a\leq a). Step 2: Every number is equal to itself, so (a\leq a) is true. Step 3: Relations defined by (\leq) are naturally reflexive on the given set.

Step 1: Reflexivity needs ((a,a)) for every (a\in A). Step 2: Every non-zero number divides itself, so (a\mid a) is true. Step 3: In divisibility relations, self-divisibility is the key reflexive check.

Step 1: The required diagonal pairs are ((1,1),(2,2),(3,3)). Step 2: The relation already has the first two but misses ((3,3)). Step 3: To make a relation reflexive, add only the missing diagonal pair.

Step 1: Put ((a,a)) in the condition to get (|a-a|=0). Step 2: Since (0\leq 1), every diagonal pair is in the relation. Step 3: For absolute-difference relations, test the diagonal by making the difference zero.

Step 1: Reflexivity would require (|a-a|<0) for every (a). Step 2: But (|a-a|=0), and (0<0) is false. Step 3: With strict inequalities, diagonal pairs often fail.

Step 1: For diagonal pairs, (a+b=2a). Step 2: For (a=1,2,3), (2a\leq6) is true, but for (a=4), (8\leq6) is false. Step 3: Only ((4,4)) must be added.

Step 1: For reflexivity, put (b=a). Step 2: The condition becomes (a+a\geq2a), i.e. (2a\geq2a), true for every (a). Step 3: Simplify the condition on the diagonal first.

Step 1: (A\times A) contains all diagonal pairs. Step 2: The pair ((2,2)) has been removed, so one required diagonal pair is missing. Step 3: A relation is not reflexive if even one diagonal pair is absent.

Step 1: (A\times A) has (n^2) ordered pairs. Step 2: A reflexive relation must include the (n) diagonal pairs, while the remaining (n^2-n) pairs are optional. Step 3: Hence the number is (2^{n^2-n}).

Step 1: For (n=3), (A\times A) has (9) pairs. Step 2: The (3) diagonal pairs are compulsory, so (6) pairs are optional. Step 3: The number of reflexive relations is (2^6=64).

Step 1: The total number of relations is (2^{4^2}=2^{16}). Step 2: The number of reflexive relations is (2^{16-4}=2^{12}). Step 3: Non-reflexive relations are (2^{16}-2^{12}).

Step 1: A reflexive relation on a four-element set must contain (4) diagonal pairs. Step 2: Since the total number of pairs is (10), the remaining pairs are (10-4=6). Step 3: Always subtract compulsory diagonal pairs first.

Step 1: (A\times A) has (16) pairs and (4) diagonal pairs are compulsory. Step 2: To get exactly (10) pairs, choose (6) more from the (12) non-diagonal pairs. Step 3: The count is (\binom{12}{6}).

Step 1: For five elements, (5) diagonal pairs are compulsory. Step 2: A relation with (9) pairs needs (9-5=4) additional non-diagonal pairs. Step 3: There are (25-5=20) non-diagonal pairs, so the answer is (\binom{20}{4}).

Step 1: The (3) diagonal pairs are compulsory. Step 2: To have (5) total pairs, choose (2) additional pairs. Step 3: There are (6) non-diagonal pairs, so the number of ways is (\binom{6}{2}).

Step 1: A reflexive relation must include every element paired with itself. Step 2: For four elements, the minimum required pairs are ((1,1),(2,2),(3,3),(4,4)). Step 3: The minimum count equals the number of elements in the set.

Step 1: Reflexivity needs all five diagonal pairs. Step 2: Since the relation has no diagonal pair, all five must be added. Step 3: Having many non-diagonal pairs does not compensate for missing diagonal pairs.

Step 1: For ((a,a)), the sum is (a+a=2a). Step 2: (2a) is always even, not odd, so no diagonal pair belongs to (R). Step 3: Parity of the diagonal sum gives a quick reflexivity check.

Step 1: To test reflexivity, put (b=a). Step 2: For every (a), (a\equiv a \pmod{2}) is true. Step 3: Congruence with the same element always includes diagonal pairs.

Step 1: On the diagonal, put (b=a), giving (a\equiv a+1 \pmod{2}). Step 2: A number and the next number do not have the same parity. Step 3: If the diagonal condition fails, the relation is not reflexive.

Step 1: For reflexivity, substitute ((a,a)) into the condition. Step 2: This gives (a^2\equiv a^2 \pmod{3}), which is always true by equality. Step 3: Even in power and modulo relations, test the same element first.