Reflexive relation

Step 1: Reflexivity requires ((a,a)) to satisfy the rule for every real (a). Step 2: This means (2a^2\le k) for every real (a). Step 3: Since (a) can be arbitrarily large, no fixed real (k) can work.

Step 1: For self-pairs, check (a^2+a^2=2a^2). Step 2: For (a=-1,0,1), the values are (2,0,2). Step 3: The largest value is 2, so the minimum (k) is 2.

Step 1: For self-pairs, the product is (a^2). Step 2: The values (1^2,2^2,3^2) are (1,4,9). Step 3: To include all self-pairs, (k) cannot exceed the smallest value, 1.

Step 1: A reflexive relation has (n) compulsory self-pairs. Step 2: To have exactly (n+2) pairs, choose 2 additional non-self pairs. Step 3: There are (n^2-n) non-self pairs, so the number is (\binom{n^2-n}{2}).

Step 1: In composition, ((1,2)) and ((2,3)) connect through the middle element 2. Step 2: This forces ((1,3)) to belong to (R\circ R). Step 3: In composition questions, first identify pairs that connect consecutively.

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}).