Reflexive relation

Step 1: (A\times A) has 16 pairs, and 4 self-pairs are compulsory in a reflexive relation. Step 2: In (P), ((4,4)) is already counted among self-pairs, while ((1,2)) and ((2,1)) 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 5 elements, 5 self-pairs are compulsory. Step 2: To have 8 total pairs, choose 3 pairs from the (25-5=20) non-self pairs. Step 3: Hence the number is (\binom{20}{3}).

Step 1: On 3 elements, the total number of reflexive relations is (2^{9-3}=64). Step 2: One of them is the identity relation and one is the universal relation. Step 3: Removing both gives (64-2=62).

Step 1: Reflexivity fixes 4 self-pairs and leaves 12 non-self pairs free. Step 2: At least 6 total pairs means at least 2 non-self pairs must be chosen. Step 3: From (2^{12}), subtract cases with 0 or 1 non-self pair: (1+12=13), so the answer is (2^{12}-13).

Step 1: The 3 self-pairs are fixed by reflexivity. Step 2: Exactly one of ((1,2)) and ((2,1)) can be chosen in 2 ways. Step 3: The remaining 4 non-self pairs are free, so the count is (2\times2^4=32).

Step 1: (R) is reflexive, so it contains all self-pairs. Step 2: (S) has no self-pair, so subtracting (S) will not remove the required pairs. Step 3: All self-pairs remain, hence (R-S) is reflexive.

Step 1: A reflexive (R) contains every ((a,a)). Step 2: In (A\times A-R), all pairs of (R) are removed, so all self-pairs are removed. Step 3: On a non-empty set, a relation without self-pairs cannot be reflexive.

Step 1: Since (R) is reflexive, ((a,a)\in R) for every (a\in A). Step 2: Reversing ((a,a)) gives the same pair ((a,a)). Step 3: Hence all self-pairs are also in (R^{-1}).

Step 1: Reflexivity of (R) gives (aRa). Step 2: Reflexivity of (S) gives (aSa). Step 3: Taking the middle element as (a), we get (a(S\circ R)a), so the composition is reflexive.

Step 1: Reflexivity fixes the 3 self-pairs. Step 2: To have 7 total pairs, 4 non-self pairs are needed, meaning 2 unordered pairs are taken in both directions. Step 3: Choose 2 of the 3 unordered pairs in (\binom{3}{2}=3) ways.

Step 1: The 4 self-pairs are fixed by reflexivity. Step 2: For 8 total pairs, 4 non-self pairs are needed, meaning 2 unordered pairs are included in both directions. Step 3: Choose 2 of the 6 unordered pairs, giving (\binom{6}{2}=15).

Step 1: The equal-remainder groups modulo 3 are ({1,4}), ({2,5}), and ({3,6}). Step 2: Pairs are formed only within each group, so each group gives (2^2=4) pairs. Step 3: Total pairs are (4+4+4=12), and the relation is reflexive.

Step 1: Square remainders form the groups ({0}), ({1,4}), and ({2,3}). Step 2: Pairs are formed between elements with the same square remainder. Step 3: Total pairs are (1^2+2^2+2^2=9).

Step 1: The equal absolute value groups are ({-3,3}), ({-2,2}), ({-1,1}), and ({0}). Step 2: The first three groups give (2^2) pairs each, and ({0}) gives one pair. Step 3: Total pairs are (4+4+4+1=13).

Step 1: For each second component (b), count the divisors present in (A). Step 2: For (1,2,3,4,6,12), the counts are (1,2,2,3,4,6). Step 3: Total pairs are (1+2+2+3+4+6=18), including all self-pairs.

Step 1: For each basic element, there are three possibilities: in (Y) only, in both, or in neither. Step 2: Being in (X) only is not allowed because (X\subseteq Y). Step 3: For 3 basic elements, this gives (3^3=27) pairs, and the relation is reflexive.

Step 1: The power set has (2^3=8) elements. Step 2: In the proper subset relation, no pair ((X,X)) is included. Step 3: To make it reflexive, add ((X,X)) for every (X), so 8 pairs must be added.

Step 1: For reflexivity, all self-pairs must satisfy the rule. Step 2: The self-sums are (2,4,6,8), and the largest is 8. Step 3: To include all self-pairs, (k\ge8), so the minimum value is 8.

Step 1: The self-pair sums are (2,6,10). Step 2: The condition (a+b<k) is strict, so even the largest sum 10 must be less than (k). Step 3: The minimum integer value is (k=11).

Step 1: For a self-pair, (|a-a|=0). Step 2: In the rule (|a-b|<k), a self-pair needs (0<k). Step 3: Therefore the necessary and sufficient condition is (k>0).