Reflexive relation

Step 1: Since (R) is reflexive, ((a,a)\in R) for every (a\in A). Step 2: In (A\times A-R), all pairs of (R) are removed, so the 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, all ((a,a)) are in it. Step 2: The inverse of ((a,a)) is also ((a,a)), so these pairs are in (R^{-1}) too. Step 3: The intersection keeps all self-pairs and (R^{-1}\cap R) is also symmetric.

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: The 4 self-pairs are fixed. Step 2: For 12 total pairs, 8 non-self pairs are needed, meaning 4 unordered pairs are chosen in both directions. Step 3: Choose 4 of the 6 unordered pairs in (\binom{6}{4}=15) ways.

Step 1: The equal-remainder groups modulo 4 are ({1,5}), ({2,6}), ({3,7}), and ({4,8}). Step 2: Each group gives (2^2=4) ordered pairs. Step 3: Total pairs are (4+4+4+4=16), including all self-pairs.

Step 1: Square remainders are grouped as ({0}) with remainder 0, ({1,5}) with remainder 1, ({2,4}) with remainder 4, and ({3}) with remainder 3. Step 2: Ordered pairs are formed within equal square-remainder groups. Step 3: The total is (1^2+2^2+2^2+1^2=10).

Option C is correct. The squares modulo 6 are: 0²≡0, 1²≡1, 2²≡4, 3²≡3, 4²≡4, and 5²≡1. Thus the elements group according to equal square residues into the classes {0}, {1,5}, {2,4}, and {3}. There are therefore four equivalence classes. The relation is based on equality of residues, so elements in the same listed group relate to one another, while elements in different groups do not.

Step 1: The power set has (2^4=16) elements. Step 2: A reflexive pair is ((X,X)) for each element (X) of the power set. Step 3: Hence the number of reflexive pairs is 16.

Step 1: For (X\subseteq Y), each basic element has three choices: in (Y) only, in both, or in neither. Step 2: Being in (X) only is not allowed. Step 3: For 4 basic elements, the total number of pairs is (3^4=81).

Step 1: For reflexivity, every ((a,a)) must satisfy the rule. Step 2: The self-sums are (4,8,12,16). Step 3: The largest self-sum is 16, so the minimum (k) is 16.

Step 1: For self-pairs, check (a^2+a^2=2a^2). Step 2: The largest value occurs at (a=\pm2), giving (2\cdot4=8). Step 3: The inequality is strict, so (8<k) is needed, and the minimum integer value is (k=9).

Step 1: Count the divisors present in (A) for each second component. Step 2: For (1,2,4,8,16), the counts are (1,2,3,4,5). Step 3: The total number of pairs is (1+2+3+4+5=15).

Step 1: The equal absolute value groups are ({-4,4}), ({-2,2}), and ({0}). Step 2: Total pairs are (2^2+2^2+1^2=9). Step 3: Since 5 are self-pairs, the number of non-self pairs is (9-5=4).

Step 1: Since (R) is reflexive, (I_A\subseteq R). Step 2: The union of a set with its subset is the larger set itself. Step 3: Therefore (R\cup I_A=R).

Step 1: A reflexive (R) contains the whole identity relation (I_A). Step 2: Intersecting with (I_A) keeps exactly those self-pairs. Step 3: Therefore (R\cap I_A=I_A).

Step 1: In composition, ((1,2)) and ((2,3)) can be joined because the middle element 2 matches. Step 2: They produce ((1,3)) in (R\circ R). Step 3: In composition questions, first look for consecutive connectable pairs.

Step 1: The 4 self-pairs are fixed. Step 2: To have 7 total pairs, choose (7-4=3) non-self pairs. Step 3: There are 12 non-self pairs, so the count is (\binom{12}{3}).

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

Step 1: The 3 self-pairs are fixed. Step 2: For ((1,2),(2,1)), valid choices are neither, only ((1,2)), or only ((2,1)); both together are not allowed. Step 3: The remaining 4 non-self pairs are free, so the count is (3\times2^4=48).

Step 1: (R) contains all self-pairs, so they remain in the union. Step 2: In (R\cup R^{-1}), every pair appears with its reverse. Step 3: Therefore the relation is both reflexive and symmetric.