Reflexive relation

Answer: A, 448. If A has three elements, then A×A has 3×3=9 ordered pairs. A relation is any subset of A×A, so the total number of relations is 2^9=512. For a reflexive relation, the three diagonal pairs (a,a) must be included. The other six ordered pairs can each be either included or excluded independently, giving 2^6=64 reflexive relations. The non-reflexive relations are all remaining relations: 512−64=448. Option B is obtained by using an incorrect count of compulsory pairs. Option C counts the reflexive relations, not the non-reflexive ones. Option D counts all relations and therefore does not exclude the reflexive cases. Memory cue: count all subsets first, then subtract the subsets that satisfy the required property.

Direct answer: C, 12. A relation on A is any subset of A×A. Since A has two elements, A×A has four ordered pairs: (1,1), (1,2), (2,1), and (2,2). Each pair may independently be included or excluded, so the total number of relations is 2^4=16. A relation is reflexive only if it contains every diagonal pair, namely (1,1) and (2,2). These two pairs are fixed as present, while the other two pairs can be chosen freely, giving 2^2=4 reflexive relations. Hence non-reflexive relations =16−4=12. Option A is the number of reflexive relations, not the requested complement. B has no valid counting basis. D is the total number of all relations, not the number that fail reflexivity. Memory cue: for an n-element set, total relations are 2^(n²), while reflexive ones are 2^(n²−n).

Step 1: (A) has 3 elements, so (A \times A) has 9 pairs. Step 2: A reflexive relation must contain 3 self-pairs, while the remaining (9-3=6) pairs are optional. Step 3: Hence the number of reflexive relations is (2^6=64).

Step 1: For 4 elements, (A \times A) has 16 pairs. Step 2: Reflexivity fixes 4 diagonal pairs, so (16-4=12) pairs are optional. Step 3: Each optional pair has two choices, giving (2^{12}).

Step 1: (A \times A) contains (n^2) pairs. Step 2: In a reflexive relation, (n) self-pairs are compulsory, so (n^2-n) pairs are free. Step 3: Therefore the number is (2^{n^2-n}).

Step 1: For (A), the required pairs are ((1,1),(2,2),(3,3),(4,4)). Step 2: Only ((3,3)) is missing in the given relation. Step 3: So adding one pair is enough.

Step 1: A reflexive relation needs self-pairs for (a,b,c,d). Step 2: ((a,a)) and ((c,c)) are present, but ((b,b)) and ((d,d)) are missing. Step 3: Identifying missing diagonal pairs is the key.

Step 1: To test reflexivity, put ((a,a)) in the rule. Step 2: Every number is equal to itself, so (a\le a) is true. Step 3: In a (\le) relation, the equality part gives reflexivity.

Step 1: Reflexivity needs ((1,1),(2,2),(3,3)). Step 2: Under (a<b), (1<1) is false, so ((1,1)) is not formed. Step 3: Strict inequality does not include self-pairs.

Step 1: For reflexivity, put (b=a). Step 2: Then (a-b=a-a=0), and (0) is divisible by (3). Step 3: In difference-based relations, quickly test (a-a).

Step 1: Reflexivity needs ((a,a)) for every (a). Step 2: For ((a,a)), the sum is (a+a=2a), which is even. Step 3: If the rule asks for an odd sum, self-pairs are not included.

Step 1: Reflexivity compares a number with itself. Step 2: Any number has the same remainder as itself when divided by (2). Step 3: Treat (a \equiv a \pmod{2}) as the basic test in congruence relations.

Step 1: Reflexivity requires all self-pairs, including ((1,1)). Step 2: The rule requires the first element to be even, but (1) is not even. Step 3: Missing even one self-pair makes the relation non-reflexive.

Step 1: Reflexivity requires every (a) to be related to itself. Step 2: Every natural number (a) divides itself because (a=a\times1). Step 3: In divisibility questions, remember self-divisibility.

Step 1: A proper divisor requires (a\ne b). Step 2: In a self-pair ((a,a)), (a=b), so self-pairs are not included. Step 3: Distinguish proper divisor from ordinary divisibility.

Step 1: Here the elements are subsets, not ordinary numbers. Step 2: Every subset (X) is a subset of itself, so (XRX) is true. Step 3: In subset relations, read the base family carefully.

Step 1: (X\subset Y) means proper subset. Step 2: No set is a proper subset of itself. Step 3: In exams, clearly distinguish (\subset) and (\subseteq).

Step 1: For reflexivity, put (b=a). Step 2: (|a-a|=0) and (0\le0) is true. Step 3: In absolute value relations, the distance from an element to itself is zero.

Step 1: For reflexivity, ((a,a)) must satisfy the rule. Step 2: (|a-a|=0), and (0>0) is false. Step 3: Self-distance is zero, so it fails a strict positive condition.

Step 1: For a self-pair, the sum becomes (a+a). Step 2: (a+a=2a) is always even, whether (a) is even or odd. Step 3: In parity-of-sum relations, first check (2a).

Step 1: For ((a,a)), (a+a=2a) is even, not odd. Step 2: So none of the self-pairs are in the original relation. Step 3: To make it reflexive, add all ((1,1),(2,2),(3,3),(4,4)).

Step 1: Both relations contain ((1,1),(2,2),(3,3)). Step 2: The intersection keeps pairs common to both, so all self-pairs remain. Step 3: The intersection of two reflexive relations is reflexive.

Step 1: Both (R) and (S) contain all self-pairs of (A). Step 2: The union contains all pairs from both relations, so the self-pairs remain. Step 3: The union of reflexive relations is always reflexive.

Step 1: (S) contains all self-pairs. Step 2: In (R-S), pairs of (S) are removed, so self-pairs may also be removed. Step 3: In difference of relations, check whether compulsory pairs are being removed.

Step 1: Since (R) is reflexive, (R) contains all self-pairs. Step 2: Because (R\subseteq T), those self-pairs are also in (T). Step 3: A super-relation of a reflexive relation within (A\times A) is also reflexive.