Symmetric relation

Step 1: Here (n=5). Step 2: The number of symmetric relations is (2^{\frac{n(n+1)}{2}}). Step 3: (2^{\frac{5\cdot6}{2}}=2^{15}=32768), so the answer is (32768).

Step 1: The four diagonal pairs are fixed as included. Step 2: The number of non-diagonal reverse-pair blocks is (\frac{4\cdot3}{2}=6). Step 3: Each block has two choices, so the total number is (2^6=64).

Step 1: Having no diagonal pair means all four diagonal choices are fixed as excluded. Step 2: There are (\frac{4\cdot3}{2}=6) non-diagonal reverse-pair blocks. Step 3: These (6) blocks are freely chosen, so the number is (2^6=64).

Step 1: Symmetry is preserved under union and intersection, but not always under composition. Step 2: The ordered composition of two symmetric relations need not preserve all reverse pairs. Step 3: For composition questions, testing with a small counterexample is often best.

Step 1: Reversing (a) and (b) changes (a^2+b^2=1) to (b^2+a^2=1). Step 2: By commutativity of addition, this is the same condition. Step 3: For algebraic conditions, check whether the equation remains unchanged after swapping variables.

Step 1: (a^2-b^2=0) means (a^2=b^2). Step 2: Reversing gives (b^2=a^2), which is also true. Step 3: Simplifying an equation often makes checking symmetry easier.

Step 1: Symmetry requires the reverse of every present pair to be present. Step 2: ((2,5)) is present, but ((5,2)) is not. Step 3: This single missing reverse pair is enough to show the relation is not symmetric.

Step 1: ({(1,2),(2,1)}) contains both reverse pairs, so it is symmetric. Step 2: Reflexivity would require ((1,1),(2,2),(3,3)), which are missing. Step 3: Understanding the difference between symmetric and reflexive is very important.

Step 1: The first option contains all diagonal pairs, so it is reflexive. Step 2: It has ((1,2)) but not ((2,1)), so it is not symmetric. Step 3: In combined-property questions, check reflexivity and symmetry separately.

Step 1: A symmetric relation is based on the reverse-pair condition. Step 2: The option states exactly that if ((a,b)) is present, then ((b,a)) must be present. Step 3: Do not mix definitions; reflexive, symmetric, and transitive are separate properties.

Step 1: In a symmetric relation, if ((a,b)) is present, then ((b,a)) must also be present. Step 2: ((2,3)) is present but ((3,2)) is missing, while ((1,2)) and ((2,1)) already form a pair. Step 3: In exams, always check the reverse of every non-diagonal ordered pair.

Step 1: If (a+b) is even, then (b+a) is also even because addition is commutative. Step 2: Hence whenever ((a,b) \in R), ((b,a) \in R) also holds. Step 3: For such questions, reverse the condition and check whether it remains true.

Step 1: (a-b>0) means (a>b). Step 2: If (a>b), then generally (b>a) is false, so the reversed pair does not satisfy the condition. Step 3: In inequality-based relations, the reverse direction often fails.

Step 1: If (|a-b|=2), then (|b-a|=|-(a-b)|=2). Step 2: So every pair with difference (2) has its reverse pair satisfying the same condition. Step 3: Absolute value conditions usually remain unchanged under reversal.

For a symmetric relation on an n-element set, each diagonal pair (a,a) can be selected independently, giving n choices of position. For each unordered pair of distinct elements, either both ordered pairs are included or both are excluded. There are n(n−1)/2 such unordered pairs, so the total number of independent choices is n + n(n−1)/2 = n(n+1)/2. For n = 3, this is 3 + 3 = 6, giving 2⁶ relations. Hence option B is correct.

Step 1: The number of symmetric relations on an (n)-element set is (2^{\frac{n(n+1)}{2}}). Step 2: Here (n=4), so the number of independent positions is (\frac{4\cdot5}{2}=10). Step 3: For counting questions, substitute (n) carefully.

Step 1: Symmetry requires ((b,a)) whenever ((a,b)) is present. Step 2: (1\mid2) is true, so ((1,2)) is in the relation, but (2\mid1) is false. Step 3: One counterexample is enough to prove a relation is not symmetric.

Step 1: If (a-b) is divisible by (5), then (b-a=-(a-b)) is also divisible by (5). Step 2: Hence ((a,b)) implies ((b,a)). Step 3: In divisibility or congruence relations, a negative sign usually preserves the condition.

Step 1: If (a+b=5), then (b+a=5) also holds. Step 2: Since addition is commutative, the reversed pair also belongs to the relation. Step 3: For sum-based conditions, reverse the order and check immediately.

Step 1: In symmetry, an equality remains true when the two sides are reversed. Step 2: If (a^2=b^2), then (b^2=a^2) is also true, so the reversed pair belongs to the relation. Step 3: For equality-based relations, write the reversed equality carefully.