Symmetric relation

Step 1: For symmetry in a relation matrix, we need (m_{ij}=m_{ji}). Step 2: Here (m_{12}=1) but (m_{21}=0), so the condition fails. Step 3: Do not decide by the diagonal alone; compare entries on both sides of it.

Step 1: The number of symmetric relations is (2^{\frac{n(n+1)}{2}}). Step 2: Here (n=5), so the count is (2^{\frac{5(6)}{2}}=2^{15}). Step 3: For larger sets, use the same rule and substitute (n) carefully.

For any relation R, the inverse R⁻¹ is formed by interchanging the coordinates of every ordered pair: (a,b) in R produces (b,a) in R⁻¹. If R is symmetric, that reversed pair already belongs to R. Consequently, every element of R⁻¹ is an element of R, and the same reasoning applied to R⁻¹ shows that every element of R is an element of R⁻¹. The two relations are therefore equal: R⁻¹ = R. Symmetry does not imply that a relation is empty or universal; it only imposes a reverse-pair condition. Also, an inverse of a relation is always a relation on the corresponding reversed sets, so option C is impossible.

Step 1: (R^{-1}) means all ordered pairs are written in reverse order. Step 2: If (R) and (R^{-1}) are equal, every pair has its reverse in (R). Step 3: This is an important alternative test for symmetry.

Step 1: Symmetry needs ((2,1)) along with ((1,2)), and option A has both. Step 2: Reflexivity needs all ((1,1),(2,2),(3,3)), which are not present. Step 3: In such questions, check both conditions separately.

Step 1: Write the reverse of every unequal ordered pair. Step 2: The reverses are ((2,1)), ((3,2)), and ((1,3)). Step 3: Symmetry requires reverse pairs, not necessarily diagonal pairs.

Step 1: If (a) divides (b), it is not necessary that (b) divides (a). Step 2: (2) divides (4), but (4) does not divide (2). Step 3: Divisibility has direction, so reverse pairs are not always present.

Step 1: For absolute value, (|a-b|=|b-a|). Step 2: If (|a-b|\le2), then (|b-a|\le2) also holds. Step 3: In absolute value relations, swapping the order often keeps the value unchanged.

Step 1: The relation is based on the same type: both even or both odd. Step 2: If ((a,b)) has the same type, then ((b,a)) also has the same type. Step 3: In same-class relations, changing order does not change the rule.

Step 1: ((1,2)) has ((2,1)), and ((1,3)) has ((3,1)). Step 2: ((2,2)) is its own reverse, so it creates no issue. Step 3: For symmetry, check reverses of existing pairs only, not pairs that are absent.

Step 1: A symmetric extension keeps all given pairs and adds only the missing reverse pairs. Step 2: The reverse of ((2,4)) is ((4,2)), and it is missing, while other unequal pairs already have reverses. Step 3: For the smallest extension, add only the compulsory reverse pairs.

Step 1: Reflexivity forces all diagonal pairs to be included. Step 2: For symmetry, we choose from (n(n-1)/2) unordered off-diagonal pair groups, so the count is (2^{\frac{6\cdot5}{2}}=2^{15}). Step 3: When reflexivity is fixed, do not count diagonal choices separately.

Step 1: No diagonal pair means every pair of the form ((a,a)) is absent. Step 2: Only the (5\cdot4/2=10) unordered off-diagonal pair groups remain optional. Step 3: In symmetric relations, off-diagonal pairs are selected together with their reverses.

Step 1: A symmetric relation can have a subset that does not contain all reverse pairs. Step 2: For example, (R={(1,2),(2,1)}) is symmetric, but (S={(1,2)}) is not symmetric. Step 3: Do not assume that every property passes to subsets.

Step 1: If ((a,b)\in R-S), then ((a,b)\in R) and ((a,b)\notin S). Step 2: Since (R) is symmetric, ((b,a)\in R), and since (S) is symmetric, ((b,a)\in S) would imply ((a,b)\in S), which is false. Step 3: In difference proofs, handle membership and non-membership together.

Step 1: For inverse relations, ((R\circ S)^{-1}=S^{-1}\circ R^{-1}). Step 2: Since (R) and (S) are symmetric, (R^{-1}=R) and (S^{-1}=S), so ((R\circ S)^{-1}=S\circ R). Step 3: If (R\circ S=S\circ R), then the composite equals its inverse and is symmetric.

Step 1: If (a+b=0), then after swapping the order, (b+a=0) is also true. Step 2: Hence ((a,b)\in R) implies ((b,a)\in R). Step 3: In sum-based rules, the order does not change the sum, so symmetry is often easier to detect.

Step 1: One counterexample is enough to disprove symmetry. Step 2: ((2,1)\in R) because (2-1>0), but ((1,2)\notin R) because (1-2<0). Step 3: In inequality relations, reversing the order may change the sign.

Step 1: Find all pairs for which (|a-b|=2). Step 2: The pairs are ((1,3),(3,1),(2,4),(4,2)), so the total is (4). Step 3: In absolute value relations, remember to count both directions.

Step 1: To check symmetry from a matrix, compare (m_{ij}) with (m_{ji}). Step 2: Here (m_{12}=m_{21}=1), (m_{13}=m_{31}=1), and (m_{23}=m_{32}=0). Step 3: If entries on both sides of the main diagonal match, the relation is symmetric.