Symmetric relation

Step 1: Symmetry requires the reverse of every ordered pair to be present. Step 2: Since (1\le2), ((1,2) \in R), but (2\le1) is false, so ((2,1) \notin R). Step 3: To disprove symmetry, one valid counterexample is enough.

Step 1: The number of symmetric relations on a set with (n) elements is (2^{\frac{n(n+1)}{2}}). Step 2: For (n=4), this becomes (2^{\frac{4(5)}{2}}=2^{10}). Step 3: Do not confuse the total number of pairs in (A\times A) with the count of symmetric relations.

Step 1: A relation is symmetric if its matrix is symmetric about the main diagonal. Step 2: Here (m_{13}=1) matches (m_{31}=1), (m_{12}=0) matches (m_{21}=0), and (m_{23}=0) matches (m_{32}=0). Step 3: In matrix questions, compare (m_{ij}) with (m_{ji}).

Step 1: In a symmetric relation, every ((a,b)) must be accompanied by ((b,a)). Step 2: The reverse of ((5,8)) is ((8,5)), so it must be present. Step 3: Symmetry alone does not force pairs like ((a,a)) to exist.

Step 1: ((1,2)) has ((2,1)) and ((2,3)) has ((3,2)), so symmetry is satisfied. Step 2: Reflexivity would need ((1,1),(2,2),(3,3)), which are missing. Step 3: Check symmetry and reflexivity separately.

Step 1: If (x-y) is odd, then (y-x=-(x-y)) is also odd. Step 2: Hence whenever ((x,y)) is in the relation, ((y,x)) will also be in it. Step 3: In difference-based parity rules, changing the sign does not change oddness or evenness.

Step 1: If ((a,b) \in R\cap S), then ((a,b)) belongs to both (R) and (S). Step 2: Since both are symmetric, ((b,a)) belongs to both, so it belongs to (R\cap S). Step 3: For intersection proofs, check membership in both relations separately.

Step 1: If ((a,b) \in R\cup S), then it belongs to (R) or (S). Step 2: The relation containing ((a,b)) is symmetric, so ((b,a)) is also in that relation and hence in (R\cup S). Step 3: In union proofs, track that the pair belongs to at least one relation.

Step 1: The pair ((1,3)) is present in the relation. Step 2: Symmetry requires its reverse ((3,1)), but it is missing. Step 3: While reading the given pairs, first identify any pair whose reverse is absent.

Step 1: If (a^2=b^2), then by equality (b^2=a^2) is also true. Step 2: Therefore, whenever ((a,b) \in R), ((b,a) \in R) too. Step 3: Rules based on equality often remain symmetric when the order is reversed.

Step 1: To test symmetry, check one pair and its reverse. Step 2: ((4,1)) belongs to the relation because (4-1=3), but ((1,4)) does not because (1-4=-3). Step 3: Relations based on a fixed directed difference are usually not symmetric.

Step 1: If (a) and (b) have the same remainder on division by (2), the same statement holds for (b) and (a). Step 2: Hence ((a,b)) implies ((b,a)). Step 3: Relations based on sameness or common class are good candidates for symmetry.

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.

Let S=(A×A)−R be the complement of R in A×A. To test symmetry, assume (a,b)∈S, so (a,b)∉R. If (b,a) were in R, then symmetry of R would imply (a,b)∈R, contradicting the assumption. Therefore (b,a)∉R and hence (b,a)∈S. Thus S is symmetric. The complement need not be empty or diagonal-only, and it can be symmetric even when R is neither universal nor empty.

Direct answer: A, n(n+1)/2. To specify a symmetric relation, treat each diagonal pair (a,a) as one position; there are n of these. For two distinct elements a and b, symmetry links (a,b) with (b,a), so the two ordered pairs form one unordered decision-position. The number of such distinct-element pairs is n(n−1)/2. Adding the diagonal positions gives n+n(n−1)/2 = [2n+n²−n]/2 = n(n+1)/2. Option C counts only the off-diagonal positions and forgets the n diagonal choices. Option B counts all ordered positions without identifying opposite pairs. Option D counts only diagonal positions. Memory cue: symmetric positions = diagonal positions + one position for every pair of distinct elements.