Symmetric relation

Step 1: If ((a,b)\in R), then (a+b=5). Step 2: Since (b+a=5) also holds, ((b,a)\in R). Step 3: In fixed-sum relations, changing the order does not change the condition, so symmetry holds.

Step 1: The pair ((1,2)) has its reverse ((2,1)). Step 2: The pair ((2,3)) has its reverse ((3,2)). Step 3: The reverse of ((1,3)), namely ((3,1)), is missing, so it must be added.

Step 1: The main condition for a symmetric relation is that every pair must have its reverse pair. Step 2: In an antisymmetric relation, reverse pairs can occur together only when the two elements are equal. Step 3: In exams, separate symmetric and antisymmetric relations by their exact conditions, not by their names.

Step 1: For (4) elements, the number of independent symmetric choices is (\frac{4(4+1)}{2}=10). Step 2: The group containing ((1,2)) is forced to be included, and the group containing ((3,4)) is forced to be excluded, so two choices are fixed. Step 3: The remaining (8) choices give (2^8=256).

Step 1: ((1,2)) is paired with ((2,1)), and ((2,3)) is paired with ((3,2)). Step 2: Diagonal pairs are their own reverses, so they do not disturb symmetry. Step 3: Since ((3,3)) is missing, reflexivity fails, but symmetry definitely holds.

Step 1: In a symmetric relation, if ((a,b) \in R), then ((b,a) \in R) must also be present. Step 2: ((1,3)) and ((3,1)) are paired, but ((2,4)) is missing its reverse ((4,2)). Step 3: In exams, always check the reverse of each unequal ordered pair.

For a set with n elements, a symmetric relation is determined by independent choices for all n diagonal pairs and for one member of each off-diagonal reverse pair. The number of such independent choices is n+n(n−1)/2=n(n+1)/2. Each chosen position may be included or omitted, giving 2^[n(n+1)/2] symmetric relations. With n=3, the exponent is 3(4)/2=6, so the answer is 2⁶. Option B counts all possible relations, not only symmetric ones.

Step 1: The condition (a+b\text{ is even}) does not change when (a) and (b) are interchanged. Step 2: If ((a,b) \in R), then ((b,a) \in R) because (b+a=a+b). Step 3: For such rules, check whether swapping the positions changes the condition.

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.