Symmetric relation

Step 1: In a symmetric relation, if ((a,b) \in R), then ((b,a) \in R) must also be present. Step 2: Here ((1,2)) has ((2,1)), and ((2,3)) has ((3,2)). Step 3: In exams, always check reverse pairs for pairs with different elements.

Step 1: ((1,2)) and ((2,1)) are already reverse pairs. Step 2: The reverse of ((3,4)) is ((4,3)), which is missing. Step 3: To make a relation symmetric, add only the missing reverse pairs.

There are n diagonal positions, each representing one self-pair, and n(n−1)/2 unordered pairs of distinct elements. Each distinct pair must be chosen as a whole: either both orientations occur or neither occurs. Thus the total independent positions are n+n(n−1)/2=n(n+1)/2, yielding the corresponding power of two relations.

Step 1: Here (n=4). Step 2: The number of symmetric relations is (2^{\frac{n(n+1)}{2}}). Step 3: (2^{\frac{4\cdot5}{2}}=2^{10}=1024), so the correct number is (1024).

Step 1: Symmetry requires the reverse of every ordered pair to be present. Step 2: ((2,3)) is present, but ((3,2)) is not present. Step 3: If even one required reverse pair is missing, the relation is not symmetric.

Step 1: A symmetric relation is identified by reverse pairs. Step 2: Since ((5,8)) is present, ((8,5)) must be present. Step 3: Remember that symmetry alone does not guarantee diagonal pairs.

Step 1: In (R^{-1}), every ordered pair is reversed. Step 2: In a symmetric relation, those reverse pairs already belong to the same relation. Step 3: Hence a useful test for symmetry is checking whether (R^{-1}=R).

Step 1: If (a-b) is even, then (b-a=-(a-b)) is also even. Step 2: Thus ((a,b) \in R) implies ((b,a) \in R). Step 3: For number-based relations, check whether the condition remains true after reversing the pair.

Step 1: Symmetry requires ((b,a)) whenever ((a,b)) is present. Step 2: (1\leq2) is true, so ((1,2)) is in the relation, but (2\leq1) is false. Step 3: One counterexample is enough to prove that a relation is not symmetric.

Step 1: The condition for symmetry applies to every pair that is present. Step 2: The empty relation has no pair, so there is no violation. Step 3: Remember that the empty relation is considered symmetric.

Step 1: (A\times A) contains all possible ordered pairs from (A). Step 2: If ((a,b)) is present, ((b,a)) is also certainly present. Step 3: The universal relation is a direct example of a symmetric relation.

Step 1: The reverse of ((a,a)) is again ((a,a)). Step 2: The identity relation contains only such diagonal pairs, so each reverse pair is already present. Step 3: Diagonal pairs never create a problem for symmetry.

Step 1: To test symmetry from a matrix, compare entries across the main diagonal. Step 2: Here (m_{13}=m_{31}=1), (m_{12}=m_{21}=0), and (m_{23}=m_{32}=0). Step 3: If (M=M^T), the relation is symmetric.

Step 1: For a symmetric relation matrix, (m_{ij}=m_{ji}) must hold. Step 2: Here (m_{12}=1), but (m_{21}=0), so the equality fails. Step 3: In matrix-based questions, compare entries on both sides of the main diagonal.

Step 1: In a directed graph, ((a,b)) is shown by an arrow from (a) to (b). Step 2: Symmetry requires the reverse arrow from (b) to (a). Step 3: In graph questions, carefully check every two-way pair.

Step 1: If (a+b) is even, then (b+a) is also even because changing the order does not change the sum. Step 2: So ((a,b)) being in the relation implies ((b,a)) is also in it. Step 3: Conditions based on commutative operations like addition or multiplication often give symmetry.

Step 1: ((a,b)) is in the relation when (a-b=3). Step 2: The reverse pair ((b,a)) would require (b-a=3), which is not generally true. Step 3: For example, ((4,1)) is in the relation, but ((1,4)) is not.

Step 1: Diagonal pairs are their own reverse pairs. Step 2: Both ((1,3)) and ((3,1)) are present, so the non-diagonal pair is balanced. Step 3: Symmetry does not require all possible pairs; it only requires reverse pairs for the pairs present.

Step 1: If ((a,b) \in R\cap S), then it belongs to both (R) and (S). Step 2: Since both are symmetric, ((b,a)) belongs to both. Step 3: Hence ((b,a) \in R\cap S), so the intersection is symmetric.

Step 1: If ((a,b) \in R\cup S), it belongs to at least one of the relations. Step 2: In that relation, symmetry gives ((b,a)). Step 3: Therefore ((b,a)) also belongs to (R\cup S), so the union is symmetric.