Symmetric relation

Step 1: In (R^{-1}), every ordered pair is reversed. Step 2: Symmetry means ((a,b)\in R) implies ((b,a)\in R), so reversing all pairs gives the same relation. Step 3: A symmetric relation can also be identified by (R=R^{-1}).

Step 1: The three diagonal pairs ((1,1),(2,2),(3,3)) can be chosen independently. Step 2: There are three unordered off-diagonal pair groups, and each group must be selected together or left together. Step 3: There are (3+3=6) independent choices, so the number of symmetric relations is (2^6=64).

Step 1: The number of symmetric relations on a set with (n) elements is (2^{\frac{n(n+1)}{2}}). Step 2: Here (n=4), so the number of independent choices is (\frac{4\cdot5}{2}=10). Step 3: Remember that diagonal pairs are chosen singly, while off-diagonal pairs are chosen in reverse-pair groups.

Step 1: Diagonal pairs are their own reverse, so they do not break symmetry. Step 2: The off-diagonal pair ((1,2)) has its reverse ((2,1)). Step 3: A relation is symmetric when every pair has its reverse in the relation.

Step 1: Symmetry needs the reverse of every off-diagonal pair. Step 2: In the first option, ((1,2)) and ((2,1)) are both present, and ((3,3)) is its own reverse. Step 3: While checking options, first look for missing reverse pairs.

Step 1: Swapping (a) and (b) in (a+b\leq5) gives (b+a\leq5). Step 2: Since addition is commutative, the condition remains true for the reverse pair. Step 3: Even in inequalities, if the expression is unchanged by swapping, symmetry holds.

Step 1: If (a+b) is odd, then (b+a) is also odd. Step 2: Hence the reverse pair ((b,a)) also belongs to the relation. Step 3: In sum-based parity relations, changing the order does not change the sum.

Step 1: Symmetry requires the condition to remain valid after reversing a pair. Step 2: ((2,1)\in R) because (2=2\cdot1), but ((1,2)\notin R) because (1\neq2\cdot2). Step 3: Directional multiplication conditions may fail after reversing the pair.

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 as well. Hence ((b,a)\in R\cap S). Step 3: The intersection of symmetric relations is symmetric.

Suppose (a,b) belongs to R ∪ S. Then it belongs to at least one of R or S. If it belongs to R, symmetry of R gives (b,a) ∈ R; if it belongs to S, symmetry of S gives (b,a) ∈ S. In either case, (b,a) belongs to the union. Therefore R ∪ S is symmetric, while the other options make unnecessarily restrictive claims.

Step 1: To show a relation is not symmetric, find one pair whose reverse is missing. Step 2: In the first option, ((2,3)) is present but ((3,2)) is missing. Step 3: The empty relation is symmetric because there is no pair that violates the condition.

Step 1: The symmetry rule is checked only when a pair ((a,b)) belongs to the relation. Step 2: The empty relation has no pairs, so no counterexample exists. Step 3: Many properties are considered true in such empty cases; keep this in mind for exams.

Step 1: (A\times A) contains all possible ordered pairs from (A). Step 2: If ((a,b)) is in it, then ((b,a)) is also definitely in it. Step 3: The universal relation is one of the simplest examples of a symmetric relation.

Step 1: ((1,2)) and ((2,1)) are already both present. Step 2: The reverse of ((1,3)), namely ((3,1)), is missing, so only one pair is needed. Step 3: Do not count already balanced reverse pairs again.

Step 1: For symmetry, the condition should remain valid after swapping (a) and (b). Step 2: In (a^2+b^2=10), swapping gives (b^2+a^2=10), which is the same condition. Step 3: Balanced sum-type conditions often produce symmetric relations.

Step 1: If (a^2=b^2), then the reversed equality (b^2=a^2) is also true. Step 2: Hence ((a,b)\in R) implies ((b,a)\in R). Step 3: Equality-based relations are symmetric when swapping the two sides keeps the statement true.

Step 1: Absolute difference satisfies (|a-b|=|b-a|). Step 2: If (|a-b|\leq2), then (|b-a|\leq2) also holds. Step 3: Distance-based bounds often produce symmetric relations.

Step 1: If (a+b\equiv 0 \pmod{3}), then (b+a\equiv 0 \pmod{3}) also holds. Step 2: Changing the order of addition does not change the remainder. Step 3: Modular conditions based on sums are usually symmetric.

Step 1: (a-b\equiv 0 \pmod{3}) means (a) and (b) have the same remainder modulo (3). Step 2: Then (b-a\equiv 0 \pmod{3}) also holds. Step 3: Congruence equality conditions remain valid after reversing the pair.

Step 1: Symmetry requires the reverse pair to satisfy the same condition. Step 2: For ((2,1)), (2-1\equiv 1 \pmod{3}), but for ((1,2)), (1-2\equiv 2 \pmod{3}), so the condition fails. Step 3: Non-zero modular difference conditions must be checked carefully.