Symmetric relation

Step 1: Symmetry depends on every pair and its reverse pair. Step 2: Therefore, all non-diagonal pairs must be checked in the reverse direction. Step 3: Do not conclude by checking only one pair in a hurry.

Step 1: In a symmetric relation, if ((a,b)) is present, then ((b,a)) must also be present. Step 2: ((2,3)) is present but ((3,2)) is missing. The pair ((1,2)) already has ((2,1)). Step 3: In exams, always check the reverse of every non-diagonal pair.

Step 1: If (a+b) is even, then (b+a) is also even because changing the order of addition does not change the sum. Step 2: So whenever ((a,b)\in R), we also get ((b,a)\in R). Step 3: If the condition remains unchanged after swapping (a) and (b), symmetry is usually satisfied.

Step 1: (a-b) being even means (a) and (b) have the same parity. Step 2: Then (b-a) is also even, so the reverse pair also belongs to the relation. Step 3: For difference-based conditions, the sign may change, but parity remains the same.

Step 1: Symmetry only requires that every ((a,b)) has ((b,a)). Step 2: Here ((1,2)) has ((2,1)), and ((2,3)) has ((3,2)). A diagonal pair is its own reverse. Step 3: All diagonal pairs are not required for symmetry.

Step 1: In a symmetric relation, reversing an ordered pair must still give a pair in the relation. Step 2: The reverse of ((4,7)) is ((7,4)), so it must be present. Step 3: Symmetry does not automatically force diagonal pairs like ((4,4)) or ((7,7)).

Step 1: One counterexample is enough to disprove symmetry. Step 2: ((1,2)\in R) because (1\leq2), but ((2,1)\notin R) because (2\leq1) is false. Step 3: Order-based relations often fail to be symmetric.

Step 1: (|a-b|=|b-a|) is always true. Step 2: So if ((a,b)) satisfies the relation, ((b,a)) also satisfies the same condition. Step 3: Conditions based on absolute distance are usually symmetric.

Step 1: Symmetry requires ((b,a)) whenever ((a,b)) is present. Step 2: ((1,2)\in R) because (1) divides (2), but ((2,1)\notin R). Step 3: In divisibility relations, direction matters.

Step 1: If (a+b=6), then (b+a=6) also holds. Step 2: Thus the reverse ((b,a)) belongs to the relation whenever ((a,b)) does. Step 3: Sum-based equality conditions usually remain unchanged after swapping the variables.

Step 1: In a symmetric relation, the reverse of each given pair is compulsory. Step 2: From ((1,3)), we need ((3,1)), and from ((2,3)), we need ((3,2)). Step 3: Only the required reverse pairs are guaranteed, not unrelated diagonal or other pairs.

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.