Symmetric relation

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

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

Step 1: If ((a,b)) is in the complement, then ((a,b)\notin R). Step 2: If ((b,a)) were in (R), symmetry would force ((a,b)\in R), a contradiction. Step 3: Therefore ((b,a)) is also in the complement.

Step 1: Both ((1,2)) and ((2,1)) are present. Step 2: Both ((2,3)) and ((3,2)) are present, and diagonal pairs are their own reverses. Step 3: Missing ((3,3)) prevents reflexivity, but not symmetry.

The key distinction is between symmetry and reflexivity. Symmetry requires that (a,b) in R imply (b,a) in R. Reflexivity requires that (a,a) belong to R for every a in the underlying set. These conditions are independent: the empty relation is symmetric because it has no counterexample, but on a non-empty set it is not reflexive. The universal relation is both symmetric and reflexive, and the identity relation is also symmetric because each pair (a,a) equals its own reverse. Since a symmetric relation can lack all self-pairs, the claim that every symmetric relation is reflexive is false. Therefore option A is correct.

Step 1: Symmetry fails only when some ((a,b)) is present but ((b,a)) is absent. Step 2: The empty relation has no pair, so there is no violation. Step 3: This is an example of a vacuously true statement.

Step 1: The universal relation contains all ordered pairs from (A\times A). Step 2: Therefore, whenever ((a,b)) is present, ((b,a)) is also present. Step 3: Having all pairs makes symmetry automatic.

Step 1: The identity relation contains only pairs of the form ((a,a)). Step 2: The reverse of such a pair is the same pair ((a,a)). Step 3: Hence the identity relation is always symmetric.

Step 1: (a\equiv b \pmod{2}) means both have the same parity. Step 2: If (a) has the same parity as (b), then (b) has the same parity as (a). Step 3: Congruence-type conditions are usually symmetric.

Step 1: (a\equiv b \pmod{3}) means (a) and (b) have the same remainder when divided by 3. Step 2: Having the same remainder works in both directions. Step 3: Swapping the order does not break congruence.

Step 1: Both ((1,2)) and ((2,1)) are present, so the relation is symmetric. Step 2: But ((1,1)) and ((2,2)) are missing, so it is not reflexive. Step 3: Symmetry and reflexivity are separate properties.

Step 1: ((1,1)) and ((2,2)) are present, so the relation is reflexive. Step 2: ((1,2)) is present but ((2,1)) is missing, so it is not symmetric. Step 3: In mixed-property questions, test each property separately.

Step 1: ((1,2)) and ((2,1)) are present. Step 2: ((2,3)) and ((3,2)) are also present, but the reverse of ((1,3)), namely ((3,1)), is missing. Step 3: Add only the missing reverse pair.

Step 1: Symmetry requires the reverse of each given pair. Step 2: The reverse of ((1,2)) is ((2,1)), and the reverse of ((2,3)) is ((3,2)). Step 3: For a minimum completion, add only the required reverse pairs.

Step 1: If (ab) is even, then (ba) is also even because multiplication is commutative. Step 2: Hence the reverse pair also belongs to the relation. Step 3: For product-based conditions, check what happens after swapping.

Step 1: If (a+b=5), then (b+a=5) as well. Step 2: Therefore the reverse of every related pair also belongs to the relation. Step 3: The key signal is that changing the order does not change the sum.

Step 1: ((3,1)) belongs to the relation because (3-1=2). Step 2: Its reverse ((1,3)) does not satisfy (1-3=2). Step 3: A fixed directed difference usually breaks symmetry.

Step 1: If (a>b), the reversed condition (b>a) is generally false. Step 2: The other conditions remain true after swapping. Step 3: Directional comparisons are a common source of non-symmetry.

Step 1: Symmetry guarantees only the reverse pairs of the given pairs. Step 2: ((2,5)) gives ((5,2)), and ((5,7)) gives ((7,5)). Step 3: Symmetry does not create a new connecting pair like ((2,7)).

Step 1: Symmetry is based on reverse ordered pairs. Step 2: Having every diagonal pair is the condition for reflexivity, not symmetry. Step 3: Mixing property names is a common exam mistake.