Symmetric relation

Step 1: (R=R^{-1}) means every pair is accompanied by its reverse pair. Step 2: This is exactly the definition of a symmetric relation. Step 3: This condition alone does not guarantee reflexivity or transitivity.

Step 1: To disprove symmetry, find one pair whose reverse is missing. Step 2: ((1,2)) is in (R) because (1<2), but ((2,1)) is not because (2<1) is false. Step 3: One counterexample is enough to show a relation is not symmetric.

A relation R on A is reflexive when every diagonal pair (a,a) belongs to R, and it is symmetric when (a,b) in R implies (b,a) in R. Here (1,1), (2,2), and (3,3) are all present, so R is reflexive. The pair (1,2) occurs together with its reverse (2,1), and diagonal pairs reverse to themselves. Hence R is symmetric as well. Therefore option A is correct; options B and C contradict the listed pairs, while D does not state the required combined property.

Step 1: ((1,2)) has ((2,1)), and ((2,3)) has ((3,2)), so the relation is symmetric. Step 2: ((1,1),(2,2),(3,3)) are missing, so it is not reflexive. Step 3: Diagonal pairs are not compulsory for symmetry alone.

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

Step 1: If ((a,b)\in R\cup S), then it belongs to (R) or (S). Step 2: The relation containing it is symmetric, so ((b,a)) is also in that relation and hence in (R\cup S). Step 3: The union of symmetric relations is also symmetric.

Step 1: (R\setminus S) keeps pairs of (R) that are not in (S). Step 2: It is possible that ((a,b)) remains but ((b,a)) is removed because it lies in (S). Step 3: Difference of relations does not always preserve symmetry.

Step 1: ((a,c)\in R\circ R) when some (b) satisfies (aRb) and (bRc). Step 2: Here (1R2) and (2R3), so ((1,3)\in R\circ R). Step 3: In composition, look for an intermediate element, not only a direct pair.

Step 1: Suppose ((a,c)\in R\circ R). Then there is some (b) such that (aRb) and (bRc). Step 2: Since (R) is symmetric, (cRb) and (bRa), so ((c,a)\in R\circ R). Step 3: In such proofs, reuse the intermediate element in the reverse direction.

The governing test for symmetry is: whenever (a,b) belongs to R, the reversed pair (b,a) must also belong to R. If |a-b|=1, then |b-a|=|-(a-b)|=|a-b|=1. Thus every pair satisfying the condition has its reverse satisfying the same condition. For example, (1,2) and (2,1), or (3,4) and (4,3), occur together. Therefore R is symmetric. It is not reflexive because |a-a|=0, and it is not universal because many pairs have distance other than 1.

Step 1: One counterexample is enough to disprove symmetry. Step 2: (1) divides (2), so ((1,2)\in R), but (2) does not divide (1). Step 3: Divisibility relations are generally not symmetric.

Step 1: If (a^2=b^2), then reversing the equality gives (b^2=a^2). Step 2: Hence ((a,b)\in R) implies ((b,a)\in R). Step 3: Equality-based conditions are often symmetric, but always verify both directions.

Step 1: The condition is (a+b\equiv 0 \pmod{3}). Step 2: Since (a+b=b+a), we also get (b+a\equiv 0 \pmod{3}). Step 3: Congruence conditions based on addition usually remain unchanged after swapping the order.

Step 1: If (a-b\equiv 1 \pmod{4}), then (b-a\equiv -1 \pmod{4}). Step 2: Since (-1\equiv 3 \pmod{4}), the reversed pair need not satisfy the original condition. Step 3: For subtraction-based congruence, check the sign carefully.

Step 1: (a-b\equiv 0 \pmod{5}) means (a-b) is divisible by (5). Step 2: Then (b-a=-(a-b)) is also divisible by (5). Step 3: Subtraction conditions with zero remainder often give symmetry.

Step 1: In the first relation, ((2,3)) is present but its reverse ((3,2)) is missing. Step 2: So at least ((3,2)) must be added to make it symmetric. Step 3: Diagonal pairs and the empty relation are already symmetric.

Step 1: The smallest symmetric relation containing (R) is formed by adding the reverse of every given pair. Step 2: The new pairs are ((2,1),(3,2),(1,3)), so the total is (3+3=6). Step 3: Add only the required reverse pairs, not extra pairs.

Step 1: ((1,1)) is self-reverse, and ((1,2),(2,1)) are already paired. Step 2: Only the reverse of ((2,3)), namely ((3,2)), is missing. Step 3: A symmetric closure does not require adding unnecessary diagonal pairs.

A relation is symmetric if, for every pair (a,b) in the relation, the reverse pair (b,a) is also in it. The empty relation contains no ordered pairs at all. Consequently, there is no pair (a,b) that could serve as a counterexample with its reverse missing; the universal condition is therefore satisfied vacuously. Hence ∅ is symmetric on every set. It is not reflexive when A is non-empty, because reflexivity would require all pairs (a,a), and it is certainly not universal or a diagonal-only relation.

Step 1: (A\times A) contains all possible ordered pairs from (A). Step 2: Therefore, if ((a,b)) is present, ((b,a)) is also definitely present. Step 3: The universal relation is always symmetric and reflexive.