Symmetric relation

Step 1: If (a-b) is divisible by (2), then (b-a=-(a-b)) is also divisible by (2). Step 2: So ((a,b)\in R) implies ((b,a)\in R). Step 3: Divisibility conditions often remain symmetric after changing the sign.

Step 1: The reverse of ((1,2)) is ((2,1)), and both are present. Step 2: The reverse of ((2,3)) is ((3,2)), and ((1,1)) is its own reverse. Step 3: Symmetry does not require all possible pairs, only reverse pairs of the chosen pairs.

Step 1: In a symmetric relation, every included ordered pair needs its reverse. Step 2: ((a,b)) requires ((b,a)), and ((c,a)) requires ((a,c)). Step 3: Diagonal pairs are required for reflexivity, not for symmetry alone.

Step 1: The number of symmetric relations on a set with (n) elements is (2^{\frac{n(n+1)}{2}}). Step 2: For (n=3), the exponent is (\frac{3\cdot4}{2}=6). Step 3: Diagonal pairs are chosen independently, while non-diagonal pairs are chosen in reverse-pair blocks.

Step 1: For four elements, the number of non-diagonal reverse-pair blocks is (\frac{4\cdot3}{2}=6). Step 2: All diagonal pairs are compulsory, so only these six blocks can be included or excluded. Step 3: When reflexive and symmetric are combined, do not count diagonal choices as free.

Step 1: There are (n) diagonal positions and (\frac{n(n-1)}{2}) non-diagonal reverse-pair blocks. Step 2: Total independent choices become (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}). Step 3: While counting symmetric relations, do not treat ((a,b)) and ((b,a)) as independent choices.

Step 1: A relation is symmetric when its matrix is symmetric about the main diagonal. Step 2: Here (m_{13}=m_{31}=1), (m_{24}=m_{42}=1), and the other matching positions also agree. Step 3: In matrix questions, compare entries across the main diagonal.

Step 1: A symmetric relation has a matrix satisfying (m_{ij}=m_{ji}). Step 2: In the first matrix, all corresponding entries across the diagonal are equal. Step 3: Even one mismatched pair is enough to make the relation non-symmetric.

Step 1: In an inverse relation, every ordered pair is reversed. Step 2: In a symmetric relation, the reversed pair is already in the same relation, so (R^{-1}=R). Step 3: Remembering the link between symmetry and inverse relation saves time in exams.

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.

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.

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.