Symmetric relation

Step 1: A symmetric extension keeps all given pairs and adds only the missing reverse pairs. Step 2: The reverse of ((2,4)) is ((4,2)), and it is missing, while other unequal pairs already have reverses. Step 3: For the smallest extension, add only the compulsory reverse pairs.

Step 1: Reflexivity forces all diagonal pairs to be included. Step 2: For symmetry, we choose from (n(n-1)/2) unordered off-diagonal pair groups, so the count is (2^{\frac{6\cdot5}{2}}=2^{15}). Step 3: When reflexivity is fixed, do not count diagonal choices separately.

Step 1: No diagonal pair means every pair of the form ((a,a)) is absent. Step 2: Only the (5\cdot4/2=10) unordered off-diagonal pair groups remain optional. Step 3: In symmetric relations, off-diagonal pairs are selected together with their reverses.

Step 1: A symmetric relation can have a subset that does not contain all reverse pairs. Step 2: For example, (R={(1,2),(2,1)}) is symmetric, but (S={(1,2)}) is not symmetric. Step 3: Do not assume that every property passes to subsets.

Step 1: If ((a,b)\in R-S), then ((a,b)\in R) and ((a,b)\notin S). Step 2: Since (R) is symmetric, ((b,a)\in R), and since (S) is symmetric, ((b,a)\in S) would imply ((a,b)\in S), which is false. Step 3: In difference proofs, handle membership and non-membership together.

Step 1: For inverse relations, ((R\circ S)^{-1}=S^{-1}\circ R^{-1}). Step 2: Since (R) and (S) are symmetric, (R^{-1}=R) and (S^{-1}=S), so ((R\circ S)^{-1}=S\circ R). Step 3: If (R\circ S=S\circ R), then the composite equals its inverse and is symmetric.

Step 1: If (a+b=0), then after swapping the order, (b+a=0) is also true. Step 2: Hence ((a,b)\in R) implies ((b,a)\in R). Step 3: In sum-based rules, the order does not change the sum, so symmetry is often easier to detect.

Step 1: One counterexample is enough to disprove symmetry. Step 2: ((2,1)\in R) because (2-1>0), but ((1,2)\notin R) because (1-2<0). Step 3: In inequality relations, reversing the order may change the sign.

Step 1: Find all pairs for which (|a-b|=2). Step 2: The pairs are ((1,3),(3,1),(2,4),(4,2)), so the total is (4). Step 3: In absolute value relations, remember to count both directions.

Step 1: To check symmetry from a matrix, compare (m_{ij}) with (m_{ji}). Step 2: Here (m_{12}=m_{21}=1), (m_{13}=m_{31}=1), and (m_{23}=m_{32}=0). Step 3: If entries on both sides of the main diagonal match, the relation is symmetric.

Step 1: Exactly two diagonal pairs can be selected in (\binom{4}{2}=6) ways. Step 2: There are (4\cdot3/2=6) unordered off-diagonal pair groups, giving (2^6) choices. Step 3: Multiply the diagonal choices and off-diagonal symmetric choices.

Step 1: All three given pairs are off-diagonal. Step 2: Symmetry forces ((2,1),(3,2),(4,3)), making (6) off-diagonal pairs in total. Step 3: For a minimum count, do not add extra diagonal or unrelated pairs.

Step 1: For a symmetric relation, (R^{-1}=R). Step 2: Therefore, (R\cup R^{-1}=R\cup R=R). Step 3: In inverse-relation questions, identify (R^{-1}) first.

The governing idea is symmetric closure. The inverse relation R⁻¹ contains the reverse pair for every pair in R. Let S = R ∪ R⁻¹. If (a,b) belongs to S, then either (a,b) belongs to R or it belongs to R⁻¹. In the first case, (b,a) belongs to R⁻¹; in the second case, (b,a) belongs to R. Thus in both cases the reverse pair belongs to S, proving that S is symmetric. The union need not be reflexive because R may contain no self-pairs, and it need not be empty. Since the union of two relations on the same sets is itself a relation, option D is also false. Hence A is always true.

Step 1: To make (R) symmetric, every pair needs its reverse. Step 2: (R^{-1}) provides exactly those reverse pairs, so (R\cup R^{-1}) contains all required pairs. Step 3: When the word smallest appears, reject options that add unnecessary pairs.

Step 1: (R\cap R^{-1}=R) implies (R\subseteq R^{-1}). Step 2: This means every pair of (R) has its reverse in (R), so (R) is symmetric. Step 3: Inclusion statements such as (R\subseteq R^{-1}) are useful for proving symmetry.

Step 1: In a relation matrix, symmetry requires (m_{ij}=m_{ji}). Step 2: Here (m_{12}=0) but (m_{21}=1), so the relation is not symmetric. Step 3: One mismatched pair of matrix entries is enough to break symmetry.

Step 1: In a symmetric relation, the number of off-diagonal pairs is always even because they appear in reverse pairs. Step 2: If there are (3) diagonal pairs out of (10), the off-diagonal count is (7), which is not even. Step 3: Checking parity of total and diagonal counts is a fast exam method.

Step 1: Swapping (a) and (b) in (a^2+b^2=25) gives (b^2+a^2=25). Step 2: The value does not change, so the reverse pair also belongs to the relation. Step 3: Rules based on sums of squares are often directly symmetric.

Step 1: (a^2-b^2=0) means (a^2=b^2). Step 2: If (a^2=b^2), then (b^2=a^2), so ((b,a)) also belongs to the relation. Step 3: Equality remains true when the two sides are interchanged.

Step 1: For ((3,1)), (3=1+2) is true, so the pair belongs to the relation. Step 2: For the reverse ((1,3)), (1=3+2) is false. Step 3: In direction-dependent rules, always test the reverse pair separately.

Step 1: The (n) diagonal pairs can be chosen independently. Step 2: Off-diagonal pairs are chosen in reverse-pair groups, and there are (\frac{n(n-1)}{2}) such groups. Step 3: The total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}).

Step 1: In a symmetric relation, the number of off-diagonal pairs is even. Step 2: Since the total is (5), the number of diagonal pairs must be odd. With (3) elements, possible diagonal counts are at most (3), so (1) is possible. Step 3: Apply parity first, then check the maximum limit.

Step 1: Since (R) is symmetric, (R^{-1}=R). Step 2: Therefore (R\cup R^{-1}=R), and (R\subseteq T\subseteq R) gives (T=R). Step 3: In inclusion-based questions, simplify the relation expressions first.

Step 1: One off-diagonal pair group means ((a,b)) and ((b,a)) together. Step 2: Therefore one group gives two ordered pairs. Step 3: In symmetric relations, count off-diagonal selections as groups, not single pairs.