Symmetric relation

Step 1: The basic condition of symmetry is based on reverse pairs. Step 2: Therefore the main check is whether the reverse of every present pair belongs to the relation. Step 3: Do not decide only by counting pairs or looking at the size.

Step 1: In a symmetric relation, whenever ((a,b) \in R), ((b,a) \in R) must also be present. Step 2: Here ((1,2)) is paired with ((2,1)), and diagonal pairs cause no problem. Step 3: In exams, always check each reverse ordered pair.

Step 1: Symmetry requires the reverse pair of every ordered pair. Step 2: ((1,2)) and ((2,1)) are both present, but the reverse of ((2,3)), which is ((3,2)), is missing. Step 3: The quickest exam method is to locate missing reverse pairs.

Step 1: If (a+b) is even, then (b+a) is also even because addition is commutative. Step 2: So whenever ((a,b)) belongs to the relation, ((b,a)) also belongs. Step 3: For condition-based relations, swap the order and test the condition.

Step 1: If (a-b) is odd, then (b-a=-(a-b)) is also odd. Step 2: Hence the reverse pair also satisfies the condition. Step 3: Remember that the negative of an odd integer is also odd.

Step 1: If (a=b), then the reverse pair also satisfies the same equality. Step 2: Conditions like (a<b), divisibility, and (a=b+1) usually fail after swapping. Step 3: Relations built on equality are commonly symmetric.

Step 1: In a symmetric relation, the reverse of every ordered pair must also be present. Step 2: The reverse of ((5,8)) is ((8,5)). Step 3: Symmetry does not force diagonal pairs to be present.

Step 1: The definition of symmetry is directly about the reverse ordered pair. Step 2: It does not necessarily require ((a,a)), ((b,b)), or (a=b). Step 3: Remember the exact definition to avoid common traps.

Step 1: ((1,2)) has ((2,1)), and ((1,3)) has ((3,1)). Step 2: ((2,2)) is its own reverse, so it creates no issue. Step 3: Diagonal pairs do not need a separate reverse pair.

Step 1: ((3,1)\in R). Step 2: For symmetry, its reverse ((1,3)) must also be in (R), but it is absent. Step 3: Missing diagonal pairs do not necessarily break symmetry.

Step 1: The 4 diagonal pairs can be chosen independently. Step 2: The remaining (\frac{4(4-1)}{2}=6) unordered off-diagonal pairs are chosen as reverse-pair blocks. Step 3: Total independent choices are (4+6=10), so the number is (2^{10}).

Step 1: There are (n) independent diagonal pairs. Step 2: Off the diagonal, there are (\frac{n(n-1)}{2}) reverse-pair blocks. Step 3: Total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}), so the answer is (2^{\frac{n(n+1)}{2}}).

Step 1: Here (n=2). Step 2: The number of symmetric relations is (2^{\frac{n(n+1)}{2}}=2^{\frac{2\cdot3}{2}}=2^3). Step 3: For small sets, applying the formula gives a quick answer.

Step 1: If (a<b), then generally (b<a) is false. Step 2: The other conditions remain true after swapping the order. Step 3: Directional inequalities often break symmetry.

Step 1: To test symmetry using a matrix, compare entries across the main diagonal. Step 2: Here the ((1,3)) and ((3,1)) positions both contain 1, and the other matching positions also agree. Step 3: A relation is symmetric when its matrix equals its transpose.

Step 1: The matrix of a symmetric relation is symmetric about the main diagonal. Step 2: In option A, the upper-right and lower-left entries are both 1. Step 3: For a (2\times2) matrix, checking those two off-diagonal entries is enough.

The governing concept is the inverse of a relation. If R is a relation from A to A, then R⁻¹ consists of all pairs (b,a) for which (a,b) belongs to R. Symmetry says precisely that whenever (a,b) is in R, its reverse (b,a) is also in R. Thus every pair placed in R⁻¹ is already in R, giving R⁻¹ ⊆ R. Applying the same argument to the reverse direction, or simply reversing the equality, gives R ⊆ R⁻¹ as well. Therefore R⁻¹ = R. The inverse need not be empty, universal, or a proper subset; those descriptions depend on the particular relation.

Step 1: (R^{-1}=R) means the relation remains the same after reversing all pairs. Step 2: Hence whenever ((a,b)) is present, ((b,a)) is also present. Step 3: This is an equivalent test for symmetry.

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

Step 1: ((1,3)) belongs to the relation because (1+2=3). Step 2: Its reverse ((3,1)) does not satisfy (3+2=1). Step 3: A one-direction equality condition is usually not symmetric.

Step 1: (a-a=0) is even, so the relation is reflexive. Step 2: If (a-b) is even, then (b-a) is also even, so the relation is symmetric. Step 3: Test each property separately in exam questions.

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.