Symmetric relation

Step 1: For any ((a,b) \in R), (a+b) is even. Step 2: Since (b+a) has the same value, ((b,a) \in R) also belongs to (R). Step 3: In exam questions based on sum conditions, first check whether changing the order changes the condition.

Step 1: In a symmetric relation, whenever ((a,b) \in R), ((b,a) \in R) must also be present. Step 2: Here ((2,3)) is present but ((3,2)) is missing. Step 3: For such questions, look only for missing reverse pairs instead of recounting all pairs.

Step 1: The (n) diagonal pairs can be chosen independently. Step 2: The off-diagonal reverse pair groups are (\frac{n(n-1)}{2}), and each group is chosen together or not chosen. Step 3: Total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}), so the number is (2^{\frac{n(n+1)}{2}}).

Step 1: The reflexive pairs ((1,1),(2,2),(3,3)) are compulsory. Step 2: The remaining off-diagonal reverse-pair groups are three. Step 3: Each group has two choices, so the total number is (2^3=8).

Step 1: To check symmetry from a matrix, compare entries across the main diagonal. Step 2: Here (m_{13}=m_{31}), (m_{12}=m_{21}), and (m_{23}=m_{32}). Step 3: If the relation matrix equals its transpose, the relation is symmetric.

Step 1: If (|a-b|=1), then (|b-a|=1) also holds. Step 2: So every pair has its reverse pair, making the relation symmetric. Step 3: Since (|a-a|=0), the relation is not reflexive.

Step 1: ((a-b)^2=0) implies (a=b). Step 2: If (a=b), then (b=a), so symmetry holds. Step 3: For every (a), ((a-a)^2=0), so reflexivity also holds.

Step 1: If ((a,b) \in R \cap S), then the pair lies in both (R) and (S). Step 2: Since both are symmetric, ((b,a)) lies in both. Step 3: Therefore, ((b,a) \in R \cap S), so intersection preserves symmetry.

Step 1: The option contains both ((1,2)) and ((2,1)), so it is symmetric. Step 2: But ((1,2)) and ((2,1)) would require ((1,1)) for transitivity, which is absent. Step 3: Symmetry and transitivity are different properties; reverse pairs do not automatically give transitivity.

Step 1: In a symmetric relation, ((a,b) \in R) implies ((b,a) \in R). Step 2: The inverse relation (R^{-1}) contains exactly the reversed pairs. Step 3: Hence, for a symmetric relation, (R^{-1}=R).

Step 1: If (a-b) is divisible by (2), then (b-a=-(a-b)) is also divisible by (2). Step 2: Thus, each pair has its reverse pair. Step 3: In divisibility-based relations, a negative sign does not break symmetry.

Step 1: The matrix test for symmetry is (M=M^T). Step 2: If the matrix is not symmetric, then for some place (m_{ij}\neq m_{ji}). Step 3: That means some pair does not have its reverse pair, so the relation is not symmetric.

Step 1: Since (T\subseteq R), the pairs of (T) come from (R). Step 2: But (T) may include a pair while excluding its reverse pair, so symmetry can fail. Step 3: Symmetry is not always preserved when taking an arbitrary subset.

Step 1: The pairs ((1,2)) and ((2,3)) must be included. Step 2: Symmetry requires their reverse pairs ((2,1)) and ((3,2)). Step 3: The smallest symmetric relation contains only these necessary pairs.

Step 1: In a symmetric relation, if ((y,x)\in R), then ((x,y)\in R) must also be present. Step 2: The question states ((x,y)\notin R). Step 3: Therefore, ((y,x)\in R) would cause a contradiction, so ((y,x)\notin R).

Step 1: Any off-diagonal pair ((a,b)) must appear with its reverse pair ((b,a)). Step 2: So off-diagonal pairs are counted in pairs. Step 3: Hence, their number must be even, even if the total number of relation pairs is odd.

Step 1: If (a\cdot b) is even, then (b\cdot a) is the same product. Step 2: Hence, ((a,b)\in R) implies ((b,a)\in R). Step 3: For product-based relations, changing order does not change the product, which helps prove symmetry.

The inverse relation R⁻¹ contains (b,a) whenever R contains (a,b). Consider any pair (a,b) in R∪R⁻¹. If it came from R, then (b,a) belongs to R⁻¹; if it came from R⁻¹, then (b,a) belongs to R. In both cases the reverse pair is in the same union, so R∪R⁻¹ is always symmetric. It need not be reflexive or transitive, and it is empty only in the special case R=∅.

Step 1: For a symmetric relation, (R^{-1}=R). Step 2: Hence, (R\cap R^{-1}=R\cap R). Step 3: The intersection of a set with itself is the same set, so the answer is (R).

Step 1: Exactly one of the three diagonal pairs must be chosen, giving (3) choices. Step 2: There are three off-diagonal reverse-pair groups, each with two choices. Step 3: Total number is (3\cdot 2^3=24).