Symmetric relation

For reflexivity on A, all diagonal pairs (1,1), (2,2), and (3,3) must be present; all three are listed, so R is reflexive. For symmetry, every non-diagonal pair must have its reverse. The pair (1,2) is accompanied by (2,1), and diagonal pairs reverse to themselves. Thus R satisfies both properties, making option A correct.

Step 1: Symmetry requires ((2,1)) whenever ((1,2)) is present. Step 2: The relation contains ((1,2)) but not ((2,1)). Step 3: Diagonal pairs do not break symmetry; the missing reverse pair does.

Step 1: All (5) diagonal pairs are fixed as excluded. Step 2: There are (\frac{5\cdot4}{2}=10) independent off-diagonal mirror groups. Step 3: Each group has two choices, so the number is (2^{10}).

Step 1: Choose (2) diagonal pairs from (4) in (\binom{4}{2}) ways. Step 2: Choose (2) off-diagonal mirror groups from (6) in (\binom{6}{2}) ways. Step 3: The choices are independent, so the total is (\binom{4}{2}\binom{6}{2}).

Step 1: Symmetry says the reverse of every present pair is also present. Step 2: If ((b,a)\in R), then ((a,b)\in R) would also have to be present. Step 3: Since ((a,b)\notin R), ((b,a)\notin R) must be true.

Step 1: ((1,2)) appears with ((2,1)), and ((2,3)) appears with ((3,2)). Step 2: Thus every present non-diagonal pair has its reverse. Step 3: The diagonal pairs are missing, so it is not reflexive.

Step 1: An arrow from (x) to (y) represents ((x,y)\in R). Step 2: Symmetry requires ((y,x)\in R) as well. Step 3: Therefore an arrow in the reverse direction is necessary.

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: Relations based on equality are often checked by simply reversing the equality.

To test symmetry, assume (a,b) ∈ R. By definition, a+b≤5. Since addition is commutative, b+a has exactly the same value as a+b, so b+a≤5 as well. Therefore (b,a) ∈ R whenever (a,b) ∈ R, which proves that R is symmetric. The relation is not empty—for example, (1,1) belongs to it—and the question does not require testing reflexivity.

Step 1: Since (1\leq2), ((1,2)) belongs to the relation. Step 2: But (2\leq1) is false, so ((2,1)) does not belong. Step 3: Order-based inequality relations are often not symmetric.

Step 1: In a symmetric (R), ((a,b)) and ((b,a)) are either both present or both absent. Step 2: Therefore, in the complement, reverse pairs also appear together. Step 3: Hence (A\times A-R) is also symmetric.

Step 1: (R) contains both ((1,2)) and ((2,1)). Step 2: (S) contains both ((2,3)) and ((3,2)). Step 3: The union keeps all reverse pairs, so it is symmetric.

Step 1: If ((a,b)\in R-S), then ((a,b)\in R) and ((a,b)\notin S). Step 2: Symmetry of (R) gives ((b,a)\in R), and symmetry of (S) ensures ((b,a)\notin S). Step 3: Thus ((b,a)\in R-S), so it is always symmetric.

Step 1: (|a-b|=|b-a|) is always true. Step 2: Therefore if (|a-b|) is even, then (|b-a|) is also even. Step 3: Absolute-difference relations usually preserve symmetry because the difference value stays the same.

Step 1: If (a-b) is even, then (b-a=-(a-b)) is also even. Step 2: So whenever ((a,b)) belongs, ((b,a)) also belongs. Step 3: The key point is that the negative of an even integer is also even.

Step 1: If (a-b) is odd, then (b-a=-(a-b)) is also odd. Step 2: Thus the condition remains true after reversing the pair. Step 3: Changing the sign does not change oddness, so the relation is symmetric.

Step 1: The condition uses addition and multiplication. Step 2: If (a+b=ab), reversing the order gives (b+a=ba), which is the same condition. Step 3: When the condition is unchanged by reversing the pair, the relation is symmetric.

Step 1: For ((4,1)), (4+2\cdot1=6), so the pair belongs to the relation. Step 2: For the reverse ((1,4)), (1+2\cdot4=9), not (6). Step 3: Unequal coefficients can break symmetry when the order is reversed.

Step 1: If (a^2+b^2=10), then (b^2+a^2=10) as well. Step 2: Hence ((a,b)) comes with ((b,a)). Step 3: The sum is unchanged when the order is reversed, so the relation is symmetric.

Step 1: For (n) elements, the number of symmetric relations is (2^{\frac{n(n+1)}{2}}). Step 2: Substituting (n=6), we get (\frac{6\cdot7}{2}=21) independent choices. Step 3: Hence the total number is (2^{21}).