Symmetric relation

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.

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}).

Step 1: For (4) elements, there are (10) independent groups. Step 2: The groups containing ((1,2)) and ((3,4)) are fixed as included. Step 3: The remaining (8) groups are free, so the number is (2^8).

Step 1: A symmetric relation must contain ((2,1)) whenever it contains ((1,2)). Step 2: The condition asks for ((1,2)) to be present but ((2,1)) absent, which violates symmetry. Step 3: Hence the number of such relations is (0).

Step 1: ((a_i,a_j)) in a relation means (m_{ij}=1). Step 2: Symmetry requires ((a_j,a_i)), meaning (m_{ji}=1). Step 3: Thus (m_{ij}=m_{ji}) for all (i,j), so (M=M^T).

Step 1: ((1,2)) needs ((2,1)). Step 2: ((2,3)) needs ((3,2)), and ((3,1)) needs ((1,3)). Step 3: All three reverse pairs are missing, so the minimum number to add is (3).

Step 1: ((1,2)) and ((2,1)) already appear as a reverse pair. Step 2: The reverse of ((3,4)), namely ((4,3)), is missing. Step 3: Adding only ((4,3)) is enough and minimal.

Step 1: (a-b=0) means (a=b). Step 2: If (a=b), then (b=a), so the reverse pair also belongs. Step 3: Equality-based relations are symmetric, while conditions like (a-b=1) change when reversed.

Step 1: Symmetry is about the presence of reverse pairs. Step 2: If ((a,b)\in R), then ((b,a)\in R) must also hold. Step 3: The diagonal condition is reflexivity, and the three-pair condition is transitivity.

Step 1: The difference of two symmetric relations is symmetric. Step 2: Therefore (R-S) and (S-R) are both symmetric. Step 3: The union of two symmetric relations is symmetric, so (R\triangle S) is symmetric.

Step 1: (a\equiv -b \pmod{5}) means (a+b\equiv0 \pmod{5}). Step 2: Reversing the order gives (b+a\equiv0 \pmod{5}), the same condition. Step 3: Therefore the reverse pair also belongs, so the relation is symmetric.

Step 1: ((1,2)) and ((2,1)) are already balanced, and diagonal pairs cause no issue. Step 2: ((2,3)) is present but ((3,2)) is missing. Step 3: If only removal is allowed, remove ((2,3)), so the minimum number is (1).

Step 1: A relation is symmetric if ((a,b)\in R) implies ((b,a)\in R). Step 2: Here the condition is (a+b\text{ is even}), and (a+b=b+a), so the reverse pair also satisfies it. Step 3: In exams, check commutative conditions like addition first.