Symmetric relation

Correct answer: A. A relation on an n-element set is a subset of A × A. There are n diagonal pairs (a,a). Each diagonal pair may independently be included or omitted, giving 2^n choices. For two distinct elements, symmetry requires that (a,b) and (b,a) are treated together: either both are included or both are omitted. The number of unordered pairs of distinct elements is C(n,2)=n(n−1)/2, so these pairs contribute 2^{n(n−1)/2} choices. Multiplying the independent choices gives 2^n × 2^{n(n−1)/2} = 2^{[2n+n(n−1)]/2} = 2^{n(n+1)/2}. Therefore option A is correct. Option B omits the diagonal choices and also does not represent the symmetric pairing correctly. Option C counts every relation without imposing symmetry. Option D is not the number of subsets of ordered pairs and has no valid basis here. Memory cue: a symmetric relation is chosen by one decision for each diagonal pair and one decision for each unordered pair of distinct elements.

Correct answer: A. A relation is symmetric when every pair appears with its reverse. In option A, (1,2) is paired with (2,1), and (1,3) with (3,1), so symmetry holds. To test transitivity, use (2,1) and (1,3). Transitivity would require (2,3), but (2,3) is not in R. Hence option A is not transitive. Option B is the identity relation: whenever two pairs can be composed, the required pair is already present, so it is transitive. Option C is not symmetric because (1,2) is present but (2,1) is absent; therefore it cannot be the answer, regardless of its other property. Option D is the universal relation, containing every ordered pair, so it is both symmetric and transitive. The key strategy is to disprove transitivity with one chain xRy and yRz for which xRz is missing; one such counterexample is enough.

Correct answer: B, 2^10. A symmetric relation must include an off-diagonal pair and its reverse together, or omit both. The diagonal pairs are forbidden in this question, so they provide no choices and must not be counted. The only available decisions concern unordered pairs of distinct elements. For five elements, their number is C(5,2)=5×4/2=10. Each of these ten pairs has two choices: include both corresponding ordered pairs or include neither. Therefore the total is 2^10. Option A, 2^5, counts the five elements rather than the ten unordered pairs. Option B correctly counts the independent choices. Option C incorrectly adds an extra factor related to diagonal or directional pairs. Option D overcounts by treating more choices as independent than symmetry permits. A helpful memory cue is that forbidding all diagonal pairs removes the n diagonal decisions; for symmetric relations on n elements, the remaining count is 2^{C(n,2)}.

Step 1: In a symmetric relation, the (4) diagonal pairs are chosen independently. Step 2: The off-diagonal pairs form (\frac{4\cdot3}{2}=6) mirror-pair groups. Step 3: Total independent choices are (10), so the number of relations is (2^{10}).

Step 1: The (5) diagonal pairs are fixed, so they give no choice. Step 2: There are (\frac{5\cdot4}{2}=10) off-diagonal mirror groups. Step 3: Each group is either included fully or excluded, so the number is (2^{10}).

Step 1: A symmetric relation must contain ((b,a)) whenever ((a,b)) is present. Step 2: ((1,2)) and ((2,1)) are already paired, but ((2,3)) needs ((3,2)). Step 3: Adding only ((3,2)) makes the relation symmetric.

Step 1: If (a+b) is even, then (b+a) is also even because addition is commutative. Step 2: So ((a,b)\in R) implies ((b,a)\in R). Step 3: For such questions, reverse the pair and check whether the condition remains true.

Step 1: For ((2,1)), (2-1=1), so it belongs to the relation. Step 2: For the reverse pair ((1,2)), (1-2=-1), which is not positive. Step 3: One counterexample is enough to prove that a relation is not symmetric.

Step 1: ((a,b)\in R\cap S) means the pair is in both relations. 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 definitely symmetric.

Step 1: In a symmetric relation, every pair appears with its reverse. Step 2: The inverse relation (R^{-1}) reverses all ordered pairs. Step 3: Reversing gives the same set of pairs, so (R^{-1}=R).

Step 1: Exactly (3) diagonal pairs can be chosen in (\binom{4}{3}=4) ways. Step 2: There are (6) independent off-diagonal mirror groups. Step 3: Therefore the total number is (4\cdot2^6).

Step 1: A relation is symmetric when its matrix is symmetric about the main diagonal. Step 2: Here (m_{13}=m_{31}=1), (m_{23}=m_{32}=1), and (m_{12}=m_{21}=0). Step 3: Hence the relation is symmetric.

Step 1: For a symmetric relation matrix, (m_{ij}=m_{ji}). Step 2: Here (m_{12}=1) but (m_{21}=0), while the other opposite entries match. Step 3: So changing only (m_{21}) to (1) makes the matrix symmetric.

Step 1: If (|a-b|=1), then (|b-a|=1) also. Step 2: Hence every pair has its reverse, so the relation is symmetric. Step 3: For ((a,a)), (|a-a|=0), so it is not reflexive.

Step 1: Numbers with the same remainder still have the same remainder after reversing their order. Step 2: Thus (a\equiv b \pmod{2}) implies (b\equiv a \pmod{2}). Step 3: Symmetry is exactly about checking this reverse pair.

Step 1: Since (1\mid2), ((1,2)\in R). Step 2: But (2\mid1) is false, so ((2,1)\notin R). Step 3: One such pair is enough to disprove symmetry.

Direct answer: A. A symmetric relation must include (a,b) and (b,a) together, or exclude them together, whenever a and b are distinct. The number of unordered pairs of distinct elements is C(n,2)=n(n−1)/2. Because all diagonal pairs (a,a) are forbidden, there are no choices for them. Each of the C(n,2) off-diagonal pair-positions has two independent choices: include both directions or include neither. Therefore the number of relations is 2^[n(n−1)/2]. Option B incorrectly includes n diagonal positions as choices, even though they are forbidden. Option C treats all n² ordered pairs as independent, ignoring symmetry. Option D gives the number of positions, not the number of possible selections. Memory cue: symmetry ties two opposite arrows into one decision; forbidden diagonals contribute zero decisions.

Step 1: For (3) elements, there are (\frac{3\cdot4}{2}=6) independent groups. Step 2: If ((1,2)) is included, symmetry forces ((2,1)), fixing one group. Step 3: The remaining (5) groups are free, so the count is (2^5).

Step 1: There are (6) independent groups in total. Step 2: The group containing ((1,2)) is fixed as included, and the diagonal pair ((1,1)) is fixed as excluded. Step 3: So (4) groups remain free, giving (2^4).

Step 1: If ((a,b)\in R\cup R^{-1}), it belongs to (R) or (R^{-1}). Step 2: In either case, the reverse pair ((b,a)) belongs to (R\cup R^{-1}). Step 3: Therefore it is always symmetric.

Step 1: Changing the order in addition does not change the sum. Step 2: If (a+b) is odd, then (b+a) is also odd. Step 3: Hence every pair has its reverse, so (R) is symmetric.

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.