Symmetric relation

A relation is symmetric if, for every pair (a,b) in the relation, the reverse pair (b,a) is also in it. The empty relation contains no ordered pairs at all. Consequently, there is no pair (a,b) that could serve as a counterexample with its reverse missing; the universal condition is therefore satisfied vacuously. Hence ∅ is symmetric on every set. It is not reflexive when A is non-empty, because reflexivity would require all pairs (a,a), and it is certainly not universal or a diagonal-only relation.

Step 1: (A\times A) contains all possible ordered pairs from (A). Step 2: Therefore, if ((a,b)) is present, ((b,a)) is also definitely present. Step 3: The universal relation is always symmetric and reflexive.

Step 1: The identity relation contains only diagonal pairs like ((a,a)). Step 2: The reverse of such a pair is again ((a,a)). Step 3: Diagonal pairs automatically satisfy symmetry.

Step 1: Symmetry says that if a pair is in the relation, then its reverse is also in it. Step 2: From the absence of one pair, we cannot always decide about the reverse pair. Step 3: Do not reverse the direction of an implication without justification.

Step 1: If ((p,q)\in R), symmetry would imply ((q,p)\in R). Step 2: Since ((q,p)\notin R), ((p,q)) cannot be in (R). Step 3: This is a valid contrapositive use of the symmetry condition.

Step 1: In (a+b=10), swapping (a) and (b) gives (b+a=10), the same condition. Step 2: The other conditions depend on direction and usually change after swapping. Step 3: To test symmetry, swap the variables and recheck the condition.

Step 1: The possible pairs are ((1,5),(2,4),(3,3),(4,2),(5,1)). Step 2: The reverse of every pair is also in the list, so the relation is symmetric. Step 3: For fixed-sum relations, listing pairs helps avoid mistakes.

Step 1: The sum is odd when one number is odd and the other is even. Step 2: There are two odd and two even numbers, so the number of pairs is (2\cdot2+2\cdot2=8). Step 3: In ordered pairs, both orders are counted separately.

Step 1: If (a+b) is odd, then (b+a) is also odd, so symmetry holds. Step 2: For ((a,a)), (a+a=2a) is even, so no diagonal pair appears. Step 3: A symmetric relation need not be reflexive.

Step 1: An even sum means the two numbers have the same parity. Step 2: If (a) and (b) have the same parity, and (b) and (c) also have the same parity, then (a) and (c) have the same parity. Step 3: This relation is both symmetric and transitive.

Step 1: In the first relation, every pair has its reverse, so it is symmetric. Step 2: ((1,2)) and ((2,3)) are present, but ((1,3)) is missing, so it is not transitive. Step 3: Symmetry and transitivity are different properties and must be checked separately.

Step 1: In a directed graph, an arrow from (a) to (b) means ((a,b)\in R). Step 2: If every arrow has its reverse arrow, then ((b,a)\in R) also holds. Step 3: Two-way arrows are a visual sign of symmetry.

Step 1: (1\to2) and (2\to1) are already reverse arrows. Step 2: The reverse of (2\to3), namely (3\to2), is missing, so it must be added. Step 3: In graph questions, first find the missing reverse arrow.

Step 1: The total number of symmetric relations on four elements is (2^{\frac{4\cdot5}{2}}=2^{10}). Step 2: Reflexive and symmetric relations require all four diagonal pairs, leaving (2^6) choices. Step 3: Hence symmetric but not reflexive relations are (2^{10}-2^6).

Step 1: The number of symmetric relations on (n) elements is (2^{\frac{n(n+1)}{2}}). Step 2: Putting (n=5), the exponent is (\frac{5\cdot6}{2}=15). Step 3: Identify counting questions on symmetric relations before applying the formula.

Step 1: In the first option, ((1,2)) and ((2,1)) are both present, so the relation is symmetric. Step 2: ((1,1)) and ((2,2)) are missing, so it is not reflexive. Step 3: Identify properties by their conditions, not by names only.

Step 1: For a symmetric relation, (R^{-1}=R). Step 2: Hence (R\cap R^{-1}=R\cap R=R). Step 3: First simplify the inverse relation using symmetry, then perform the set operation.

Step 1: In a symmetric relation, (R^{-1}=R). Step 2: Therefore, (R\cup R^{-1}=R\cup R=R). Step 3: This idea is useful for both union and intersection questions involving inverses.

Step 1: (a^2+b^2=b^2+a^2), so swapping (a) and (b) does not change the condition. Step 2: If ((a,b)) belongs to the relation, then ((b,a)) also belongs to it. Step 3: Conditions involving sums often preserve symmetry after swapping terms.

Step 1: Take a counterexample: (a=2), (b=1). Then (a+2b=4) is even, so ((2,1)\in R). Step 2: For the reverse pair ((1,2)), (1+4=5) is odd, so ((1,2)\notin R). Step 3: One counterexample is enough to reject symmetry.

Step 1: In a distance function, the distance between two points does not change when direction is reversed. Step 2: If (d(a,b)=2), then (d(b,a)=2) also holds. Step 3: Distance-based relations are usually symmetric because of this property.

A relation R on a set A is symmetric exactly when every ordered pair in R is accompanied by its reverse ordered pair. In formal notation, for all a,b in A, (a,b) in R implies (b,a) in R. Option A states this condition without adding any unnecessary restriction. Option B is the definition of a reflexive relation, which requires all self-pairs. Option C describes transitivity, where two linked pairs force a third pair. Option D describes a relation contained in the identity relation; it does not define symmetry and is stronger than necessary. Therefore the first statement is the only correct definition.

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}}).