Symmetric relation

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.