Relations that are reflexive and symmetric

Answer: B, 8. Reflexivity fixes all three diagonal pairs: (1,1), (2,2), and (3,3) must be present. For symmetry, an off-diagonal pair must occur together with its reverse. The three independent unordered pairs are {1,2}, {1,3}, and {2,3}. For each one, we have two choices: include both ordered versions or include neither. Thus the number of choices is 2×2×2=2^3=8. Option A misses one independent choice, while option C counts four independent choices that do not exist. Option D treats all six off-diagonal ordered pairs as independent, but symmetry links them in reverse-direction pairs. Memory cue: diagonal pairs are fixed by reflexivity; only one decision is made for each unordered pair of different elements.

A relation R is symmetric when every ordered pair (a,b) in R is accompanied by its reverse pair (b,a) in R. The relation contains (1,2), and its reverse (2,1) is also present, so that pair causes no problem. However, R contains (2,3), while the required reverse pair (3,2) is absent. This single failure is enough to show that R is not symmetric, making option A correct. The absence of (1,1) is irrelevant because a self-reverse pair would be its own reverse, and symmetry does not require every possible pair to exist. Option B and C are false because both pairs are actually listed.

The governing concept is a symmetric relation. A relation R is symmetric when, whenever (a,b) belongs to R, the reversed pair (b,a) also belongs to R. For the pair (1,4), its reverse (4,1) is explicitly present. For (4,1), its reverse (1,4) is also present. The pair (2,2) reverses to itself, so it automatically satisfies the condition. Every pair in R therefore has its required reverse, and R is symmetric. Hence option A is correct. The relation need not contain every possible pair, and symmetry is different from reflexivity: reflexivity would require all relevant self-pairs, which is not the condition being tested here.

A relation on A is reflexive when every element is related to itself; that is, (a,a) must belong to R for every a in A. Here (1,1), (2,2), and (3,3) are all present, so R is reflexive. It is not symmetric, because (1,2) belongs to R but its reverse pair (2,1) does not. It is also transitive in this particular case, but the question asks for a valid type and option A states the required property. Option B is false because the reverse pair is missing, option C is incomplete and misleading, and option D is false because reflexivity is clearly established.

To test reflexivity, check whether every diagonal pair is present. The pairs (1,1), (2,2), and (3,3) are all in R, so the relation is reflexive. To test symmetry, every pair (a,b) must be accompanied by (b,a). The non-diagonal pair (1,2) is accompanied by (2,1), and diagonal pairs are automatically their own reverses. Thus R is symmetric as well. Therefore option A is correct. Option B is false because no diagonal pair is missing, and option C is false because both directions of the non-diagonal pair are present. Option D does not provide the correct classification and incorrectly focuses on a negated property.

The relation defined by a≤b is reflexive because every number is less than or equal to itself: a≤a for all a in A. It is also transitive: whenever a≤b and b≤c, combining the inequalities gives a≤c. Thus it must be both reflexive and transitive, so option A is correct. It is not generally symmetric; for example, 1≤2 is true, but 2≤1 is false. Therefore option B cannot be right, and option C is also false. Option D contradicts both properties that follow directly from the order relation. These are standard properties of the usual less-than-or-equal relation.

The governing concept is an equivalence class under congruence modulo 3. Two integers are congruent modulo 3 when their difference is divisible by 3, or equivalently when they leave the same remainder after division by 3. The class of 1 contains all elements of A that have the same remainder as 1. Since 1 leaves remainder 1 and 4 also leaves remainder 1, and 4-1=3 is divisible by 3, both numbers belong to the class. The numbers 2 and 5 leave remainder 2, while 3 and 6 leave remainder 0. Hence [1]={1,4}, so option A is correct. The other options represent different residue classes or combine incompatible remainders.

The identity relation on a set A is defined as I_A={(a,a):a∈A}; it contains exactly the diagonal ordered pairs and no pairs connecting different elements. Since A={1,2,3} and R contains precisely (1,1), (2,2), and (3,3), R is the identity relation. Therefore option C is correct. It is not empty because it contains three pairs, and it is not universal because A×A would contain all nine possible ordered pairs. It is also not asymmetric: every relation containing a pair (a,a) fails asymmetry, since asymmetry prohibits self-related pairs. In fact, the identity relation is both reflexive and symmetric.

A relation R is symmetric when, for every ordered pair (a, b) in R, the reversed pair (b, a) is also in R. In the given relation, the reverse of (1, 2) is (2, 1), and both pairs are present. The pair (3, 3) is its own reverse because interchanging equal coordinates leaves it unchanged. Thus every pair has its reverse in R, so the relation is symmetric and option A is correct. Option B is false because (1, 2) has unequal components. Option C describes a universal relation containing every pair of A × A, which is not required for symmetry. Option D is false because three pairs are explicitly listed.

Reflexivity follows because for every a in A, |a − a| = 0 ≤ 1, so (a,a) belongs to R. Symmetry follows from absolute value: |a − b| = |b − a|, so membership of (a,b) always gives membership of (b,a). Transitivity fails. In particular, (1,2) belongs to R because |1−2|=1, and (2,3) belongs because |2−3|=1, but (1,3) does not belong because |1−3|=2>1. Thus the relation is reflexive and symmetric but not transitive. It cannot be an equivalence relation, since transitivity is required, and it is not antisymmetric because both (1,2) and (2,1) are present while 1 ≠ 2. Therefore A is correct.

The identity relation on a set A is defined by I_A = {(a,a): a ∈ A}. It relates every element to itself and to no different element. For A = {1,2,3}, substituting each element gives exactly the three ordered pairs (1,1), (2,2), and (3,3), so option A is the identity relation. Option B includes cross-pairs such as (1,2) and (2,1), which are not allowed in an identity relation. Option C relates 1 to other elements and therefore also fails the definition. The empty relation in option D contains none of the required self-pairs. The identity relation is reflexive, symmetric, and transitive, but those properties alone do not define its complete list; the decisive feature is that it contains precisely all diagonal pairs and no off-diagonal pair.

The relation compares the squares of two elements. It is reflexive because a² = a² for every a in A. It is symmetric because a² = b² immediately implies b² = a². For transitivity, if a² = b² and b² = c², then both a² and c² equal b², so a² = c²; hence aRc. Therefore all three conditions for an equivalence relation hold. Its equivalence classes are {-2,2} and {-1,1}, since numbers with opposite signs can have the same square. The relation is not antisymmetric: for example, 2R(-2) and (-2)R2, but 2 ≠ -2. Thus option B is false, while C and D contradict the direct proofs of reflexivity and transitivity. Option A is correct.

The condition 4 divides a − b is equivalent to saying that a and b leave the same remainder when divided by 4. Congruence modulo 4 is an equivalence relation: it is reflexive because a−a=0, symmetric because divisibility of a−b implies divisibility of b−a, and transitive because sums of multiples of 4 are multiples of 4. The possible remainders are 0, 1, 2, and 3. Within A the classes are {0,4}, {1,5}, {2,6}, and {3,7}. Thus there are exactly four classes, so option B is correct. The number 8 is the number of elements, not classes; 2 and 6 do not represent the distinct residue classes modulo 4.

The governing concepts are reflexivity, symmetry, and transitivity. A relation is reflexive when every element is related to itself, symmetric when aRb implies bRa, and transitive when aRb and bRc imply aRc. Here every element divides itself, so a | a for all a in A; hence R is reflexive. If a | b and b | c, then b = am and c = bn for some integers m,n, so c = a(mn), proving a | c and therefore transitivity. However, symmetry fails because 1 | 2 is true but 2 does not divide 1. Thus option A is correct. Option C is false because an equivalence relation must also be symmetric, while option D incorrectly denies reflexivity and transitivity.

A relation on a three-element set is a subset of A × A, whose ordered pairs can be represented by a 3 × 3 matrix. Reflexivity requires all three diagonal pairs, (a,a), to be present; these choices are compulsory. For symmetry, each off-diagonal pair must occur together with its reverse: (a,b) and (b,a). There are 3 unordered off-diagonal pairs, namely one for each pair of distinct elements. Each such pair can either be included or omitted independently, giving 2 choices for each and therefore 2 × 2 × 2 = 2^3 relations. Thus option A is correct. Option B, 2^6, counts the six off-diagonal ordered positions as independent, which violates symmetry; 2^9 ignores the compulsory and paired conditions.

The greatest common divisor is commutative: gcd(a,b) = gcd(b,a). Therefore, whenever (a,b) belongs to R, the reversed pair (b,a) also belongs to R, so R is symmetric. Reflexivity would require gcd(a,a) = 1 for every a in A. But gcd(a,a) = a, and, for example, gcd(2,2) = 2 rather than 1. Thus (2,2) is absent and R is not reflexive. Since an equivalence relation must at least be reflexive, option D cannot be correct. Option C is also ruled out by the same counterexample, while option B reverses the actual properties. Hence option A is the unique correct answer.

The relation contains the pairs (1,4), (2,3), (3,2), and (4,1). It is symmetric because if (a,b) satisfies a+b=5, then reversing the order gives b+a=5 as well; hence (b,a) also belongs to R. It is not reflexive, because reflexivity would require (a,a) for every a. For example, (1,1) is not in R since 1+1=2, not 5; likewise no diagonal pair satisfies 2a=5 for an integer element of A. Since reflexivity is necessary for both a reflexive-transitive relation and an equivalence relation, options C and D are false. Option B also contradicts the established symmetry. Therefore option A is correct.