Relations that are reflexive and symmetric

Correct answer: B. 8

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.

Correct answer: A. Reflexive

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.

Correct answer: A. Reflexive and symmetric

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.

Correct answer: A. Reflexive and transitive

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.

Correct answer: A. {1,4}

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.

Correct answer: A. The reverse of every pair is also present

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.

Correct answer: A. {(1,1),(2,2),(3,3)}

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.

Correct answer: B. 4

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.

Correct answer: A. 2^3

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.

Correct answer: A. Symmetric but not reflexive

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.