Relations that are reflexive and symmetric

Correct answer: A. Reflexive and symmetric but not transitive

Explanation: 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.

Correct answer: A. Equivalence relation

Explanation: 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.

Correct answer: A. Reflexive and transitive but not symmetric

Explanation: 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.

Correct answer: A. Symmetric but not reflexive

Explanation: 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.