Relations

Correct answer: A. Irreflexive and transitive

Explanation: Since (a-a=0), (aRa) is never true, so it is irreflexive. If (a>b) and (b>c), then (a>c), so it is transitive.

Correct answer: A. (a\leq b)

Explanation: The function (x^3) is strictly increasing, so (a^3\leq b^3) is exactly equivalent to (a\leq b). Use monotonicity to identify relation properties quickly.

Correct answer: A. Symmetric but not reflexive

Explanation: Addition is commutative, so the relation is symmetric. But (a+a=2a) is always even, so (aRa) never holds.

Correct answer: A. Symmetric but not transitive

Explanation: If (a\neq b), then (b\neq a), so it is symmetric. But (1R2) and (2R1) hold while (1R1) does not, so it is not transitive.

Correct answer: A. (a\leq b)

Explanation: The given pairs are exactly those where the first element is less than or equal to the second. Thus it is the usual order (\leq) on (A).

Correct answer: A. ((1,4))

Explanation: By transitivity, ((1,2)) and ((2,4)) force ((1,4)). A closure adds the minimum necessary pairs.

Correct answer: A. (3)

Explanation: The pairs to add are ((1,3),(2,4),(1,4)). Note that newly added ((1,3)) with ((3,4)) also requires ((1,4)).

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

Explanation: In a reflexive closure, ((a,a)) must be present for every (a\in A). Therefore all three diagonal pairs are added.

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

Explanation: In a symmetric closure, ((b,a)) is added for every ((a,b)). Thus ((2,1)) is needed for ((1,2)), and ((3,2)) for ((2,3)).

Correct answer: A. Yes, because no distinct reverse pair appears together

Explanation: Antisymmetry is not affected by diagonal pairs. It fails only when ((a,b)) and ((b,a)) both occur for (a\neq b).

Correct answer: A. Equivalence relation

Explanation: Having the same greatest prime factor behaves like equality, so it is reflexive, symmetric, and transitive. Such same invariant conditions often give equivalence relations.

Correct answer: A. ([a]={a+q:q\in\mathbb{Q}})

Explanation: (xRa) means (x-a\in\mathbb{Q}), so (x=a+q) where (q\in\mathbb{Q}). When writing a class, solve the relation condition for the variable.

Correct answer: A. Symmetric but not reflexive

Explanation: Since (a+b=b+a), the relation is symmetric. But ((4,4)\notin R) because (4+4\leq6) is false, so it is not reflexive.

Correct answer: A. (R\cap S) is also an equivalence relation

Explanation: The intersection retains common diagonal pairs, and symmetry and transitivity are also preserved. Hence the intersection of equivalence relations is again an equivalence relation.

Correct answer: A. ({1,3}) and ({2,4})

Explanation: (a+b) is even exactly when (a) and (b) have the same parity. Hence the equivalence classes are the odd class ({1,3}) and the even class ({2,4}).