Relations

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

For symmetry, ((3,2)) must accompany ((2,3)). The other required reverse pairs are already present.

Correct answer: A. (R) is transitive but not reflexive

From ((1,2)) and ((2,3)), ((1,3)) is present, so the needed transitive condition holds. But ((1,1),(2,2),(3,3)) are absent, so it is not reflexive.

Correct answer: A. Symmetric but not transitive

Every present pair has its reverse, so the relation is symmetric. But ((1,2)) and ((2,3)) are present while ((1,3)) is not, so it is not transitive.

Correct answer: A. Equivalence relation

Since (a-a=0\in\mathbb{Z}), (a-b\in\mathbb{Z}) implies (b-a\in\mathbb{Z}), and the sum of integers is an integer. Hence it is an equivalence relation.

Correct answer: A. ({x\in\mathbb{Z}:x\equiv2\pmod{5}})

Because (7\equiv2\pmod{5}), all integers with the same remainder are in its class. Always form an equivalence class from the relation condition.

Correct answer: A. (4)

The remainders (0,1,2,3) form (4) different classes. For a modulo relation, the number of classes equals the modulus.

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

By same parity, the odd class is ({1,3}) and the even class is ({2,4}). A partition is the collection of equivalence classes.

Correct answer: A. It is an equivalence relation

Since (a^2=a^2), equality is symmetric, and (a^2=b^2), (b^2=c^2) imply (a^2=c^2). Hence (R) is an equivalence relation.

Correct answer: A. ({-3,3})

From (|x|=|-3|=3), (x=-3) or (x=3). In an absolute value relation, opposite signs may lie in the same class.

Correct answer: B. $R$ is symmetric but not transitive.

Symmetry: since $\gcd(a,b)=\gcd(b,a)$, if $aRb$ then $bRa$, so $R$ is symmetric. Not transitive: take $a=2$, $b=3$, $c=4$. We have $\gcd(2,3)=1$ so $2R3$, and $\gcd(3,4)=1$ so $3R4$. But $\gcd(2,4)=2\ne 1$, so $2\not R4$. Hence transitivity fails. Therefore, $R$ is symmetric but not transitive.

Correct answer: A. Symmetric

Because (a+b=b+a), reversing the pair keeps the same sum. But not every (a+a) is prime, so reflexivity is not guaranteed.

Correct answer: A. Reflexive and transitive but not symmetric

Every (a\leq a) is true, and (a\leq b\leq c) implies (a\leq c). But (2\leq3) does not imply (3\leq2).

Correct answer: A. Partial order relation

Every (A\subseteq A), and the subset relation is antisymmetric and transitive. Do not treat it like equality because it is generally not symmetric.

Correct answer: A. Reflexive

A relation on $A$ is reflexive if every $(a,a)$ is in $R$. Here, $(1,1)$, $(2,2)$ and $(3,3)$ are not present in $R$ (for example, $(2,2)\notin R$). Hence reflexive property is false.

Correct answer: A. None

If ((a,b)) with (a\neq b) occurs, symmetry forces ((b,a)), contradicting antisymmetry. Hence only diagonal pairs can occur.

Correct answer: A. (4)

In the order ( \leq ), the greatest element is one that every element is less than or equal to. Here (a\leq4) for every (a\in A), so (4) is greatest.

Correct answer: A. None

A least element must divide every element. Here neither (2\mid3) nor (3\mid2), so no least element exists.

Correct answer: A. (1) is least and (8) is greatest

(1) divides every element, and every element divides (8). In divisibility order, identify least and greatest by the direction of divisibility.

Correct answer: A. It is reflexive, symmetric, and transitive

In the universal relation, every possible ordered pair is present. Hence reflexive, symmetric, and transitive properties all hold automatically.

Correct answer: A. It is symmetric and transitive but not reflexive

The empty relation has no counterexample, so symmetry and transitivity are vacuously true. But ((1,1)\notin R), so it is not reflexive.

Correct answer: A. Both equivalence relation and partial order

The identity relation is reflexive, symmetric, and transitive, so it is an equivalence relation. It is also reflexive, antisymmetric, and transitive, so it is a partial order.

Correct answer: A. (R^{-1}) is also reflexive

A reflexive (R) contains every ((a,a)), and the inverse of ((a,a)) is again ((a,a)). Therefore (R^{-1}) remains reflexive.

Correct answer: A. (R=R^{-1})

In a symmetric relation, ((a,b)\in R) implies ((b,a)\in R), so the inverse has the same pairs. Hence (R=R^{-1}).

Correct answer: A. (6)

The classes are ({1,4},{2},{3}), so the number of pairs is (2^2+1^2+1^2=6). Add the squares of the sizes of the equivalence classes.

Correct answer: A. Reflexive and symmetric but not transitive

Since (|a-a|=0\leq1) and (|a-b|=|b-a|), it is reflexive and symmetric. But (1R2) and (2R3) hold while (1R3) does not.