Relations

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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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}})

Explanation: 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)

Explanation: 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}})

Explanation: 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

Explanation: 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})

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

Correct answer: A. Symmetric but not reflexive

Explanation: The direct answer is A: symmetric but not reflexive. The rule is
aRb\iff\gcd(a,b)=1
. The greatest common divisor does not change when the two numbers are exchanged:
\gcd(a,b)=\gcd(b,a)
. Thus, whenever
aRb
, we also have
bRa
, so the relation is symmetric. Reflexivity would require
\gcd(a,a)=1
for every
a
in the set. But
\gcd(2,2)=2
, and also
\gcd(3,3)=3
, so the relation is not reflexive. Option A is correct. Option B is wrong because reflexivity fails. Option C, equivalence relation, is wrong because an equivalence relation must be reflexive, symmetric and transitive; it is not reflexive, and transitivity also fails, for example
\gcd(2,3)=1
and
\gcd(3,4)=1
, but
\gcd(2,4)=2\ne1
. Option D is wrong because the relation is not reflexive. Memory cue: coprime relation is symmetric, but a number is not coprime with itself unless it is 1.

Correct answer: A. Symmetric

Explanation: 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

Explanation: 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

Explanation: 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. Symmetric

Explanation: The direct answer is A: symmetric is the false property. A relation is reflexive when every diagonal pair
(a,a)
is present. Here
(1,1),(2,2),(3,3)
are all listed, so it is reflexive. It is antisymmetric because the only non-diagonal pair is
(1,2)
, while
(2,1)
is absent; there is no distinct pair occurring in both directions. It is transitive: diagonal pairs cause no problem, and the only possible chain involving the extra pair is
(1,1),(1,2)
or
(1,2),(2,2)
, both leading to
(1,2)
, which is present. However, it is not symmetric because
(1,2)\in R
but
(2,1)\notin R
. Therefore A is the false statement. Option B is true due to all diagonal pairs. Option C is true by the definition of antisymmetry. Option D is true after checking the possible chains. A common mistake is to assume that having
(1,2)
automatically requires
(2,1)
; that requirement belongs to symmetry, not every relation.

Correct answer: A. None

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

Correct answer: A. (4)

Explanation: 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

Explanation: 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

Explanation: (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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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})

Explanation: 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)

Explanation: 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

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