Relations

Correct answer: A. 5

Explanation: For a = 1, the condition 1² < b means b > 1, giving b = 2, 3, 4, 5 and hence 4 ordered pairs. For a = 2, the condition 4 < b is satisfied only by b = 5, giving 1 pair. For a = 3 and 4, a² is 9 and 16 respectively, so no element of B satisfies the condition. Therefore, |R| = 4 + 1 = 5. Exam tip: Count the valid values of b in B separately for each value of a.

Correct answer: C. (5,2)

Explanation: The pair (2,5) belongs to R because 2+5=7. In the inverse relation R⁻¹, the coordinates of every ordered pair are interchanged; hence (5,2)∈R⁻¹. The pairs (2,4) and (4,5) do not satisfy the required sum of 7, while (1,6) is not in A×B because 6∉B. Exam tip: To form an inverse relation, interchange the coordinates of each ordered pair.

Correct answer: B. 6

Explanation: The even values of \(a\) in \(A\) are 2 and 4. For \(a=2\), the condition gives \(b\le4\), so there are 4 choices: 1, 2, 3, and 4. For \(a=4\), it gives \(b\le2\), so there are 2 choices. Therefore, the total number of ordered pairs is \(4+2=6\). Exam tip: Count the permitted values of \(b\) separately for each allowed value of \(a\).

Correct answer: A. 5

Explanation: For a = 1, the pairs are (1, 2) and (1, 3); for a = 2, they are (2, 4) and (2, 6); and for a = 3, only (3, 6) is valid. For a = 4, 2a = 8 and 3a = 12, neither of which belongs to B. Thus, R = {(1, 2), (1, 3), (2, 4), (2, 6), (3, 6)}, so |R| = 5. Although (2, 6) and (3, 6) have the same second component, they are distinct ordered pairs. Exam tip: Check whether each resulting b belongs to B before counting the ordered pair.

Correct answer: C. 5

Explanation: For every \(a\in A\), there is exactly one residue \(b\in B\) congruent to \(a^2\pmod{5}\). The residues for \(a=0,1,2,3,4\) are \(0,1,4,4,1\), respectively, so five ordered pairs are formed. Although some values of \(b\) repeat, the pairs remain distinct because their first components are different. Hence, \(|R|=5\). In an exam, count complete ordered pairs rather than only the distinct second components.

Correct answer: B. 7

Explanation: For \(b-a=0\), the ordered pairs are \((1,1),(2,2),(3,3),(4,4)\), giving 4 pairs. For \(b-a=2\), the valid pairs are \((1,3),(2,4),(3,5)\), giving 3 pairs. These cases do not overlap, so \(|R|=4+3=7\). A common mistake is to count only the two possible differences or to include pairs outside \(A\times B\); always list the valid ordered pairs within the given sets.

Correct answer: A. (R) is an equivalence relation

Explanation: Since (a-a=0), symmetry and transitivity also hold. In exams, always test all three properties for divisibility relations.

Correct answer: C. Transitivity

Explanation: Direct answer: Option C, transitivity fails. A relation is reflexive when every element is related to itself, symmetric when reversing every ordered pair keeps it in the relation, and transitive when
(a,b)
and
(b,c)
together require
(a,c)
. Here (1,1), (2,2), and (3,3) are present, so reflexivity holds. Each non-diagonal pair has its reverse: (1,2) with (2,1), and (2,3) with (3,2), so symmetry holds. But (1,2) and (2,3) are present while (1,3) is absent; therefore transitivity fails. Option A is wrong because all self-pairs occur. Option B is wrong because all listed unequal pairs have reverses. Option C is correct because of the missing link (1,3). Option D is wrong because it says both reflexivity and symmetry fail, whereas both actually hold. Memory cue: for transitivity, join the first member of the first pair to the second member of the second pair.

Correct answer: C. \(2^{25}\)

Explanation: Every relation on set \(A\) is a subset of \(A\times A\). Since \(|A|=5\), we have \(|A\times A|=5^2=25\). A set with 25 elements has \(2^{25}\) subsets, so the total number of relations is \(2^{25}\). The value \(5^2\) counts the ordered pairs in \(A\times A\), not the relations themselves. Exam tip: for an \(n\)-element set, the number of relations on it is always \(2^{n^2}\).

Correct answer: A. Reflexive and transitive but not symmetric

Explanation: The direct answer is A: reflexive and transitive but not symmetric. The relation contains
(a,b)
exactly when
a\le b
. For reflexivity, every element satisfies
a\le a
, so
(1,1),(2,2),(3,3),(4,4)
are present. For transitivity, if
a\le b
and
b\le c
, then
a\le c
, so the relation is transitive. It is not symmetric because
1\le2
, so
(1,2)
is present, but
2\le1
is false and
(2,1)
is absent. Thus A is correct. Option B is wrong because the relation is not symmetric and it is actually reflexive. Option C, an equivalence relation, is wrong because equivalence requires reflexivity, symmetry and transitivity together; failure of symmetry prevents it. Option D is wrong because both reflexivity and transitivity have been proved. The finite set does not change these conclusions. Exam cue: an order relation such as
\le
is usually reflexive and transitive, but generally not symmetric.

Correct answer: B. (2^6)

Explanation: There are (3) diagonal pairs and (3) unordered off-diagonal pairs. Hence the total count is (2^{3+3}=2^6).

Correct answer: B. 3

Explanation: Two numbers belong to the same equivalence class when they leave the same remainder on division by 3. Thus the classes are \(\{1,4\}\), \(\{2,5\}\), and \(\{3,6\}\), giving a total of 3 equivalence classes. Choosing 6 confuses the number of elements with the number of classes. Exam tip: for congruence modulo \(m\), group elements according to their possible remainders modulo \(m\).

Correct answer: A. Equivalence relation

Explanation: It relates integers of the same parity, so it is reflexive, symmetric, and transitive. Thinking in remainders is very useful here.

Correct answer: A. Symmetric but not reflexive

Explanation: Direct answer: Option A, symmetric but not reflexive. The relation contains pairs whose sum is 5: (1,4), (4,1), (2,3), and (3,2). Since addition is commutative, if a+b=5 then b+a=5, so reversing a valid pair always gives another valid pair; hence symmetry holds. Reflexivity would require (1,1), (2,2), (3,3), and (4,4). Their sums are 2, 4, 6, and 8, none of which is 5, so the relation is not reflexive. Option A is exactly correct. Option B is wrong because reflexivity fails and symmetry does hold. Option C is wrong because an equivalence relation must be reflexive, symmetric, and transitive; this relation is not reflexive. Option D is wrong for the same reason: it is neither reflexive nor transitive as a complete claim. Exam cue: for a+b=k, symmetry is automatic, but reflexivity requires 2a=k for every element.

Correct answer: A. It is an equivalence relation

Explanation: The universal relation contains all possible ordered pairs. Hence reflexivity, symmetry, and transitivity all hold.

Correct answer: A. Symmetry

Explanation: Since (|a-b|=|b-a|), the relation is symmetric. But ((a,a)) is absent, so it is not an equivalence relation.

Correct answer: A. Reflexivity

Explanation: (a+a=2a) is always even, so ((a,a)) is not in the relation. The relation is symmetric but not reflexive.

Correct answer: A. Equivalence relation

Explanation: All diagonal pairs are present, reverse pairs are present, and the needed diagonal pairs from ((1,3),(3,1)) are present. So it is an equivalence relation.

Correct answer: A. ((2,12))

Explanation: Direct answer: Option A, (2,12). Here aRb means a divides b, written a|b. The given facts are 2|6 and 6|12. By transitivity of divisibility, if the first number divides the middle number and the middle number divides the last number, the first divides the last: 2|12, since 12=2×6. Option A is correct. Option B, (12,2), reverses the direction; 12 does not divide 2. Option C, (6,2), is also reversed and false because 6 does not divide 2. Option D, (4,12), is not forced by the given chain; in fact 4 does divide 12, but it is not the pair obtained by transitivity from 2, 6, and 12. The required rule is (a,b) and (b,c) imply (a,c). Memory cue: in a divisibility chain, connect the first divisor to the final multiple.

Correct answer: A. Symmetric but not reflexive

Explanation: If \((a,b)\in R\), then \(a+b\le 6\). Since \(a+b=b+a\), it follows that \((b,a)\in R\) as well; hence, \(R\) is symmetric. However, it is not reflexive because \((4,4)\notin R\), as \(4+4=8\not\le 6\). It is also not transitive: \((4,1)\in R\) and \((1,4)\in R\), but \((4,4)\notin R\). Exam tip: To test reflexivity, check whether \((a,a)\) belongs to the relation for every \(a\in A\).

Correct answer: A. ({-2,2})

Explanation: Since ((-2)^2=2^2), (-2) and (2) are in the same class. While forming a class, choose elements giving the same value.

Correct answer: A. Reflexive and symmetric but not transitive

Explanation: Since (|a-a|=0) and (|a-b|=|b-a|), it is reflexive and symmetric. But ((1,2)) and ((2,3)) are present while ((1,3)) is absent.

Correct answer: A. Symmetric but neither reflexive nor transitive

Explanation: From (a\ne b), we get (b\ne a), so it is symmetric. But ((1,2)) and ((2,1)) would need ((1,1)), which is absent.

Correct answer: A. Symmetry

Explanation: Since (\gcd(a,b)=\gcd(b,a)), the relation is symmetric. But all ((a,a)) are not included because (\gcd(2,2)=2).

Correct answer: A. (2^5)

Explanation: For reflexivity, (3) diagonal pairs are compulsory and ((1,2)) is forbidden. The remaining (9-3-1=5) pairs are optional.