Relations

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.

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.

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 \(a\).

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.

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.

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.

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

Check symmetry. For symmetry, whenever $(a,b)\in R$ we must have $(b,a)\in R$. Here $(1,2)\in R$ but $(2,1)\notin R$ (it is not listed). So symmetry fails. The other properties hold: $R$ contains $(1,1),(2,2),(3,3)$ so it is reflexive; since all pairs implied by $(1,2)$ and $(2,3)$ lead to $(1,3)$ which is present, transitivity holds; and antisymmetry holds because there is no pair with both $(a,b)$ and $(b,a)$ for $a\ne b$.

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 \(2^{n^2}\).

For any $a\in A$, we have $a\le a$, so $(a,a)\in R$; hence $R$ is reflexive. If $(a,b)\in R$ and $(b,a)\in R$, then $a\le b$ and $b\le a$, so $a=b$; hence $R$ is antisymmetric. For transitivity: if $(a,b)\in R$ and $(b,c)\in R$, then $a\le b$ and $b\le c$, so $a\le c$, thus $(a,c)\in R$. Therefore $R$ is reflexive, antisymmetric and transitive.

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

The correct answer is 3. Numbers are equivalent modulo 3 when they have the same remainder on division by 3. The classes in A are {1, 4}, {2, 5}, and {3, 6}, corresponding to remainders 1, 2, and 0, respectively.

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

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.

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

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

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

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.

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.

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 \(a\in A\).

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

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.

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.

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

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