Relations

By transitivity, if \((a,b)\in R\) and \((b,c)\in R\), then \((a,c)\in R\). Here, \((2,6)\in R\) and \((6,12)\in R\), so \((2,12)\in R\) must hold; indeed, \(2\mid12\). Option B reverses the divisibility relation, since \(12\nmid2\). Exam tip: in a transitive relation, when the second element of the first pair matches the first element of the second pair, combine the two outer elements.

(A\times A) has (n^2) ordered pairs. Each relation can be any subset of it, so there are (2^{n^2}) relations.

In a reflexive relation, (n) diagonal pairs are compulsory. The remaining (n^2-n) pairs are free, so the number is (2^{n^2-n}).

A relation on an n-element set is a subset of A × A. For a symmetric relation, every off-diagonal pair (a,b) must be selected together with (b,a), or both must be omitted. The number of unordered pairs of distinct elements is C(n,2) = n(n−1)/2, and each such pair gives two independent choices: include both ordered pairs or include neither. Diagonal pairs (a,a) are automatically compatible with symmetry, and each of the n diagonal pairs can independently be included or excluded. Therefore the total number of independent binary choices is n + C(n,2) = n + n(n−1)/2 = n(n+1)/2. The number of symmetric relations is consequently 2^(n(n+1)/2), so option A is correct.

The (n) diagonal pairs are free. Each off-diagonal unordered pair has three choices, so the count is (2^n3^{\frac{n(n-1)}{2}}).

For any \(a,b\in A\), \(\gcd(a,b)=\gcd(b,a)\), so the relation is symmetric. However, reflexivity requires \((a,a)\in R\) for every \(a\in A\). Since \(\gcd(2,2)=2\neq1\), we have \((2,2)\notin R\), so the relation is not reflexive. Therefore, option A is correct. Options B and D are false because reflexivity fails. Exam tip: For a gcd-based relation, test symmetry by interchanging the two elements and test reflexivity using \(\gcd(a,a)\).

(\operatorname{lcm}(2,4)=4), so ((2,4)) is in the relation. In the other options, the least common multiple is not (4).

For every real number a, |a| = |a|, so R is reflexive. If |a| = |b|, then |b| = |a|, so it is symmetric. Further, |a| = |b| and |b| = |c| imply |a| = |c|, so it is transitive as well. Therefore, R is an equivalence relation. Exam tip: To identify an equivalence relation, verify all three properties—reflexivity, symmetry, and transitivity.

The condition (a-b\geq0) means (a\geq b). This order relation is reflexive, antisymmetric and transitive.

For the condition \(a+b=6\), the ordered pairs in \(A\times A\) are \((1,5),(2,4),(3,3),(4,2),(5,1)\). Hence, \(R\) contains 5 ordered pairs. The distractor 4 may result from overlooking \((3,3)\); also, reversing the entries gives a different ordered pair unless both entries are equal. Exam tip: For each \(a\in A\), calculate \(b=6-a\) and retain the pair only when \(b\in A\).

The direct answer is option A, symmetric but not reflexive. The rule is that (a,b) belongs to R when the product ab is even. For symmetry, compare ab and ba. They are equal, so if ab is even, then ba is also even. Thus whenever (a,b) is present, (b,a) is present, and R is symmetric. For reflexivity, every (a,a) would have to be present. Check a = 1: 1 times 1 = 1, which is odd, so (1,1) is missing. Therefore R is not reflexive. Option A is correct. Option B is wrong because reflexivity fails at 1. Option C, equivalence relation, is wrong because an equivalence relation must be reflexive, symmetric, and transitive; failure of reflexivity is enough to reject it. Option D, antisymmetric, is wrong: for example (2,4) and (4,2) both belong to R, while 2 and 4 are distinct. Memory cue: multiplication is commutative, giving symmetry; test (1,1) for reflexivity.

The sum is even when both elements have the same parity. With (3) odd and (2) even elements, the number of pairs is (3^2+2^2=13).

If (a\neq b), then (b\neq a), so it is symmetric. But ((1,2)) and ((2,1)) would require ((1,1)), which is not in the relation.

An equivalence relation must be reflexive, symmetric and transitive. Antisymmetry is a property related to partial order.

Reflexivity and antisymmetry do not automatically imply transitivity. In exams, check each property separately from its definition.

((1,2)) and ((2,3)) are present, but ((1,3)) is absent. Hence transitivity fails, so it is not an equivalence relation.

Since \(7\equiv1\pmod{3}\), the class \([7]\) contains all elements of \(A\) that leave remainder 1 when divided by 3. These elements are 1, 4, and 7, so \([7]=\{1,4,7\}\). Option B contains elements with remainder 0, option C contains elements with remainder 2, and option D is incomplete because it omits 1 and 4. Exam tip: To find \([a]\), collect every element of the set that is congruent to \(a\) modulo the given number.

The direct answer is 220, so option B is correct. A relation from A to B is any subset of A×B. Since A has 3 elements and B has 4 elements, the Cartesian product has 3×4=12 ordered pairs. To make a relation containing exactly 3 pairs, choose any 3 of these 12 pairs: C(12,3) = (12×11×10)/(3×2×1) = 220. Option A, 120, uses the wrong count. Option B, 220, is correct because it counts all selections of exactly three distinct ordered pairs. Option C, 440, is double the correct result and has no basis here. Option D, 2¹², counts every possible relation, because each of the 12 pairs may be selected or not; it does not restrict the relation to exactly 3 pairs. Memory cue: exactly k pairs means choose k from the total, using C(n,k), not 2ⁿ.

The (4) diagonal pairs are compulsory and ((1,2)) is forbidden. The remaining (11) pairs are free, so the count is (2^{11}).

Reflexivity makes all (5) diagonal pairs compulsory. Each off-diagonal unordered pair has (3) choices, so the count is (3^{10}).

(a^2\leq a^2) gives reflexivity and the inequality gives transitivity. But ((-1,1)) and ((1,-1)) are both true while (-1\neq1).

(a-a=0\in\mathbb{Q}), and rational numbers are closed under negatives and addition. Hence it is reflexive, symmetric and transitive.

If (a-b) is irrational, then (b-a) is also irrational. But (a-a=0) is not irrational and the sum of irrational differences can be rational.

For ((1,1)), (1+2\cdot1=3), which is divisible by (3). Substituting each option is the safest method.

Each element divides itself and is equal to itself. Divisibility together with (\leq) gives reflexive, antisymmetric and transitive properties.