Relations

Every (a) divides itself. Divisibility is antisymmetric and transitive, but generally not symmetric.

An ordered pair \((a,b)\) belongs to \(R\) exactly when \(a\) divides \(b\). Since \(3\mid12\), \(4\mid12\), and \(6\mid12\), the first three pairs are in \(R\). However, \(12\nmid6\), so \((12,6)\) does not belong to \(R\). Exam tip: in \(a\mid b\), the first element is the divisor and the second is the dividend.

The (5) diagonal pairs are independent and the off-diagonal (\binom{5}{2}=10) unordered pairs are independent. So the number is (2^{5+10}=2^{15}).

The (4) diagonal pairs may be included or excluded freely. Each off-diagonal unordered pair has three choices, so (2^4\cdot3^{\binom{4}{2}}=2^4\cdot3^6).

Reflexivity makes the (3) diagonal pairs compulsory. For the remaining (\binom{3}{2}=3) off-diagonal pairs, each pair is either included together or excluded.

(a-a=0\in\mathbb{Z}), and integers are closed under negatives and addition. Hence the relation is reflexive, symmetric and transitive.

For reflexivity, a² = a² holds for every real number a, so aRa. If aRb, then a² = b², which implies b² = a²; hence bRa, so the relation is symmetric. If aRb and bRc, then a² = b² and b² = c², giving a² = c²; therefore aRc, so the relation is transitive as well. Thus, R is an equivalence relation. Remember that a² = b² means a = b or a = −b, but this does not invalidate any equivalence property.

(a+b) is even when both integers have the same parity. So the two classes are even integers and odd integers.

([3]) contains elements having the same remainder as (3) modulo (2). Hence all odd elements ({1,3,5}) are included.

All diagonal pairs are present and ((3,1)) is present with ((1,3)). From ((1,3)) and ((3,1)), ((1,1)) already exists, so transitivity is not broken.

Here, \(x-y>0\) means \(x>y\). For every \(x\in\mathbb{R}\), \(x-x=0\), and \(0>0\) is false; hence, \(R\) is irreflexive. If \(x>y\) and \(y>z\), then \(x>z\), so the relation is also transitive. It is not symmetric because \(x>y\) does not generally imply \(y>x\); therefore, it is not an equivalence relation. Exam tip: Relations defined by strict inequalities such as ‘>’ or ‘<’ are generally irreflexive and transitive.

If (|a-b|=1), then (|b-a|=1), so it is symmetric. But ((1,2)) and ((2,3)) do not give ((1,3)), so it is not transitive.

For symmetry, if ((a,b)) is present, then ((b,a)) must also be present. Hence ((2,1)) must be added because of ((1,2)).

(A\times A) contains all possible ordered pairs. Hence reflexive, symmetric and transitive properties always hold.

The empty relation has no counterexample, so symmetry and transitivity are vacuously true. For non-empty (A), ((a,a)) is missing, so it is not reflexive.

If (a+b=5), then (b+a=5), so the relation is symmetric. But diagonal pairs like ((1,1)) are absent, so it is not reflexive.

Symmetry follows because addition is commutative. But for ((4,4)), (4+4\leq6) is false, so reflexivity fails.

If (a-b) is odd, then (b-a) is also odd. But (a-a=0) is not odd, and ((1,2),(2,3)) do not imply ((1,3)).

The condition that the sum is odd is symmetric. But (1R2) and (2R1) are true, while (1R1) is false, so transitivity fails.

If ((a,b)\in R\cap S), then ((a,b)) belongs to both relations. Since both are symmetric, ((b,a)) also belongs to both.

The direct answer is option A, reflexive. Since R is reflexive, for every element a in A, the pair (a,a) belongs to R. Since S is also reflexive, the same pair (a,a) belongs to S. Therefore it belongs to both R and S, which means it belongs to their intersection R ∩ S. This is true for every a in A, so the intersection is reflexive. Option A is correct. Option B, irreflexive, is wrong because the self-pairs are definitely present, not absent. Option C, must be symmetric, is not guaranteed: the intersection of reflexive relations remains reflexive, but symmetry has not been stated for R or S. Option D, empty relation, is impossible on a nonempty A because every self-pair remains in the intersection. The key idea is that common elements survive an intersection, and all self-pairs are common to both relations.

The identity relation contains exactly the pairs (a,a) for elements a of A. Thus every element is related to itself, which proves reflexivity. If (a,b) is in the relation, then a=b, so (b,a) is the same pair; therefore the relation is symmetric. It is also antisymmetric because whenever both (a,b) and (b,a) occur, the equality a=b already holds.

Finally, if (a,b) and (b,c) are in the identity relation, then a=b and b=c, so a=c; hence (a,c) is also in the relation. The relation is therefore transitive as well. All four stated properties hold simultaneously, so option A is correct. “Only reflexive” is incomplete, and the other options incorrectly deny properties that the identity relation has.

All diagonal pairs are present because of (a=b), and both conditions are symmetric. However, transitivity must be checked carefully; this relation actually stays transitive within paired blocks ({1,4}) and ({2,3}).

(a=b) gives reflexivity and (|a-b|=1) gives symmetry. But ((1,2)) and ((2,3)) are true, while ((1,3)) is false.

The relation is not symmetric because \(1R2\) is true (\(1\leq2\)), whereas \(2R1\) is false (\(2\nleq1\)). It is reflexive since \(a\leq a\) for every integer \(a\), antisymmetric because \(a\leq b\) and \(b\leq a\) imply \(a=b\), and transitive. Exam tip: For the relation \(\leq\), check reflexivity, antisymmetry, and transitivity first; symmetry fails for unequal comparable elements.