Relations

A relation on a set is reflexive only when every element is related to itself. Here, pairs such as (1,1), (2,2), and (3,3) are missing, so the relation is not reflexive. It is symmetric when each pair has its reversed pair in the relation. The reverse of (1,2) is (2,1), and the reverse of (2,3) is (3,2); all required reverse pairs are present.

It is not antisymmetric because both (1,2) and (2,1) occur even though 1 and 2 are different, and similarly for 2 and 3. It is also not an equivalence relation because an equivalence relation must be reflexive, symmetric, and transitive. For example, (1,2) and (2,3) are present, but (1,3) is absent, so transitivity fails. Thus option B is correct.

The reverse pair for distinct elements is not present. In antisymmetry, if ((a,b)) and ((b,a)) both occur, then (a=b) must hold.

Every number divides itself and divisibility is transitive. But it is not symmetric since (1 \mid 2) while (2 \nmid 1).

A pair (a, b) belongs to R exactly when a + b = 6. In that case, b + a = 6 as well, by the commutative property of addition; hence (b, a) also belongs to R, so R is symmetric. It is not reflexive because, for example, (1, 1) ∉ R. It is not antisymmetric because both (1, 5) and (5, 1) belong to R although 1 ≠ 5. Exam tip: To test symmetry, check whether every pair (a, b) is accompanied by (b, a).

Because (A \times A) has (9) pairs and a relation is any subset of it. Remember the formula (2^{n^2}).

For a set A with n elements, the Cartesian product A × A contains n² ordered pairs. A relation is any subset of this product. For n = 4, there are 4² = 16 possible ordered pairs in all. Reflexivity requires the four diagonal pairs (1,1), (2,2), (3,3), and (4,4), or their corresponding pairs for any labelled four-element set, to be included. The remaining 16 − 4 = 12 off-diagonal pairs may be selected independently: each is either included or excluded. Therefore the number of choices is 2^12. Option A counts all relations without imposing reflexivity, option C counts only choices among the mandatory diagonal pairs, and option D is not the relevant counting expression. Hence option B is correct.

Self-pairs give (2^3) choices and each distinct unordered pair gives (3) choices. Hence the number is (2^3\cdot 3^3).

For transitivity, whenever (a,b) and (b,c) belong to R, (a,c) must also belong to R. Here, (1,2) and (2,3) imply (1,3), which is present; all other possible chains satisfy the same condition. In fact, R behaves like the ≤ relation on A, so it is transitive. Option B is false because (1,2) is present but (2,1) is not. Option C is false because (1,1), (2,2), and (3,3) are all present. Exam tip: to test transitivity, look for chains of the form (a,b), (b,c) and check whether (a,c) is included.

The required ((1,3)) is present for ((1,2)) and ((2,3)). In transitivity, check only actual chains.

The condition a − b divisible by 3 means that a and b have the same remainder upon division by 3. It defines congruence modulo 3. It is reflexive because a − a = 0, and 0 is divisible by 3. It is symmetric because if 3 divides a − b, then 3 also divides its negative b − a. It is transitive because if 3 divides a − b and 3 divides b − c, then 3 divides their sum a − c. Since reflexivity, symmetry, and transitivity all hold, R is an equivalence relation. Option B is incomplete and option D contradicts the first two checks. The relation is certainly not empty, as every integer is related to itself.

(5) is odd, and all integers differing from it by an even number are odd. An equivalence class contains all elements with the same property.

Two numbers are related when they leave the same remainder on division by 2. The even numbers \(\{2,4,6\}\) form one equivalence class, and the odd numbers \(\{1,3,5\}\) form the other. Hence, there are 2 equivalence classes. Option 6 gives the number of elements in the set, not the number of classes. Exam tip: For modulo 2, group the elements into even and odd numbers.

For every a ∈ A, a ≤ a, so R is reflexive. If a ≤ b and b ≤ a, then a = b, so it is antisymmetric. Also, a ≤ b and b ≤ c imply a ≤ c; hence it is transitive. Therefore, R is a partial order relation. It is not symmetric because (1, 2) ∈ R but (2, 1) ∉ R, so it cannot be an equivalence relation either. Exam tip: For a relation defined by ≤, check reflexivity, antisymmetry, and transitivity to identify a partial order.

For reflexivity, \((a,a)\) must belong to \(R\) for every \(a\in A\). Here, \((a,a)\in R\) would require \(a<a\), but no number is less than itself. Therefore, \(R\) is not reflexive; it is irreflexive. Option B is also incorrect because \(a<b\) is not symmetric, and symmetry is a different property from reflexivity. Exam tip: To test reflexivity, check whether every pair \((a,a)\) is present in the relation.

There is no element in the empty set, so the condition is automatically true. On an empty base set, many properties are vacuously true.

(A \times A) contains all pairs, so it is both reflexive and symmetric. It is also transitive but generally not antisymmetric.

There is no violating pair or chain, so symmetry and transitivity are vacuously true. But on non-empty (A), it is not reflexive.

If \((a,b)\in R\), then \(a+b\) is odd. Since \(b+a=a+b\), \((b,a)\in R\) as well; hence the relation is symmetric. However, for every \(a\in A\), \(a+a=2a\) is even, so \((a,a)\notin R\), and the relation is not reflexive. It is therefore not an equivalence relation; in fact, \((1,2),(2,3)\in R\), but \((1,3)\notin R\), so it is not transitive either. Exam tip: A relation defined by an odd sum links elements of opposite parity and cannot be reflexive.

Elements with the same parity are related, so the relation is reflexive, symmetric, and transitive. It forms even and odd classes.

If \((a,b)\in R\), then \(|a-b|=2\). Since \(|b-a|=|a-b|=2\), we also have \((b,a)\in R\); hence the relation is symmetric. It is not reflexive because \(|a-a|=0\ne2\), so \((a,a)\notin R\) for every \(a\). It is also not transitive: \((1,3)\in R\) and \((3,5)\in R\), but \((1,5)\notin R\). Exam tip: for relations involving absolute differences, use \(|a-b|=|b-a|\) to test symmetry first.

Each ordered pair \((a,b)\) must have both entries in \(A\) and satisfy \(a+2b=6\). For \(b=1\), \(a=4\), giving \((4,1)\); for \(b=2\), \(a=2\), giving \((2,2)\). For \(b=3\), \(a=0\), and for \(b=4\), \(a=-2\); neither value belongs to \(A\). Therefore, the relation contains 2 ordered pairs. Exam tip: when counting relation pairs, verify that both components belong to the specified set, not just that the equation is satisfied.

For every \(x\), \(x-x=0\in\mathbb Q\), so the relation is reflexive. If \(x-y\) and \(y-z\) are rational, then \(x-z=(x-y)+(y-z)\) is rational. It is therefore an equivalence relation. Exam tip: use closure of rational numbers under addition.

Since \(\gcd(a,b)=\gcd(b,a)\), whenever \((a,b)\in R\), we also have \((b,a)\in R\); hence the relation is symmetric. It is not reflexive because \(\gcd(2,2)=2\neq1\). It is not transitive: \(\gcd(2,3)=1\) and \(\gcd(3,4)=1\), but \(\gcd(2,4)=2\). It is also not antisymmetric because both \((2,3)\) and \((3,2)\) belong to \(R\), although \(2\neq3\). Exam tip: for relations defined by \(\gcd(a,b)=1\), first use the symmetry \(\gcd(a,b)=\gcd(b,a)\).

A first element \(a\) is related to \(12\) precisely when \(a\mid 12\). Every element of \(A\)—namely \(1,2,3,4,6,12\)—divides \(12\). Thus the related ordered pairs are \((1,12),(2,12),(3,12),(4,12),(6,12),(12,12)\), giving 6 first elements. Exam tip: when counting divisors, include the number itself; counting only proper divisors would incorrectly give 5.

By definition of a symmetric relation, if \((a,b)\in R\), then \((b,a)\in R\) as well. Since \((2,5)\in R\), its reversed pair \((5,2)\) must belong to \(R\). The given information does not require \((2,2)\), \((5,5)\), or \((1,2)\) to be in the relation. Exam tip: For a symmetric relation, interchange the two entries of any given ordered pair.