Relations

A pair \((x,y)\) belongs to \(R\) only when \(x\) is an even element of \(A\) and \(y\) is an element of \(B\). Since \(4\in A\) is even and \(b\in B\), \((4,b)\in R\). The pair \((3,b)\) fails because 3 is odd, while \((a,2)\) has the elements in the wrong sets and order. Exam tip: check the condition on the first component first, then verify that the second component belongs to \(B\).

The governing concept is that a relation from A to B is any subset of the Cartesian product A × B. Since |A| = 2 and |B| = 3, the product contains |A × B| = 2 × 3 = 6 ordered pairs. Every one of these six pairs may either be included or excluded independently when forming a relation. Consequently, the number of possible subsets, and hence the number of relations, is 2⁶ = 64. Thus option B is correct. Option C gives only the number of product pairs, not the number of their subsets. Option D counts 2 × |A × B| incorrectly, while option A uses the wrong exponent because the product has six rather than five pairs.

The governing concept is the range of a relation: it is the set of all second coordinates that actually occur in valid ordered pairs. Both x and y must belong to A, and y must equal 2x. Test the possible x-values in A. For x = 1, y = 2, which is in A; for x = 2, y = 4, also in A. For x = 3, 4, or 5, the values 6, 8, and 10 are not in A, so those pairs are not allowed. Thus R = {(1,2),(2,4)} and its range is {2,4}. Option A is correct. Option C ignores the requirement y ∈ A, while B gives an incorrect range and D confuses the whole set A with the actual outputs.

The governing counting principle is that every relation from A to B is an arbitrary subset of the Cartesian product A × B. Since |A| = 2 and |B| = 4, the product contains |A × B| = 2 × 4 = 8 ordered pairs. For each of these eight pairs, there are two independent choices: include it in the relation or leave it out. Therefore the number of subsets, and hence the number of possible relations, is 2^8. Option C is correct. The expression 2^4 incorrectly uses only the size of B, while 2^6 does not represent the product size. Option D, 8^2, is a different calculation and is not the subset-counting rule.

The governing concept is counting ordered pairs satisfying a condition. Since A has four elements, the Cartesian product A × A contains 4 × 4 = 16 ordered pairs in total. The condition a ≠ b excludes exactly the four diagonal pairs (1,1), (2,2), (3,3), and (4,4), because their components are equal. Every other ordered pair has unequal components and is included in R. Therefore the number of pairs is 16 − 4 = 12, so option C is correct. A common error is to count only unordered pairs, which would ignore order; here (1,2) and (2,1) are distinct and both qualify. Option D counts all pairs without removing the diagonal, while options A and B remove too many pairs.

Since (A\times B) has (3\times2=6) elements, the number of relations is (2^6=64). In exams, count the elements of (A\times B) first.

The domain contains all first components, so it is ({2,4,6}). In exams, look at the first place in each ordered pair.

The range of a relation is the set of second components of its ordered pairs. Here the second components are 3, 3, 5, and 7; since a set does not repeat elements, the range is \(\{3,5,7\}\). Option A is the set of first components, called the domain. Exam tip: to find the range, list the second entry of each ordered pair and remove duplicates.

For a relation to be reflexive, \((a,a)\in R\) must hold for every element \(a\) of the set. Here, \((1,1)\) and \((2,2)\) are present, but \((3,3)\) is missing. Therefore, the relation is not reflexive. Exam tip: To test reflexivity, check all diagonal pairs \((a,a)\).

For every ((a,b)), ((b,a)) is also present, so the relation is symmetric. For reflexivity, ((1,1),(2,2),(3,3)) are also needed.

Transitivity needs ((a,b)) and ((b,c)) to imply ((a,c)). Here ((1,2)) and ((2,3)) give ((1,3)).

The direct answer is option A: R is an equivalence relation. The rule is that \(aRb\) when \(a-b\) is even. Two integers have an even difference exactly when they have the same parity, meaning both are odd or both are even. Reflexive: for every a, \(a-a=0\), and 0 is even, so \(aRa\). Symmetric: if \(a-b\) is even, then \(b-a=-(a-b)\) is also even, so \(bRa\). Transitive: if \(a-b\) and \(b-c\) are even, their sum \(a-c=(a-b)+(b-c)\) is even, so \(aRc\). Therefore all three properties hold. Option A is correct. Option B is wrong because the relation is not merely reflexive; it is also symmetric and transitive. Option C is wrong for the same reason. Option D is false because reflexivity was proved. The set divides into odd elements \(\{1,3,5\}\) and even elements \(\{2,4\}\), which are the equivalence classes. Memory cue: same parity means equivalent.

Direct answer: option C, (3,1), is not in R. The relation is defined by a divides b, meaning b must be exactly divisible by a with no remainder. Check the ordered pair in the stated direction: the first number is the divisor and the second is the number being divided. A: (2,4) works because 4 ÷ 2 = 2. B: (1,3) works because every integer is divisible by 1, so 3 ÷ 1 = 3. C: (3,1) fails because 1 ÷ 3 is not an integer; 3 does not divide 1. D: (4,4) works because 4 ÷ 4 = 1. Thus only C is absent. Do not reverse the pair: (1,3) and (3,1) have different meanings. Memory cue: in “a divides b,” read from left to right as “b is divisible by a.”

The ordered pairs satisfying x + y = 5 are (1, 4), (2, 3), (3, 2), and (4, 1). Therefore, R contains 4 ordered pairs. Choosing 3 would be incorrect because reversed pairs such as (2, 3) and (3, 2) are distinct. Exam tip: In an ordered pair, changing the order changes the pair.

The relation requires a < b. Its ordered pairs are (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), and (3, 4). The second components appearing in these pairs are 2, 3, and 4, so the range is {2, 3, 4}. Option A consists of the first components and represents the domain, not the range. Exam tip: To find the range of a relation, collect the second components of all its ordered pairs.

In the inverse relation, the components of each ordered pair are reversed. Thus ((1,2)) gives ((2,1)) and ((2,3)) gives ((3,2)).

By definition of a symmetric relation, if \((a,b)\in R\), then \((b,a)\in R\) must also hold. Here, \(a=5\) and \(b=8\), so \((8,5)\in R\) is definitely true. The pairs \((5,5)\) and \((8,8)\) would require reflexivity, which is not implied by symmetry, while option D contradicts the given condition. Exam tip: For a symmetric relation, every ordered pair must be accompanied by its reversed pair.

The governing property is transitivity. A relation R is transitive when, whenever (a, b) ∈ R and (b, c) ∈ R, the pair (a, c) must also belong to R. Here the common middle element is 4: the first pair gives 1 related to 4, and the second gives 4 related to 7. Applying the definition with a = 1, b = 4, and c = 7 forces (1, 7) to be in R. Thus option A is correct. The other choices either reverse the direction or use the common middle element as an endpoint; transitivity does not require those pairs. No additional assumption such as symmetry is provided.

A reflexive relation must contain a self-related ordered pair \((a,a)\) for every element \(a\) of the set. Thus, \((1,1),(2,2),(3,3)\) are compulsory, giving a minimum of 3 ordered pairs. One or two pairs cannot cover all three elements, while 9 is the total number of pairs in \(A\times A\), not the minimum. Exam tip: a reflexive relation on an \(n\)-element set contains at least \(n\) ordered pairs.

(A\times A) has (16) pairs and (4) diagonal pairs are compulsory. The remaining (12) pairs are free, so the number is (2^{12}).

Set \(A\) has 2 elements, so \(A\times A\) contains \(2^2=4\) ordered pairs. Every relation on \(A\) is a subset of \(A\times A\). A set with 4 elements has \(2^4=16\) subsets, so the total number of relations is 16. Option 4 is only the number of ordered pairs, not the number of relations. Exam tip: For a set with \(n\) elements, the number of relations on it is \(2^{n^2}\).

The empty relation has no pair that violates symmetry. But it is not reflexive because ((1,1),(2,2),(3,3)) are absent.

The universal relation on a set \(A\) is \(A\times A\). For \(A=\{1,2\}\), \(A\times A=\{(1,1),(1,2),(2,1),(2,2)\}\), so option A is correct. Option B is only the identity relation, containing pairs of the form \((a,a)\). Exam tip: a universal relation contains every possible ordered pair formed from the set.

In this relation, every element of the set \(A\) is related only to itself; that is, \((x,x)\in R\) for every \(x\in A\). Hence, it is called the identity relation. An empty relation contains no ordered pairs, whereas a universal relation contains every ordered pair in \(A\times A\). Exam tip: whenever a relation is written as \(\{(x,x):x\in A\}\), identify it as the identity relation.

Since the condition a=b allows only the diagonal pairs (1,1), (2,2), (3,3), and (4,4), relation R contains exactly 4 ordered pairs. Exam tip: the identity relation on a set with n elements has n ordered pairs, not n².