Relations

The empty relation has no ordered pair, so it has no first component. Hence the domain is (\varnothing).

The first components are (1,2,3), and the second components are (2,3,1), which as a set is ({1,2,3}). Be careful with set order.

For the pair \((2,4)\), \(2+4=6\), which is even; therefore, \((2,4)\in R\). A sum is even when both numbers have the same parity—both even or both odd. In the other options, one number is even and the other is odd, so their sums are odd. In such questions, checking parity first is a quicker method than performing every addition.

The second components are (1,2,2), and after removing repetition we get ({1,2}). The range uses only second components.

In an identity relation, each element gives one pair ((x,x)). Therefore (3) elements give (3) pairs.

To find the inverse relation, reverse the entries in each ordered pair. The reverse of (1, 2) is (2, 1), and the reverse of (2, 1) is (1, 2). Both pairs are already in R, so R⁻¹ = {(1, 2), (2, 1)} = R.

The direct answer is option D: \((2,9)\) is not in R. The rule is \(y=x^2\), so for each proposed pair, square the first coordinate and compare the result with the second coordinate. For \((1,1)\), \(1^2=1\), so it belongs to R. For \((2,4)\), \(2^2=4\), so it belongs. For \((3,9)\), \(3^2=9\), so it belongs. For \((2,9)\), however, \(2^2=4\), not 9, so it does not belong. Option D is correct. Options A, B, and C are all valid because each exactly follows the rule. The fact that 9 is in B is not enough; the pair must satisfy the formula as well. Memory cue: in a relation defined by a formula, substitute the first coordinate and check the second one.

In the inverse relation, second components become first components. Hence the domain of (R^{-1}) is the range of (R), which is ({a,b,c}).

For every element \(a\in A\), the pair \((a,a)\) is present in \(R\); hence \(R\) is reflexive. It is also symmetric and transitive, but option A gives the required classification. Exam tip: check all diagonal pairs \((a,a)\) to test reflexivity.

The empty relation has (0) pairs, and the universal relation (A\times A) has (2^2=4) pairs. Remember these two extreme relations.

The element (1) appears as the first component in ((1,2)) and ((1,3)). So it has (2) images, (2) and (3).

The ordered pairs satisfying a > b are (2, 1), (3, 1), and (3, 2). Thus, R = {(2, 1), (3, 1), (3, 2)}, so it contains 3 ordered pairs. The answer 6 is incorrect because reversed pairs such as (1, 2) do not satisfy the relation. Exam tip: For an inequality relation, check each ordered pair according to the stated condition.

The smallest subset is (\varnothing), so the empty relation is the smallest relation. It is a subset of every (A\times B).

Every relation from \(A\) to \(B\) is a subset of \(A\times B\). Therefore, the largest relation is the complete Cartesian product \(A\times B=\{(1,3),(1,4),(2,3),(2,4)\}\). Option \(D\) reverses the order and represents a relation from \(B\) to \(A\), not from \(A\) to \(B\). Exam tip: The largest relation from \(A\) to \(B\) is always \(A\times B\).

The domain of a relation is the set of first components of its ordered pairs. Here, the first components are 1, 2, and 4, so the domain of \(R\) is \(\{1,2,4\}\); therefore, 3 is not in the domain. Exam tip: To find the domain, list only the first element from each ordered pair.

The domain of a relation is the set of first components of its ordered pairs, while the range is the set of second components. Here, the first components are 1, 2, and 3, so the domain is {1, 2, 3}; all the second components are 2, so the range is {2}. Therefore, option B is correct. Remember that the codomain cannot generally be determined from the relation alone.

The condition \(x-y=0\) gives \(x=y\). Thus each element of \(A\) must be paired with itself, producing \((1,1),(2,2),(3,3)\). Hence \(R=\{(1,1),(2,2),(3,3)\}\), the identity relation on \(A\). Option A contains pairs of unequal elements, so it does not satisfy the condition. Exam tip: whenever \(x-y=0\) appears, replace it with \(x=y\) and list the equal ordered pairs from the given set.

Since ((b,2)\in R), reversing it gives ((2,b)\in R^{-1}). Changing the order is the key point in inverse relations.

The pairs whose sum is (4) are ((1,3),(2,2),(3,1)). In such questions, check all possible pairs systematically.

The pairs satisfying (x<y) are ((1,2),(1,3),(2,3)). In exams, check all possible ordered pairs systematically.

In an inverse relation, the order of components in every pair is reversed. So ((2,1)) becomes ((1,2)), and similarly for the other pairs.

The governing definition is that a relation on A is any subset of A×A. The empty relation is the particular relation containing no ordered pairs at all, and it is denoted by R=∅. Thus option B is correct. For A={a,b}, the product A×A consists of (a,a), (a,b), (b,a), and (b,b). A relation may contain all of them, some of them, or none of them. Option A contains every possible pair, so it is the universal relation rather than the empty relation. Options C and D each contain one ordered pair, so neither can be empty. The empty-set symbol means that the relation has zero elements.

The universal relation contains all ordered pairs of (A\times A). In exams, think of it as the total relation.

In an identity relation, every element of the set is related only to itself. Here, 1 is related only to 1 and 2 only to 2, so R = {(1,1), (2,2)} is the identity relation. It is not a universal relation, because a universal relation on A must contain all ordered pairs: (1,1), (1,2), (2,1), and (2,2). Exam tip: The identity relation on A is generally written as {(a,a) : a ∈ A}.

The relation is explicitly listed as R={(1,1),(2,2),(3,3)}. To identify a pair that is definitely in R, we only need to compare each option with this list. An ordered pair includes both its entries and their order, so a pair such as (1,2) is different from (2,1), even if the same numbers occur.

The pair (2,2) appears exactly in the given relation, so option B is correct. The pairs (1,2), (3,1), and (2,3) do not appear in the list. This relation is the identity relation on A: each element is paired with itself and no different elements are paired. The word “definitely” is satisfied because membership is directly stated rather than inferred from an additional rule.