Relations

A relation on (A) is a subset of (A\times A), and (n(A\times A)=16). Therefore total relations are (2^{16}).

The pairs are ((1,2),(1,3),(1,4),(2,3),(2,4),(3,4)). For such questions, list pairs systematically using the condition.

For every (a\in A), ((a,a)\in R), so the relation is reflexive. Symmetry would also require ((2,1)).

For every ((a,b)\in R), ((b,a)\in R) is also present, so it is symmetric. To be reflexive, it needs ((1,1),(2,2),(3,3)).

Since ((1,2)) and ((2,3)) imply ((1,3)), the relation shows transitivity. In exams, check matching middle elements.

An ordered pair \((a,b)\) belongs to \(R\) when \(a-b\) is even. For option C, \(1-3=-2\), which is even; therefore, \((1,3)\in R\). In contrast, option D gives \(4-1=3\), which is odd. Exam tip: two numbers have an even difference exactly when they have the same parity.

The number (4) does not divide (2), so ((4,2)\notin R). In divisibility, treat the first element as the divisor.

The domain of a relation is the set of first components of its ordered pairs. The first components here are 1, 2, and 3, so the domain is \(\{1,2,3\}\). Option B is the range, consisting of the second components, while option C combines both sets. Exam tip: To find the domain, collect the first entry from every ordered pair.

The direct answer is option B, {1,3,7}. In an ordered pair (first, second), the domain is formed from first components, while the range is formed from second components. Read the relation pair by pair: (2,1) contributes 1 to the range; (2,3) contributes 3; (4,3) contributes 3 again; and (5,7) contributes 7. Sets do not repeat an element, so the range is {1,3,7}. Option A, {2,4,5}, is the set of first components, so it is the domain, not the range. Option B correctly lists the second components. Option C, {1,2,3,4,5,7}, mixes first and second components and includes values that are not all distinct range values. Option D, {3}, leaves out 1 and 7, although both occur as second components. The repeated 3 is written only once. Memory cue: domain means first positions; range means second positions.

In the inverse relation, the components of each ordered pair are interchanged. Thus ((1,3)) becomes ((3,1)).

The direct answer is option A, {(1,3),(2,2),(3,1)}. Both a and b must belong to A={1,2,3}, and they must satisfy a+b=4. Test the possible first values: if a=1, then b=4-1=3, giving (1,3). If a=2, then b=4-2=2, giving (2,2). If a=3, then b=4-3=1, giving (3,1). These are all possibilities because A has only 1, 2, and 3. Option A contains exactly these three pairs. Option B contains (1,2) and (2,1), but each sum is 3, not 4. Option C contains equal pairs whose sums are 2, 4, and 6; only (2,2) satisfies the condition. Option D contains (3,3), whose sum is 6, so it fails. Notice that addition is commutative, so both (1,3) and (3,1) occur, but only values from A are allowed. Memory cue: find b=4-a for every a in A.

In ((4,5)), (5\notin A), so it is not a pair of (A\times A). While forming a relation, both components must belong to the set.

By definition, a relation from \(A\) to \(B\) consists of ordered pairs \((a,b)\) where \(a\in A\) and \(b\in B\). All such ordered pairs form the Cartesian product \(A\times B\). Any collection of them is therefore a subset of \(A\times B\).

Total relations are (2^{5\times 3}=2^{15}), and removing the empty relation gives (2^{15}-1). For non-empty relations, remember to subtract (1).

There is no ordered pair in (\varnothing), so it is the empty relation. Since (A\neq\varnothing), it is not reflexive.

When a relation contains all pairs of (A\times A), it is the universal relation. It contains every possible ordered pair.

In ((4,2)), (4\geq 2) is true, so it belongs to (R). In inequalities, changing the order can change the answer.

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 in \(R\), but \((3,3)\) is missing, so the relation is not reflexive. The pairs in options A and C are not required for reflexivity because their two components are different. Exam tip: To test reflexivity, check only pairs of the form \((a,a)\) for every element of the set.

An ordered pair \((a,b)\) belongs to \(R\) only when \(a+b\) is odd. For \((2,3)\), \(2+3=5\), which is odd; therefore, \((2,3)\in R\). The sums for the other options are 4, 6, and 6, all of which are even. Exam tip: the sum of two numbers is odd exactly when one number is even and the other is odd.

Option A is reflexive because every pair \((a,a)\) is present. It is symmetric since \((1,3)\) and \((3,1)\) both occur, and it is transitive. Option B lacks \((3,1)\). Exam tip: check reflexivity, symmetry and transitivity separately.

The governing concept is counting ordered pairs that satisfy an absolute-difference condition. The equation |a−b|=1 means that the entries are adjacent integers, and either order is allowed. In A={1,2,3,4}, the adjacent pairs are 1 and 2, 2 and 3, and 3 and 4. Since the relation contains ordered pairs, each adjacency contributes two: (1,2),(2,1), (2,3),(3,2), and (3,4),(4,3). Thus R contains six pairs, so option D is correct. Counting only the forward direction gives 3 and misses the reverse pairs. The values 4 and 5 cannot result from the complete symmetric list. No pair such as (1,1) is included because its absolute difference is zero, not one.

For each \(a\in A\), the relation gives \(b=2a\). Substituting \(a=1,2,3\) gives \(b=2,4,6\), so \(R=\{(1,2),(2,4),(3,6)\}\). Therefore, the range of \(R\) is \(\{2,4,6\}\). Option A is the domain, not the range. Exam tip: To find the range of a relation, collect the second components of its ordered pairs.

In every ordered pair \((x,y)\), the second component is 3 greater than the first: \(4=1+3\), \(5=2+3\), and \(6=3+3\). Therefore, the rule is \(y=x+3\). For example, \(y=2x\) would give \((1,2)\), so it does not match the given pairs. Exam tip: substitute the coordinates of each ordered pair into a proposed rule to verify it.

It is reflexive because ((1,1),(2,2),(3,3)) are present, and symmetric because the reverse of ((1,2)) is ((2,1)). Hence it satisfies both.

Since ((1,2)) and ((2,3)) are present, transitivity requires ((1,3)). It is missing, so transitivity fails.