Relations

Odd whole numbers less than 8 are 1, 3, 5, 7. Listing exactly these elements gives the limited roster form {1, 3, 5, 7}.

A relation from (A) to (B) is always a subset of (A\times B). In exams first identify the Cartesian product.

A relation on (A) is a subset of (A\times A), and (|A\times A|=4), so total relations are (2^4=16). Remember subset counting for exams.

The domain of a relation is the set of all first coordinates appearing in its ordered pairs. It is not the set of second coordinates, and it is not the relation itself as a collection of complete pairs. Repeated first coordinates would be written only once because a domain is a set.

The relation contains (1,2), (2,3), and (3,4). Reading the first entry from each pair gives 1, 2, and 3. Therefore the domain is {1,2,3}, which is option A. The set {2,3,4} is the range, since it contains the second coordinates. The complete list of ordered pairs is R itself, not its domain. Thus the distinction between first and second components determines the answer.

The range is the set of second components, and repeated values are written once. So the range is ({5,6}).

In the ordered pair ((3,7)), (3) is the preimage and (7) is the image. While reading relations, the first component forms the domain.

For a relation from (A) to (B), the codomain is the whole set (B). The range may be a subset of the codomain.

The direct answer is option A: \(R=\varnothing\). A relation from A to B is any subset of \(A\times B\), so it may contain some pairs, all pairs, or no pairs. The empty relation is the relation containing zero ordered pairs. Therefore it is written \(R=\varnothing\). Option A is correct. Option B, \(R=A\times B\), is the universal or full relation here; it contains all four possible pairs \((1,x),(1,y),(2,x),(2,y)\), so it is not empty. Option C contains one pair, \((1,x)\), and option D contains one pair, \((2,y)\); each is therefore non-empty. A common mistake is to think a relation must contain at least one pair. It does not: the empty set is a valid subset of every set, including \(A\times B\).

A universal relation contains all ordered pairs of (A\times A). Therefore the answer is (A\times A).

The governing definition is that the identity relation on a set A contains exactly every ordered pair (a, a) for a ∈ A. For A = {1, 2, 3}, those pairs are (1,1), (2,2), and (3,3), so option B is the identity relation. Option A cycles distinct elements, C gives several images for 1, and D contains no required diagonal pair.

In the inverse relation, the two components of each ordered pair are interchanged. Thus ((1,2)) becomes ((2,1)).

Since the condition \(a=b\) permits only ordered pairs with equal components, \(R=\{(1,1),(2,2),(3,3)\}\). This is the identity relation on \(A\). It is not the universal relation, which would contain all 9 ordered pairs of \(A\times A\). Exam tip: In an identity relation, each element \(a\in A\) is related only to itself, forming the pair \((a,a)\).

In ((1,4)), (1<4) is true and both components are from the correct sets. Check both the condition and the sets in a relation.

The governing concept is the identity relation on a set. For A = {1,2,3}, the identity relation is I_A = {(1,1), (2,2), (3,3)}; every pair has equal first and second coordinates and represents an element related to itself. Comparing R with this set, the first three listed pairs are precisely identity pairs. The remaining pair is (1,2), whose coordinates are unequal, so it is not part of the identity relation. Therefore option D is correct. This does not mean that (1,2) is invalid as an ordered pair; it is valid in A × A, but it is an extra non-identity pair in R. Options A, B, and C are already contained in the identity relation.

The first components give domain ({1,2}), and the second components give range ({5}). Therefore the first option is correct.

A pair belongs to the relation exactly when both coordinates are from {1,2,3,4} and their sum is 5. To find the pair that does not belong, substitute each option into the condition x+y=5. This direct check avoids relying on the appearance of the numbers alone.

For (1,4), the sum is 5, so it belongs. For (2,3), the sum is 5, so it belongs. For (3,2), the sum is also 5, so it belongs. For (4,2), the sum is 6, not 5, so it does not belong. Therefore option D is correct. Although both coordinates of (4,2) are in the stated set, the defining equation is not satisfied.

The range is formed only from second components actually appearing in the relation. (5) is in the codomain, but not in the range.

The governing concept is the domain of a relation. The domain is the set of all first components of the ordered pairs in that relation, not the set of complete pairs. In R, the pairs are (1, 1), (2, 2), and (3, 3), whose first components are 1, 2, and 3. Therefore Domain(R) = {1, 2, 3}, which is the whole set A, so option B is correct. This relation is also the identity relation on A because each element is related to itself. Option A includes only one first component, option C repeats the original ordered pairs rather than extracting their first components, and option D would be appropriate only for an empty relation.

The direct answer is option C: 4 pairs. The relation contains pairs \((a,b)\) from \(A\times B\) satisfying \(a\le b\). List all possible pairs: \((1,2),(1,3),(2,2),(2,3)\). Check them one by one: \(1\le2\) is true, \(1\le3\) is true, \(2\le2\) is true because equality is allowed, and \(2\le3\) is true. Thus all four possible pairs belong to R. Option C is correct. Option A, 2, misses two valid pairs. Option B, 3, also omits one valid pair. Option D, 5, is impossible because \(A\times B\) itself has only \(2\cdot2=4\) pairs, so a relation contained in it cannot have five pairs. A common mistake is reading \(\le\) as strictly less than; the equal pair \((2,2)\) must be included. Memory cue: the symbol \(\le\) means “less than or equal to,” so equality qualifies.

In the inverse relation \(R^{-1}\), the two components of every ordered pair in \(R\) are interchanged. Thus, because \((1,2)\in R\), we have \((2,1)\in R^{-1}\). In fact, \(R^{-1}=\{(2,1),(1,2),(3,3)\}\). Options B and C do not arise from the given pairs, while option D would be the inverse of \((1,3)\), which is not in \(R\). Exam tip: To find an inverse relation, swap the first and second elements of every ordered pair.

The ordered pair \((2,3)\) belongs to \(R\) because \(b=3\) and \(a+1=2+1=3\), and both elements are in \(A\). In option A, \(1\neq 1+1\); in option C, \(2\neq 3+1\); and in option D, 5 is not an element of \(A\). Exam tip: For each ordered pair, first verify \(b=a+1\), then check that both elements belong to \(A\).

When (R\subseteq A\times B), it is a relation from the first set (A) to the second set (B). Never change the order.

The governing concept is the image of an element under a relation. In an ordered pair (x,y), the first coordinate is the input and the second coordinate is its related image. To find which elements have image 4, inspect the pairs whose second coordinate equals 4. The pairs (1,4) and (2,4) show that 1 and 2 are mapped or related to 4. The pair (3,5) instead gives image 5 for input 3. Therefore the set of elements whose image includes 4 is {1,2}, so option A is correct. Option B confuses the pair producing 5, option C lists output values rather than input elements, and option D includes 3 even though its second coordinate is 5.

The pair ((2,3)) is wrong because (2) does not divide (3). In divisibility relations, the remainder must be zero.

A relation containing all pairs of (A\times A) is the universal relation. It is the largest possible relation.