Cartesian product of sets

Every element of (A) is in (B), so every pair of (A\times D) will be in (B\times D). In subset questions, check the related component.

Order matters in a Cartesian product. In \(A\times B\), the first element comes from \(A\) and the second from \(B\), so \(A\times B=\{(6,9)\}\). Similarly, \(B\times A=\{(9,6)\}\). Option B is incorrect because \((6,9)\) and \((9,6)\) are different ordered pairs. Exam tip: In \(X\times Y\), the first component comes from \(X\) and the second from \(Y\).

Each ordered pair can be treated as a point and there are (3\times 2=6) pairs. Use the same counting rule in coordinate questions.

There are (3) breakfast options and (2) drink options, so (3\times 2=6) combinations are formed. Real-life choices can be counted using Cartesian product.

A Cartesian product lists ordered pairs whose first member comes from the first set and whose second member comes from the second set. The number of pairs is found by multiplying the number of elements in the two sets. The order of the sets changes the pairs themselves, but it does not change their total number.

Here, set A has four elements: 2, 4, 6, and 8. Set B has one element: 1. In B × A, the only possible first component is 1, and it can be matched with each of the four elements of A. Thus the pairs are (1,2), (1,4), (1,6), and (1,8), giving 1 × 4 = 4 elements. Therefore option B is correct; option A would count only the elements of B.

The first component (2) is fixed and the second component is (7) or (8) from (B). With a fixed first component, all elements of the second set are used.

The second component (5) is fixed and the first component is (6) or (8) from (A). In such questions, do not change the position of the fixed component.

The direct answer is option A: ordered pairs. By definition, the Cartesian product A×B consists of all pairs (a,b) such that a belongs to A and b belongs to B. Here A={0,2} and B={1,4}, so the product is {(0,1),(0,4),(2,1),(2,4)}. Each item has two positions, and the position matters: the first position is from A and the second is from B. Thus the elements are ordered pairs. Option A is correct. Option B, ordinary numbers, is wrong because a product element is written as a pair, not as a single number; for example, (0,1) is not the same object as 1. Option C, only elements of A, ignores the required second component from B. Option D, only elements of B, ignores the required first component from A. The notation itself gives a clue: A×B means pairs with an A-choice first and a B-choice second. Remember the order A then B.

The governing concept is the definition of Cartesian-product membership: (u, v) belongs to A × B exactly when the first component u belongs to A and the second component v belongs to B. For (3,4), 3 is an element of A = {1,3}, and 4 is an element of B = {2,3,4}. Thus both required conditions are satisfied, so option A is correct. Option B reverses the roles of the two sets, even though 3 is also in B; importantly, 4 is not in A. Option C is false because 3 and 4 are different, and option D is false because A and B are not equal. Membership depends on position, not on equality of the sets.

This question uses two governing ideas: interpreting set-builder notation and applying the cardinality rule for a Cartesian product. With the standard school convention that N denotes the positive natural numbers, the condition x ≤ 3 gives A = {1,2,3}, so n(A) = 3. The set B = {0,1} has n(B) = 2. Every element of A can be paired independently with every element of B, producing 3 × 2 possible ordered pairs. Hence n(A × B) = n(A)n(B) = 3 × 2 = 6, and option C is correct. Option A counts only A, option B incorrectly adds the cardinalities, and option D does not use the sizes of the two different factors correctly. If a convention included 0 in N, the question would need clarification; under the stated school convention, C is unambiguous.

The answer is option B, 3. A pair in A×B has equal components when its first and second entries are the same. Since both sets are {2,4,6}, the qualifying pairs are (2,2), (4,4), and (6,6), giving three pairs. Option A counts too few and omits two valid pairs. Option C, 6, is not the number of equal-component pairs; it may result from confusing the condition with another count. Option D, 9, is the total number of all pairs in A×B, since 3×3=9, but most of those pairs have unequal components. Therefore B is correct. Look specifically for diagonal pairs (x,x).

In ((-1,3)), the first component (-1\in A) and the second (3\in B). A negative number is checked like any ordinary element.

The first component (\text{night}) is from (A) and the second component (\text{winter}) is from (B). Therefore, it is an element of (A\times B).

The second component (7) is fixed and the first component comes from (2) or (5) in (A). With a second-component condition, the first position changes.

Elements of (A) come in the first position and (0) from (B) comes in the second position. Do not treat Cartesian product as ordinary union.

The largest element of (A) is (6), and it pairs with the (3) elements of (B). For a fixed first component, the answer is (n(B)).

The smallest element of (B) is (4), and the first component can come from the (3) elements of (A). For a fixed second component, the answer is (n(A)).

The governing concept is conditional counting in a Cartesian product. An ordered pair (a,b) belongs to A × B only when a ∈ A and b ∈ B; here it must also satisfy a + b = 4. Check each permitted value of the second component. If b = 2, the equation gives a = 2, and 2 is in A, so (2,2) is valid. If b = 4, the equation gives a = 0, and 0 is in A, so (0,4) is valid. No other value of b is available, so there are exactly two acceptable pairs. Thus option B is correct. Although the complete product contains 3 × 2 = 6 pairs, most do not satisfy the sum condition. For example, (1,3) has sum 4 but is invalid because 3 is not in B.

The direct answer is option C: 4 pairs. A Cartesian product contains every ordered pair
(a,b) with a from A and b from B. Here A has the even numbers 2 and 4, while B has the odd numbers 1 and 5. Pair them systematically: (2,1) has sum 3, (2,5) has sum 7, (4,1) has sum 5, and (4,5) has sum 9. Every sum is odd because even plus odd is always odd. There are 2 choices for the first component and 2 choices for the second, so the total is 2 times 2 = 4. Option A, 0, is wrong because no pair fails the odd-sum condition. Option B, 2, counts only part of the pairs. Option C, 4, correctly counts all pairs. Option D, 6, is impossible because the product has only four pairs. Memory cue: even plus odd always gives odd; count pairs by multiplying the set sizes.

The ordered pairs in \(A\times B\) are \((1,2),(1,6),(3,2),(3,6)\). In every pair, the second component is even, so every product is even. The total number of pairs is \(|A|\times|B|=2\times2=4\). Therefore, the correct answer is 4. Exam tip: A product is even if at least one of its factors is even.

The direct answer is option C, 3 pairs. In A × B, the first entry comes from A={1,2,4} and the second from B={1,3}. We check each first entry against each second entry, keeping the order. For first entry 1, neither (1,1) nor (1,3) has 1 greater than the second entry. For first entry 2, (2,1) works but (2,3) does not. For first entry 4, both (4,1) and (4,3) work. Therefore the required pairs are (2,1), (4,1), and (4,3), making 3 pairs. Option A, 1, misses two valid pairs. Option B, 2, counts only the pairs beginning with 4 and misses (2,1). Option C is correct. Option D, 6, counts all pairs in A × B, since 3×2=6, but the question asks only for pairs satisfying the inequality. Always preserve the coordinate order.

The equal-component pairs are ((2,2)) and ((3,3)). In such questions, look at common elements of both sets.

The governing idea is enumeration of ordered pairs under a coordinate equation. For (a,b) ∈ A × B, the first coordinate must come from A, the second from B, and b − a must equal 1. Test the possible values of b. When b = 1, the equation 1 − a = 1 gives a = 0; since 0 ∈ A, (0,1) is valid. When b = 3, the equation 3 − a = 1 gives a = 2; since 2 ∈ A, (2,3) is valid. These exhaust B, so exactly two pairs satisfy the condition and option B is correct. The pair (4,3) fails because 3 − 4 = −1, not 1. The product has six pairs in all, but the question asks for the subset satisfying the stated difference.

The pairs are ((2,6)) and ((5,6)), whose products are greater than (10). In condition-based questions, quickly check all possible pairs.

The governing definition is B × A = {(b,a) : b ∈ B and a ∈ A}. Therefore the first component must be selected from B = {1,3}, while the second component must be selected from A = {2,4,6}. Option A, (1,4), satisfies both requirements because 1 ∈ B and 4 ∈ A. Thus option A is correct. Option B, (4,1), reverses the required order and belongs to A × B. Option C, (2,3), also has its first coordinate from A and its second from B, so it belongs to A × B rather than B × A. Option D similarly places 6, an element of A, first. This demonstrates that Cartesian products are ordered: in general, B × A is not the same set as A × B.