Cartesian product of sets

For B × A, the first coordinate must be an element of B and the second coordinate must be an element of A. Since B = {2,4}, the first entry must be 2 or 4. Since A = {1,3,5}, the second entry must be 1, 3, or 5. The pair (4,3) meets both requirements: 4 ∈ B and 3 ∈ A, so it belongs to B × A. Therefore option A is correct. The pair (3,4) reverses the factor order and belongs to A × B. The pair (5,2) also has an A-element first and a B-element second, while (1,4) is invalid because 1 is not in B. The question therefore tests ordered membership, not merely whether both numbers occur somewhere in the two sets.

The governing definition says that every pair in A × B has the form (a,b), where a ∈ A and b ∈ B. Therefore the possible second coordinates come from B. Since A is non-empty, each element of B actually occurs as a second component: (m,7), (m,8), (m,9), (n,7), (n,8), and (n,9) are all in the product. The distinct second components are consequently {7,8,9}, so option A is correct. Option B lists the possible first components, not second components. Option C combines elements from both sets instead of identifying one coordinate position, and option D gives only two ordered pairs rather than the set of coordinate values. The result is a projection onto the second coordinate.

The defining concept is the Cartesian product of two sets. By definition, A × B = {(x, y) : x ∈ A and y ∈ B}; therefore, the first coordinate is chosen from A and the second coordinate from B. In this example the product contains (1,3), (1,4), (1,5), (2,3), (2,4), and (2,5). Option B reverses the sets and describes B × A, not A × B. Options C and D produce numerical sums or products, so they describe sets of numbers rather than sets of ordered pairs. The use of parentheses and the separate membership conditions are essential. Hence option A gives the correct set-builder representation.

The governing definition of a Cartesian product is B × A = {(b, a) : b ∈ B and a ∈ A}. The order is essential: the first component must come from B, while the second component must come from A. For option B, 3 belongs to B and 2 belongs to A, so (3, 2) is an element of B × A. Option A has first component 2, which is not in B. Option C has first component 5 in B, but its second component 1 is not in A. Option D has first component 1 in B, but second component 3 is not in A. Notice that reversing the product to A × B would change the membership rule and could change which pairs are valid.

The governing concept is the Cartesian product: A × B contains ordered pairs whose first component is selected from A and whose second component is selected from B. Test each possible y in B using x = 3y. For y = 1, x = 3, giving (3, 1), which is valid. For y = 2, x = 6, giving (6, 2), also valid. For y = 3, x = 9, giving (9, 3), also valid. Thus the complete set of satisfying pairs is {(3,1), (6,2), (9,3)}, containing three pairs. The distractor 2 omits one valid pair, 6 confuses the answer with the total size of A × B, and 1 counts only one trial.

The governing idea is equality of sets and the definition of a Cartesian product. If A = B, then A and B have exactly the same elements, so replacing B by A in A × B gives A × A. Similarly, replacing A by B in B × A gives B × B. Since A and B are equal sets, A × A and B × B contain exactly the same ordered pairs; therefore A × B = B × A. This conclusion is valid for every set, including non-empty and empty sets. Option A is incorrect because equality follows from the premise. Option C is too restrictive, and option D gives an irrelevant condition: the result depends on equality of the sets, not merely on their cardinalities.

The defining rule for A × B is that every ordered pair has its first component from A and its second component from B. The question specifies both positions: the first must be 3 and the second must be b. Therefore the required ordered pair is (3, b). The order is essential; (b, 3) belongs to B × A, not generally to A × B, and it reverses the requested components. The pair (3, a) has the correct first component but the wrong second component, while (2, b) has the correct second component but the wrong first component. This illustrates that ordered pairs are position-sensitive: changing either coordinate changes the pair. Thus option A follows directly from the definition and the stated component requirements.

The governing rule for the cardinality of a Cartesian product of finite sets is |A × B| = |A| × |B|. Set A has two distinct elements, 0 and 1, so |A| = 2. Set B has three distinct elements, 2, 4, and 6, so |B| = 3. Each element of A can be paired with every element of B, giving the ordered pairs (0,2), (0,4), (0,6), (1,2), (1,4), and (1,6). Thus the total is 2 × 3 = 6, so option B is correct. Option D counts only the elements of B, option A does not follow from the product rule, and option C overcounts the possible ordered pairs. Order matters in a Cartesian product, although here only the number of pairs is required.

Membership in A × B must be checked coordinate by coordinate: the first coordinate must lie in A and the second must lie in B. For (2,2), 2 ∈ A and 2 ∈ B, so it is present. For (3,4), 3 ∈ A and 4 ∈ B, so it is also present. For (4,3), the second coordinate 3 is in B, but the first coordinate 4 is not in A, because A contains only 1, 2, and 3. Therefore (4,3) is not an element of A × B. Option D is false because two pairs are present. The example also shows why the order of coordinates matters: even though 4 occurs in B, it cannot occupy the first position in A × B.

The cardinality rule for Cartesian products is |X × Y| = |X| × |Y|. Here A has three elements, so |A| = 3, and B has two elements, so |B| = 2. Although the order B × A means that the first coordinate comes from B and the second from A, the number of choices is still 2 × 3 = 6. Explicitly, the pairs are (1,a), (1,b), (1,c), (2,a), (2,b), and (2,c). Therefore option B is correct. Option A adds the cardinalities instead of multiplying them, option C has no valid product basis, and option D would incorrectly treat both sets as having three elements.

The governing concepts are set union and Cartesian-product cardinality. In a union, an element appearing in both sets is counted only once. Combining B = {2,3,4} and C = {3,4,5} gives B ∪ C = {2,3,4,5}, because 3 and 4 are repeated but are retained once. Thus |B ∪ C| = 4. Set A = {1,2,3} has |A| = 3. Applying the Cartesian-product rule, |A × (B ∪ C)| = |A| × |B ∪ C| = 3 × 4 = 12. Therefore option B is correct. Option A incorrectly treats the union as having three elements. Options C and D result from incorrect multiplication or from counting repeated elements more than once. Each of the three elements of A pairs with all four elements of the union.

The governing concepts are set difference and the Cartesian product. In B − C, we retain elements that belong to B but not to C. Since B = {1,2,3} and C = {2,3,4}, the common elements 2 and 3 are removed from B, leaving B − C = {1}. The Cartesian product A × {1} contains one ordered pair for every element of A, with 1 fixed as the second component. Therefore it is {(0,1),(1,1),(2,1)}, so option A is correct. Option B incorrectly retains 2, option C incorrectly retains 3, and option D would apply only if every element of B also belonged to C.

The governing concept is the membership rule for a Cartesian product. A pair (a,b) belongs to A × B if and only if its first component is in A and its second component is in B. For (3,6), 3 belongs to A and 6 belongs to B, so the pair is present. For (5,6), the second component 6 is in B, but the first component 5 is not in A; therefore this pair is absent. For (4,7), 4 is in A and 7 is in B, so the pair is present. Thus only option B is not in the product. Option D is false because it claims that all three pairs are present. The positions cannot be interchanged when testing membership.

For finite sets, the cardinality of a Cartesian product is the product of the cardinalities: n(A × B) = n(A) × n(B). Substituting the given values gives 72 = 8 × n(B). Dividing both sides by 8 yields n(B) = 72/8 = 9. The value 8 is n(A), not n(B), while 64 and 80 do not satisfy the product equation. Thus option A is the only correct answer.

The Cartesian product contains one ordered pair for every choice of an element from A and an element from B. Therefore, \(|A\times B|=|A||B|=3\times2=6\). A set with six elements has \(2^6\) subsets, because each element can independently be included or excluded. Thus \(|\mathcal{P}(A\times B)|=2^6=64\). Option D is correct; 32 would correspond to a five-element set, not this product.

A Cartesian product A × B consists of ordered pairs whose first coordinate comes from A and whose second coordinate comes from B. The relation is defined by b = 2a, so calculate b for each a in A. For a = 1, b = 2 and the pair is (1, 2). For a = 2, b = 4 and the pair is (2, 4). For a = 3, b = 6 and the pair is (3, 6). Each resulting second coordinate belongs to B, so the complete relation is the set shown in option A. Option B reverses the coordinates, which changes an ordered pair. Option C omits a valid pair, and option D contains pairs that do not consistently satisfy b = 2a.

The governing concept is membership in a Cartesian product together with a rule-defined relation. For every a in A, calculate b = a² and retain the pair only if that value belongs to B. The calculations are: (-2)² = 4, (-1)² = 1, 0² = 0, 1² = 1, and 2² = 4. Each resulting value is in B, so the valid ordered pairs are (-2,4), (-1,1), (0,0), (1,1), and (2,4). Although the second and fifth values repeat, the first coordinates differ, so the ordered pairs remain distinct. Thus there are five pairs, making option C correct. Options A and B omit valid inputs, while D counts an extra pair that does not exist.

The governing idea is that A × A contains every ordered pair (a,b) whose two coordinates are chosen independently from A. We need to list all choices for which the sum is 5. Taking a = 1 gives b = 4, taking a = 2 gives b = 3, taking a = 3 gives b = 2, and taking a = 4 gives b = 1. Therefore the complete set is {(1,4), (2,3), (3,2), (4,1)}, containing four ordered pairs. Order matters: (1,4) and (4,1) are different pairs even though both have the same sum. Hence option C is correct. Option A misses two valid pairs, option B misses one, and option D includes a pair with no possible first coordinate in A.

The direct answer is 2, so option B is correct. The Cartesian product A×B contains every ordered pair (x,y) in which the first entry x comes from A and the second entry y comes from B. Set A has two elements, a and b, while B has only one element, 1. Pairing each element of A with the only element of B gives (a,1) and (b,1), and no other pairs are possible. The cardinality rule confirms this: |A×B|=|A|×|B|=2×1=2. Option A counts only one first-coordinate choice and omits one pair. Option C overcounts by one, and option D incorrectly behaves as if both sets had two elements. Order also matters: (1,a) is not in A×B because its first coordinate does not come from A and its second coordinate does not come from B. A useful cue is “first set gives first coordinate, second set gives second coordinate.”

In the Cartesian product \(A\times A\), there are 3 choices for the first component and 3 choices for the second component. Hence, \(|A\times A|=|A|\times|A|=3\times3=9\). The value 6 misses some ordered pairs; for example, \((1,2)\) and \((2,1)\) are distinct. Exam tip: Remember the formula \(|A\times B|=|A|\,|B|\).

The set \\(A\\) has 2 elements. Therefore, the number of ordered pairs in \\(A\\times A\\) is \\(n(A)\\times n(A)=2\\times2=4\\). The pairs are \\( (1,1),(1,2),(2,1),(2,2)\\). Hence, 4 is correct. Exam tip: If \\(n(A)=n\\), then \\(n(A\\times A)=n^2\\).

The Cartesian product 8A\times A9 contains all ordered pairs whose two components belong to 8A9. Here, 8A\times A=\{(1,1),(1,2),(2,1),(2,2)\}9, so 8(2,1)9 is correct. Every other option contains 3, which is not an element of 8A9. Exam tip: In 8A\times A9, both positions must be filled only with elements of 8A9.

The Cartesian product A × A consists of every ordered pair (a, b) in which both a and b are selected from A. Since A has 3 elements, there are 3 choices for the first coordinate and independently 3 choices for the second coordinate. By the multiplication principle, |A × A| = |A| × |A| = 3 × 3 = 9. Therefore R, which is the entire Cartesian product, contains 9 ordered pairs, so option C is correct. Explicitly, the pairs are (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), and (3,3). Counting only the elements of A gives 3, while 6 and 12 do not follow from the product rule.

For finite sets, the Cartesian product A × B contains all ordered pairs (a, b) with a from A and b from B. The number of pairs is obtained by multiplying the number of choices for each coordinate: |A × B| = |A| × |B|. Here |A| = 2 and |B| = 3, so |A × B| = 2 × 3 = 6. The pairs are (1,3), (1,4), (1,5), (2,3), (2,4), and (2,5), confirming the count. Thus option B is correct. Adding the set sizes would give 5, but addition counts elements across a union rather than combinations of two coordinates; 8 and 9 have no basis in the product rule.

The number of ordered pairs in a Cartesian product is the product of the cardinalities of the two sets. Here, \(|A\times B|=2\times3=6\). For any finite set containing \(n\) elements, the total number of subsets is \(2^n\), since every element independently has two choices: included or not included. Therefore, \(A\times B\) has \(2^6=64\) subsets. Option C is correct; option A is only the number of product elements.