Cartesian product of sets

In (A\times B), the first component comes from (A) and the second from (B). In exams, always check the order of the pair.

The Cartesian product A × B contains all ordered pairs (a, b) such that a belongs to A and b belongs to B. Since A has 1 and 2 while B has only 3, the pairs are (1, 3) and (2, 3). The first coordinate must come from A, so option A is correct. Option B reverses the order, option C lists elements rather than ordered pairs, and option D forms an incorrect pair.

The governing concept is the Cartesian product. For finite sets A and B, every element of A is paired with every element of B, so the number of ordered pairs is n(A×B)=n(A)×n(B). Since A={5}, it has one element and n(A)=1. Since B={7,8}, it has two elements and n(B)=2. Therefore n(A×B)=1×2=2. Listing the product confirms the calculation: A×B={(5,7),(5,8)}. Thus option B is correct. Option A counts only one element of B, while options C and D overcount the possible pairings. The order within each pair matters, and the first coordinate must come from A while the second must come from B.

The direct answer is option B: 20. The Cartesian product A×B is the set of all ordered pairs (a,b) where the first element comes from A and the second comes from B. If A has 4 elements, each of those 4 elements can be paired with every one of the 5 elements of B. Therefore the total number of pairs is n(A×B)=n(A)×n(B)=4×5=20. Option B is correct. Option A, 9, is obtained by adding 4 and 5, but Cartesian-product counting uses multiplication, not addition. Option C, 1, does not represent the number of possible choices and has no basis here. Option D, 45, is not the product 4×5 and therefore cannot be the count. The order also matters in a Cartesian product: (a,b) is generally different from (b,a), though the counting formula remains the product of the two set sizes. Memory cue: for ‘each element with every element’, multiply.

There is no element in (A) for the first component, so no ordered pair is formed. Pay special attention to the empty set.

In ((2,3)), the first component (2) is in (A) and the second (3) is in (B). Check both positions separately for membership.

In (A\times B), the first component of an ordered pair is from (A) and the second is from (B). This rule solves most membership questions.

Order matters in Cartesian product, so (A\times B) is generally different from (B\times A). Reversed order is a common exam mistake.

(A\times B={(1,1),(1,2)}) and (B\times A={(1,1),(2,1)}). Similar looking numbers do not remove the importance of order.

In ((5,7)), the first component is (5) and the second component is (7). In Cartesian product, the first component comes from the first set.

In an ordered pair \((a,b)\), the first component is \(a\) and the second component is \(b\). Therefore, the second component of \((5,7)\) is 7. Option C reverses the order of the components, while option D represents a set. In exams, pay close attention to the position of each component.

(A) has (2) elements and (B) has (2) elements, so (2\times 2=4). For small sets, you may also list all pairs to check.

In ((3,5)), (3\in A) and (5\in B), so it belongs to (A\times B). Other options have wrong position or membership.

In ((3,1)), the first component (3) is not in (A). To find the wrong option, check the first component first.

The only element (0) of (A) stays in the first position and all elements of (B) come in the second position. Do not reverse order even for singleton sets.

For a Cartesian product, \(n(A\times B)=n(A)\times n(B)\). Since both sets here are \(A\), \(n(A\times A)=3\times3=9\). Option 6 is incorrect because the cardinalities are multiplied, not added. Exam tip: the Cartesian product of an \(n\)-element set with itself contains \(n^2\) ordered pairs.

For two finite sets, n(A × B) = n(A) × n(B). Thus, 12 = 3 × n(B), so n(B) = 12 ÷ 3 = 4. Option 9 is incorrect because the unknown cardinality is found by dividing the product by the known cardinality, not by multiplying them. Exam tip: The number of ordered pairs in a Cartesian product equals the product of the cardinalities of the two sets.

If (A) is not empty and still no pair is formed, then (B) must be empty. If a Cartesian product is empty, at least one set is empty.

In (A\times B), the first position is filled from (A) and the second from (B). Confusing it with (A\cup B) is a common mistake.

Changing positions changes an ordered pair. Equality holds only when corresponding components are equal.

In (A\times A), every element of (A) pairs with every element of (A). Taking only equal component pairs is incomplete.

There is no element in (B), so no second component is available. Since the number is asked, the answer is (0).

Both sets have one element each, so only ((4,6)) is formed. The product of singleton sets gives one ordered pair.

The direct answer is B: a=2 and b=3. Ordered pairs are equal only when their corresponding components are equal. Thus, from (a,3)=(2,b), compare first components: a=2. Compare second components: 3=b, so b=3. Option B is correct. Option A reverses the positions and treats the pair as unordered; an ordered pair is not like an ordinary set. Option C gives the wrong second component. Option D gives the wrong first component. The comma order matters: first coordinate matches first coordinate, and second matches second. This rule is also used in Cartesian products and coordinate geometry.

By definition, (A\times B={(x,y):x\in A,\ y\in B}). In set-builder form, keep the order of components unchanged.