Cartesian product of sets

In ((4,1)), the first element is not from (A) and the second is not from (B). Position checking is necessary in ordered pairs.

The greatest first component is (2) and the greatest second component is (5), so the maximum sum is (7). ((5,2)) is not in (A\times B).

The condition x=y requires the same value to occur in both sets, because x must come from A and y must come from B. The common elements are A∩B={1,2}. Each common element produces exactly one equal-component pair: (1,1) and (2,2). Therefore the required number is |A∩B|=2. Although A contains 3 elements, the value 3 is absent from B, so (3,3) cannot belong to A × B. Option C is the full product size, |A|·|B|=3·2=6, without applying the equality condition. Option D overlooks one of the two common elements. This illustrates that diagonal pairs in a product are counted through the intersection of the two sets.

Using \(x=2y\), check each value of \(y\) in \(B\): \(y=1\) gives \(x=2\), \(y=2\) gives \(x=4\), and \(y=3\) gives \(x=6\). Thus, the valid ordered pairs are \((2,1),(4,2),(6,3)\), so there are 3 pairs. Option 2 misses one valid pair. Exam tip: in \(A\times B\), the first element comes from \(A\) and the second from \(B\).

(A\times B={(x,0):1\le x\le3}), so it is a line segment on the (x)-axis. In interval questions, write the coordinate form.

By definition, A × B contains all ordered pairs (a,b) with a∈A and b∈B. Since A has only the element 0, the first coordinate must always be 0. The second coordinate may be any real number y in the interval [−2,2]. Therefore A × B={(0,y):−2≤y≤2}, which is option A. Geometrically, these points form the vertical line segment on the y-axis from (0,−2) to (0,2). Option B places the variable in the first coordinate and reverses the product order. Option C also reverses the coordinates, and option D keeps only the endpoint-like pair (0,0), omitting infinitely many other valid pairs such as (0,1) and (0,−1.5).

Here, \(A=\{1,2\}\) and \(B=\{1,2,3\}\), so \(n(A)=2\) and \(n(B)=3\). In a Cartesian product, every element of the first set is paired with every element of the second set; therefore, \(n(A\times B)=n(A)\times n(B)=2\times3=6\). Option 3 gives only the number of elements in \(B\), so it is incorrect. Exam tip: remember the formula \(n(A\times B)=n(A)n(B)\).

(A\times B) is a set of ordered pairs, so its form is ({(x,y):x\in A,,y\in B}). Sum or product of entries is not Cartesian product.

In (A\times B), all first components come from (A), and (B) is non-empty. Therefore the set of first components is (A).

Second components always come from (B). If (A) is non-empty, every element of (B) appears as a second component.

In Cartesian product, every element of (A) pairs with each element of (B). That is why total pairs are (n(A)n(B)).

The listed valid distinct pairs are ((1,2),(1,4),(2,4),(3,4)), so the count is (4). Avoid counting the same pair twice.

The sum is even when both components are even or both are odd. Here the valid pairs are ((2,2),(2,4),(4,2),(4,4)).

A pair in A × B has its first component in A and its second component in B. We need products equal to 2, so each possible value of x must be tested with values of y from B. The zero value cannot work because multiplying by zero gives zero, not 2.

If x=1, y=2 gives 1×2=2, so (1,2) is valid. If x=2, y=1 gives 2×1=2, so (2,1) is also valid. No other value of y works for either x, and x=0 gives no solution. Thus there are exactly two ordered pairs, making option A correct. Although the same numbers occur, their reversed orders are different ordered pairs.

Use the condition y−x=2 by testing each possible x from A and calculating y=x+2. For x=1, y=3, and 3 belongs to B, giving (1,3). For x=2, y=4, and 4 belongs to B, giving (2,4). For x=3, y=5, and 5 belongs to B, giving (3,5). All three choices of x produce a valid second component, so there are 3 qualifying ordered pairs. Option B would miss one valid pair, while option C overcounts possibilities that do not satisfy the equation or membership requirement. Option D counts only one match. Both the numerical equation and the membership in A × B must be checked.

The valid pairs are ((2,1),(3,1),(3,2),(5,1),(5,2),(5,3)). In inequalities, reversing order can change the answer.

First perform the set operation inside the parentheses. Since B={3,4} and C={5,6} have no common elements, B∪C={3,4,5,6}, whose cardinality is 4. The set A has cardinality 2. Applying the Cartesian-product cardinality rule |X×Y|=|X|·|Y| gives |A×(B∪C)|=2·4=8. Thus option A is correct. Option B uses only three elements for the second factor, and option C appears to multiply the sizes of A and one original set rather than the union. Option D overcounts the union. If B and C had overlapped, their common elements would be counted only once; here there is no overlap, so the union has four distinct elements.

(B\cap C={3,4}), so (n(A\times(B\cap C))=3\times2=6). Find the intersection before counting the Cartesian product.

Evaluate the union before forming the Cartesian product. A∪B combines all distinct elements from A and B, so A∪B={1,2,3}; the repeated element 2 is written only once. Therefore |A∪B|=3. Since C={3,4}, |C|=2. The product rule gives |(A∪B)×C|=3·2=6. The six pairs would have first components 1, 2, or 3 and second components 3 or 4. Option B incorrectly counts 2 and 3 twice when forming the union. Option C uses only the size of C multiplied by an incomplete first set, and option D has no valid cardinality calculation. This example emphasizes that union removes duplication before multiplication, while the Cartesian product then creates every possible ordered pairing.

The valid pairs are ((1,2),(1,4),(1,6),(2,2),(2,4),(2,6),(3,6),(4,4)). In divisibility, note carefully which number divides the other.

The valid pairs are ((3,6),(5,4),(5,6)). In boundary inequalities, check whether equality is included or not.

We need ordered pairs \\((x,y)\\) with \\(x\\in A\\), \\(y\\in B\\), and \\(x-y=1\\). Rearranging gives \\(x=y+1\\). Check each possible value of \\(y\\) from \\(B=\\{1,2\\}\\). If \\(y=1\\), then \\(x=2\\), which belongs to \\(A\\). If \\(y=2\\), then \\(x=3\\), which also belongs to \\(A\\).

Thus the valid ordered pairs are \\((2,1)\\) and \\((3,2)\\). No other value of \\(y\\) is available in \\(B\\), so no additional pair can satisfy the condition. There are therefore 2 pairs, making option A correct. Order matters in a Cartesian product: \\((2,1)\\) and \\((1,2)\\) are different, and only the first satisfies the required equation.

The valid pairs are ((1,1),(2,4),(3,9)). In a Cartesian product, applying a condition forms a subset relation.

The direct answer is option A: 4. We need ordered pairs (x,y) from A×B, where both coordinates belong to {1,2,3,4}, and x+y=5. Check each possible first coordinate: x=1 requires y=4, giving (1,4); x=2 requires y=3, giving (2,3); x=3 requires y=2, giving (3,2); x=4 requires y=1, giving (4,1). All four second coordinates are in B, so all four pairs are valid. Therefore the count is 4 and option A is correct. Option B, 3, misses one pair. Option C, 5, includes an impossible extra pair because there are only four possible x values and each determines exactly one y. Option D, 2, counts only part of the valid list. Notice that (2,3) and (3,2) are different ordered pairs because their positions are reversed, even though their sums are equal. Exam cue: list pairs systematically from the first coordinate to avoid omissions.

(A-B={1,3,5}), so (n((A-B)\times C)=3\times3=9). First find the set difference and then count the Cartesian product.