Cartesian product of sets

The governing concept is the definition of a Cartesian product followed by a strict inequality test. In A × B, the first component must be selected from A and the second from B. The complete product is {(1,1), (1,2), (2,1), (2,2), (3,1), (3,2)}. We require the first component to be strictly greater than the second, so equality does not qualify. The valid pairs are (2,1), (3,1), and (3,2), giving three pairs in total. The pair (1,2) has the first component smaller, while (1,1) and (2,2) have equal components. Thus option C is correct. Option D counts every product pair without applying the condition.

In the Cartesian product \(A\times B\), each element of \(A\) is paired with every element of \(B\). Hence, \(n(A\times B)=n(A)\times n(B)=2\times3=6\). Option 3 is only the number of elements in \(B\), and option 2 is only the number of elements in \(A\); neither gives the total number of ordered pairs. Exam tip: multiply the cardinalities of the two sets to find the size of their Cartesian product.

In (A\times B), the first element comes from (A) and the second from (B). Order is very important in ordered pairs.

The governing definition of a Cartesian product is A × B={(a,b): a∈A and b∈B}. The first component must therefore be selected from A, while the second component must be selected from B. Here, 1 belongs to A={0,1}, and y belongs to B={x,y,z}; hence the ordered pair (1,y) satisfies both membership conditions and is in A × B. Option B reverses the roles of the two sets, which is not allowed even if the symbols look similar. Option C is irrelevant because the components need not be equal, and option D is false because A and B are different sets. The order of components is essential in every Cartesian product.

The Cartesian product of any set with an empty set is (\varnothing). If one set is empty, no ordered pair is formed.

The direct answer is option A: 5. For finite sets, the number of ordered pairs in a Cartesian product is n(A × B) = n(A)n(B). We are told n(A) = 4 and n(A × B) = 20. Substitute these values: 20 = 4n(B). Divide both sides by 4: n(B) = 20/4 = 5. Thus B must contain five elements. Option A, 5, matches the calculation. Option B, 16, could result from subtracting 4 from 20, but subtraction is not the rule here. Option C, 24, comes from adding 4 and 20, which is also irrelevant. Option D, 80, comes from multiplying 4 and 20 instead of finding the unknown factor. The safest method is to remember that every element of A can pair with every element of B. Therefore, total pairs equal first-set size times second-set size. Exam cue: when one set size and product size are known, divide the product size by the known set size.

For finite sets, the cardinality rule is |A × B|=|A|·|B|. Since A has three elements, |A|=3. In the self-product A × A, the first component has three choices and, independently, the second component also has three choices. Therefore |A × A|=3·3=9. The nine ordered pairs are (p,p),(p,q),(p,r),(q,p),(q,q),(q,r),(r,p),(r,q), and (r,r). Option B, 6, may result from counting only distinct-element pairs or treating order as irrelevant, but neither is appropriate here. Option C counts only the elements of A, and option D does not follow from the product rule. Equal-component pairs such as (p,p) are included, and reversed pairs are generally different.

Both entries are from their respective sets, so ((2,2)) is included. An ordered pair with equal entries is also valid.

Having the same number of elements is not enough. Usually (A\times B=B\times A) only when (A=B) or one set is empty.

The first element is taken from (A) and the second from (B). Since (B) has only (5), the second entry in both pairs is (5).

Since (-1\in A) and (4\in B), ((-1,4)\in A\times B) is true. Check first and second positions separately.

In (A\times B), the first entry must be from (A) and the second from (B). Here both positions are wrong.

The integers from \(-1\) to \(1\) are \(-1,0,1\), so \(A\) has 3 elements, while \(B\) has 2 elements. Each element of \(A\) forms an ordered pair with every element of \(B\), giving \(|A\times B|=|A|\times|B|=3\times2=6\). Hence, the correct answer is 6. Exam tip: If two sets have \(m\) and \(n\) elements, respectively, their Cartesian product has \(mn\) ordered pairs.

(n(A\times B)=4) and (n(C)=1), so (n((A\times B)\times C)=4). Remember that ((A\times B)) itself is a set.

In A × B, every element of A is paired with every element of B. To keep the first component fixed as 1, pair 1 successively with each element of B: (1,a), (1,b), and (1,c). Thus there are exactly three valid ordered pairs. The general principle is that a fixed first component from A appears |B| times, because the second component can vary through all elements of B. Option B incorrectly counts the elements of A, while option C counts only the fixed value itself rather than the pairs it generates. Option D is the total size of A × B, 2·3=6, not the number having first component 1. The order condition makes the first position decisive.

In an ordered pair belonging to A × B, the first component must come from A and the second component must come from B. To count how often a particular second component appears, fix that element in the second position and pair it with every element of A. Each element of A gives one distinct ordered pair.

The element 8 belongs to B. Since A has three elements, m, n, and o, the pairs with second component 8 are (m,8), (n,8), and (o,8). Therefore 8 appears as the second component three times, so option A is correct. The number 2 would incorrectly count the elements of B, while 6 is the total number of pairs in A × B, not the number having second component 8.

The Cartesian product A × B consists of ordered pairs whose first component belongs to A and whose second component belongs to B. The order matters, so (3,4) and (4,3) are different pairs. We must check both membership and the required sum, rather than checking the numbers alone.

For option A, 3 belongs to A and 4 belongs to B. Also, 3 + 4 = 7, so (3,4) satisfies every condition. Although (4,3) also has sum 7, it is not in A × B because 4 is not in A as the first component and 3 is not in B as the second component. Therefore, option A is correct.

The ordered pairs satisfying \(x<y\) are \((1,2), (1,3), (2,3)\), so there are 3 pairs. The answer 9 is the total number of pairs in the Cartesian product, without applying the inequality. Exam tip: for three distinct elements, the number of pairs satisfying \(x<y\) is \(\binom{3}{2}=3\).

The direct answer is option A: 2. In A×B, the first coordinate x must come from A={0,1,2}, and the second coordinate y must come from B={0,1}. We need x+y=2. Test the possible y values: if y=0, then x=2, which is in A, giving (2,0). If y=1, then x=1, which is in A, giving (1,1). No other y values are available in B. Therefore the valid pairs are (1,1) and (2,0), and there are 2. Option A is correct. Option B, 1, misses one valid pair. Option C, 3, may result from counting all possible x values without checking whether the required y belongs to B; for x=0, y would be 2, but 2 is not in B. Option D, 4, is the total number of pairs in A×B, not the number satisfying the equation. Always check membership and the coordinate order.

A pair in \(A\times B\) has its first entry from A and its second entry from B. The condition that y divides x means that \(x\) must be an exact multiple of \(y\), written \(y\mid x\). Since B contains only 2 and 4, test each possible x from A against these divisors rather than reversing their positions.

For \(x=1\), neither 2 nor 4 divides it. For \(x=2\), only \(y=2\) works, giving \((2,2)\). For \(x=3\), neither works. For \(x=4\), both 2 and 4 work, giving \((4,2)\) and \((4,4)\). There are therefore three pairs. The pair \((2,4)\) is not valid because 4 does not divide 2, so option A is correct.

A relation from (A) to (B) is any subset of (A\times B). Only ({(1,3),(2,5)}) has both pairs in (A\times B).

(n(A\times B)=3\times2=6), and every relation is a subset of (A\times B). Therefore the number of relations is (2^6).

(n(A\times B)=3\times1=3), so the number of subsets is (2^3=8). For counting relations, remember (2^{n(A\times B)}).

Both have (2\times2=4) pairs, but changing order changes the pairs. Cartesian product is not commutative.

((2,2)) is in both (A\times B) and (B\times A). For intersection, the pair must belong to both sets.