Cartesian product of sets

The direct answer is option B: \(A=\varnothing\) or \(B=\varnothing\). A Cartesian product \(A\times B\) contains ordered pairs \((a,b)\), where \(a\in A\) and \(b\in B\). If both sets are nonempty, at least one such pair can be formed, so the product cannot be empty. Therefore an empty product means at least one set must be empty. Both sets do not have to be empty. Option A is too strong because, for example, if \(A=\varnothing\) and \(B=\{1,2\}\), then the product is empty although B is not empty. Option B is exactly the required condition. Option C, \(A=B\), says the sets are equal, which has no connection with emptiness. Option D, \(A\cap B=\varnothing\), only says they are disjoint; two nonempty disjoint sets can still have a nonempty Cartesian product. Exam cue: an empty Cartesian product means at least one factor is empty.

First find the cardinality of the union: \(n(A\cup B)=n(A)+n(B)-n(A\cap B)=4+5-2=7\). Since \(n(A\cap B)=2\), use the Cartesian-product formula \(n(X\times Y)=n(X)n(Y)\): \(n\big((A\cap B)\times(A\cup B)\big)=2\times7=14\). Therefore, option C is correct. Exam tip: when finding the cardinality of a union, subtract the intersection once to avoid double-counting.

For (a=1,2,3,4,5), the numbers of choices are (4,3,3,2,2). Total is (14), so count (b) for each (a).

For each value of \(a\), calculate \(b=a^2\): \(0^2=0\), \(1^2=1\), \(2^2=4\), and \(3^2=9\). Every resulting value belongs to \(B\), giving the ordered pairs \((0,0),(1,1),(2,4),(3,9)\). Therefore, the total number of pairs is \(4\). Exam tip: Apply the given rule to every element of \(A\) and count only those results that lie in \(B\).

The pairs are ((1,8)), ((2,4)), and ((4,2)). Values may multiply the same way, but changing positions changes an ordered pair.

The direct answer is option A. A relation here consists of ordered pairs from A × B that satisfy a + b = 4. Test each a in A. If a = 1, then b = 3, which belongs to B, giving (1,3). If a = 2, then b = 2, giving (2,2). If a = 3, then b = 1, giving (3,1). These are all possibilities because A has only 1, 2, and 3. Therefore R = {(1,3),(2,2),(3,1)}. Option A lists exactly these pairs. Option B lists pairs whose sums are 3, not 4. Option C contains (3,3), whose sum is 6. Option D lists equal-coordinate pairs with sums 2, 4, and 6, so only one of them works. Always check both membership and the stated condition.

(n(A\times B)=16), so the number of ways to choose exactly (2) pairs is ({}^{16}C_2=120). Use subset selection when counting relations.

(A\times B) has (3\cdot2=6) pairs, and each relation is a subset of it. Hence total relations are (2^6=64).

The sum is odd when one number is odd and the other is even. (A) has (2) odd and (1) even elements, while (B) has (2) even and (1) odd elements, so (2\cdot2+1\cdot1=5).

For \(\frac{a}{b}\) to be an integer, \(b\) must be a divisor of \(a\). For \(a=2\), the possible values of \(b\) are \(1,2\), giving 2 pairs; for \(a=4\), they are \(1,2,4\), giving 3 pairs; and for \(a=6\), they are \(1,2,3\), giving 3 pairs. Thus, the total is \(2+3+3=8\). Since \((a,b)\) is an ordered pair, count the valid values of \(b\) separately for each value of \(a\).

Since both sets contain only 1, 2, 3, and 4, the equation \(a^2+b^2=25\) is satisfied by \(3^2+4^2=25\). Thus, the ordered pairs are \((3,4)\) and \((4,3)\), giving a total of 2. These are distinct because order matters in a Cartesian product. Exam tip: when counting ordered pairs, treat reversed pairs as different unless the question concerns unordered pairs.

Since \(|A|=|B|=3\), the Cartesian product contains \(3\times3=9\) ordered pairs. Exactly 3 pairs have equal components: \((1,1),(2,2),(3,3)\). Therefore, the number with \(a\ne b\) is \(9-3=6\). Option 9 counts all pairs, not just those with unequal components. Exam tip: for two identical sets of \(n\) elements, the number of pairs with unequal components is \(n^2-n=n(n-1)\).

The non-coprime pairs are ((2,2),(2,4),(3,3),(4,2),(4,4),(5,5)). Thus (25-6=19) pairs are relatively prime.

In (A\times B), the first coordinate must be from (A) and the second from (B). In ((x,1)), the order is reversed, so it is not a member.

The direct answer is option C: 8 elements. First form the union. Since A contains 1,2 and B contains 3,4, there is no repetition, so \(A\cup B=\{1,2,3,4\}\), which has 4 elements. Set C has 2 elements, namely 5 and 6. In a Cartesian product, every element of the first set is paired with every element of the second set. Therefore \(n((A\cup B)\times C)=n(A\cup B)n(C)=4\cdot2=8\). Option C is correct. Option A, 4, counts only the elements in the union and forgets pairing with C. Option B, 6, does not follow the product rule and may result from adding set sizes. Option D, 16, incorrectly squares 4 or otherwise overcounts the pairs. The elements are ordered pairs such as \((1,5)\) and \((1,6)\), so each of the four first coordinates gives two pairs. Memory cue: for Cartesian products, multiply cardinalities; for unions, remove duplicates first.

First find the intersections: \(A\cap C=\{1,2,4\}\), which has 3 elements, and \(B\cap C=\{2,4\}\), which has 2 elements. The number of ordered pairs in a Cartesian product is \(|X\times Y|=|X|\times|Y|\). Hence, \(3\times2=6\). Exam tip: multiply the cardinalities of the two sets for a Cartesian product; do not add them.

(B\setminus A={4}), so (4) becomes the second coordinate with each element of (A). Find the difference before multiplying.

The direct answer is option B: 3 pairs. We need \(a-b>0\), which is the same as \(a>b\). List the values of A and compare them with values of B. For \(a=1\), neither 1 nor 2 is smaller, so there are no pairs. For \(a=2\), only \(b=1\) works, giving \((2,1)\). For \(a=3\), both values work, giving \((3,1)\) and \((3,2)\). Thus there are exactly three pairs. Option A, 2, misses one valid pair. Option B, 3, is correct. Option C, 4, counts a pair where the first coordinate is not greater, such as \((1,2)\), or otherwise overcounts. Option D, 5, overcounts still further. The order matters: the first coordinate comes from A and the second from B. Memory cue: positive \(a-b\) means first number is bigger than second.

For each fixed value of \(a\), count the elements of \(B\) that are greater than it. For \(a=1,2,3,4\), the respective numbers of choices are \(3,3,3,2\). Hence, the total is \(3+3+3+2=11\). The distractor 10 results from missing one valid choice in one of these cases. Exam tip: Since \(A \times B\) contains ordered pairs \((a,b)\), fix each value of \(a\) and count the permissible values of \(b\).

Prime sums possible are (3,5,7); the valid pairs are ((1,2),(2,3),(2,5),(3,2)), and ((1,4)) is not allowed, so the correct total is (4). Use only available elements while checking options.

Count the possible values of b for each value of a such that a+b≤5. For a=1, there are 3 choices of b; for a=2, 3 choices; for a=3, 2 choices; and for a=4, 1 choice. Thus, the total number of ordered pairs is 3+3+2+1=9. Exam tip: In a Cartesian product, order matters, so each pair is counted as (a,b).

For a=1, all 4 elements of B are multiples of 1, giving 4 pairs. For a=2, only 6 and 12 are multiples of 2, giving 2 pairs. For a=3, all four elements 3, 6, 9 and 12 are multiples of 3, giving 4 pairs. Therefore, the total number of pairs is 4+2+4=10. In an exam, count the suitable values of b separately for each value of a to avoid missing or repeating a pair.

For an even sum, both numbers must have the same parity. (A) has (2) even and (1) odd elements, while (B) has (2) even and (2) odd elements, so (2\cdot2+1\cdot2=6).

Every element of \(A\) is odd, while every element of \(B\) is even. The sum of an odd number and an even number is always odd, so \(a+b\) is odd for every ordered pair \((a,b)\) in \(A\times B\). ‘Prime’ is not guaranteed because \(3+6=9\), which is composite. Exam tip: the sum or difference of an odd and an even number is always odd.

If A×B = B×A and both sets are non-empty, choose b∈B. From (a,b)∈A×B=B×A, we get a∈B, so A⊆B; similarly B⊆A. Hence A=B. Exam tip: order matters in ordered pairs.