Cartesian product of sets

The governing concept is counting ordered pairs in a Cartesian product by using congruence modulo 3. For a=1, the sum 1+b must be divisible by 3, so b must leave remainder 2; only b=2 works, giving (1,2). For a=2, b must leave remainder 1, and b=1 or b=4 work, giving (2,1) and (2,4). For a=3, which has remainder 0, b must also have remainder 0; only b=3 works, giving (3,3). Thus the complete list contains four pairs. Therefore option B is correct. The other options result from omitting a valid pair or counting a pair whose sum is not divisible by 3.

The governing concept is that Cartesian products are generally not commutative: A × B and B × A may contain different ordered pairs because the positions of the components matter. A single counterexample is enough to disprove equality. The pair (1,4) belongs to A × B because 1 ∈ A and 4 ∈ B. However, (1,4) does not belong to B × A because 1 ∉ B. Therefore the two products are not equal, and option B, False, is correct. This does not mean the products never have related elements; rather, equality requires every ordered pair in one product to occur in the other in the same order.

The governing concept is a relation defined as a subset of the Cartesian product A × B. We count ordered pairs for which the first coordinate is less than or equal to the second. If a = 1, b may be 1, 2, or 3, giving three pairs. If a = 2, only b = 2 or 3 work, giving two pairs. If a = 3, only b = 3 works, giving one pair. Therefore |R| = 3 + 2 + 1 = 6. Explicitly, R = {(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)}. Option C is correct. Option D counts all nine pairs in A × B and ignores the condition, while smaller options omit valid pairs, especially those with equality.

The governing concept is counting a relation inside a Cartesian product using a strict inequality. Test each possible first coordinate. For a = 1, the condition becomes b > 2, so b = 3 and 4 are valid; this gives (1,3) and (1,4). For a = 2, it becomes b > 3, so only b = 4 is valid; this gives (2,4). For a = 3, it becomes b > 4, and no member of B satisfies the condition. Hence S = {(1,3),(1,4),(2,4)} and |S| = 3, making option B correct. Equality is not accepted because the symbol is >, not ≥; this is the key reason boundary pairs such as (1,2) are excluded.

Two positive integers are coprime when their greatest common divisor is 1. We count ordered pairs from {1,2,3,4,5}. Every pair beginning with 1 contributes four additional pairs after (1,1), because gcd(1,n)=1 for every n. A direct row count is clearer: for a=1, all 5 values of b work; for a=2, b=1,3,5 work, giving 3; for a=3, b=1,2,4,5 work, giving 4; for a=4, b=1,3,5 work, giving 3; and for a=5, b=1,2,3,4 work, giving 4. Therefore the total is 5 + 3 + 4 + 3 + 4 = 19. Thus option C is correct; the other values result from missing or incorrectly including a non-coprime pair.

The Cartesian product supplies ordered pairs whose two entries must both lie in {1,2,3,4,5,6}. First list the pairs satisfying a+b=7: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). Now impose the second condition a<b. The first three pairs satisfy it because 1<6, 2<5, and 3<4. The remaining three have a>b and must be rejected. The pair (3.5,3.5) is not relevant because the sets contain only integers, and it would not satisfy a<b anyway. Therefore exactly three ordered pairs qualify, so option B is correct. Option D counts all sum-seven pairs without applying the inequality.

For finite sets, the cardinality of a Cartesian product is the product of the cardinalities: |A×B|=|A||B|. Given 24=4|B|, divide by 4 to obtain |B|=6. Now apply the same rule to B×C: |B×C|=|B||C|=6×7=42. Hence option C is correct. The value 28 uses |A| instead of the derived size of B, while 35 multiplies the wrong quantities. The value 56 would result from incorrectly using 8 for |B| or from another arithmetic error. The sets do not need to be listed; their cardinalities are sufficient, because every element of B can be paired with every element of C.

The condition b/a ∈ N means that b/a is a natural number, equivalently a divides b. We count suitable b values for each a in A, remembering that pairs are formed with b from B. For a=1, all four values 2,4,6,8 work, giving 4 pairs. For a=2, the quotients are 1,2,3,4, so all four values again work. For a=3, only b=6 gives the natural quotient 2, giving 1 pair. For a=4, b=4 and b=8 give quotients 1 and 2, giving 2 pairs. Thus |R|=4+4+1+2=11, so option C is correct. Values such as b=2 for a=3 are excluded because 2/3 is not a natural number.

The governing concept is inclusion–exclusion for the union of two conditions. Let P contain pairs with a+b=6: (1,5),(2,4),(3,3),(4,2),(5,1), so |P|=5. Let Q contain pairs with a=b: (1,1),(2,2),(3,3),(4,4),(5,5), so |Q|=5. The pair (3,3) satisfies both conditions, so |P∩Q|=1. Therefore |P∪Q|=|P|+|Q|−|P∩Q|=5+5−1=9. Option B is correct. Adding the two counts without subtraction gives 10 because it counts (3,3) twice. The alternatives therefore reflect double-counting, omission, or addition of an invalid pair.

The governing rule is |A×B|=|A|×|B|, but the cardinalities must first be found from the defining equations. Solving x²=4 gives x=−2 and x=2, so A={−2,2} and |A|=2. Factor the second equation as y²−5y+6=(y−2)(y−3)=0. Thus B={2,3} and |B|=2. Every element of A can be paired with every element of B, producing (−2,2),(−2,3),(2,2), and (2,3). Therefore |A×B|=2×2=4, so option B is correct. The other choices arise from adding the cardinalities, counting repeated numerical values incorrectly, or treating one equation as if it had more roots.

The defining property of a Cartesian product is that every ordered pair has its first coordinate from A and its second coordinate from B, with all combinations included. Looking at the given pairs, the distinct first coordinates are 1 and 2, so A={1,2}. The distinct second coordinates are 3 and 4, so B={3,4}. Their product is exactly {(1,3),(1,4),(2,3),(2,4)}, confirming the reconstruction. The order matters: reversing the sets would produce pairs such as (3,1), not the displayed pairs. Therefore option A is correct; the other options confuse coordinate positions or group the numbers incorrectly.

The governing idea is to count the required subset of A×B row by row, fixing x and then testing possible y-values. If x=0, all three elements 1,2,3 of B are greater than x, giving (0,1),(0,2),(0,3): three pairs. If x=1, only y=2 and y=3 satisfy x<y, giving two pairs, (1,2) and (1,3). If x=2, only y=3 works, giving one pair, (2,3). Hence the total is 3+2+1=6, so option C is correct. Although A×B has 3×3=9 pairs, the three pairs with y≤x fail the inequality. Option D ignores the condition, while A and B omit valid pairs.

The symbol x|y means that x divides y, or equivalently y is an integer multiple of x. Count the valid pairs for each possible first coordinate. For x=1, all four values y=1,2,3,4 work, giving 4 pairs. For x=2, y=2 and y=4 work, giving 2 pairs. For x=3, only y=3 works, giving 1 pair. For x=4, only y=4 works, giving 1 pair. Thus the total is 4+2+1+1=8, so option C is correct. The diagonal pairs are valid because every nonzero integer divides itself. Counting only those diagonal pairs would give 4, while the additional pairs (1,2), (1,3), (1,4), and (2,4) must also be included.

The Cartesian product A × B consists of ordered pairs (a,b), where a belongs to A and b belongs to B. Because order matters in an ordered pair, (a,b) and (b,a) are generally different. Thus A × B and B × A are not usually equal; they can coincide only in special cases, such as suitable identical sets. The first identity follows by checking membership in both products: the second coordinate must belong to both B and C. The second identity follows because the second coordinate may belong to B or C. The third identity also follows because the first coordinate remains in A and the second lies in B but not C. Hence option D is the statement that is generally not true.

First evaluate the two intersections: \(A\cap C=\{2,3\}\) and \(B\cap C=\{3,4\}\). Each resulting set contains two elements. For finite sets \(X\) and \(Y\), the Cartesian product has \(|X\times Y|=|X||Y|\) ordered pairs. Hence the required number is \(2\times2=4\), so option B is correct.

Let \(n(B)=m\). Then \(n(A)=m+2\). Since \(n(A\times B)=n(A)n(B)\), we obtain \(m(m+2)=24\), or \(m^2+2m-24=0\). Factoring gives \((m+6)(m-4)=0\). A cardinality cannot be negative, so \(m=4\), and consequently \(n(A)=4+2=6\). Thus option B is correct.

For finite sets, the cardinality of a Cartesian product is given by \(n(X\times Y)=n(X)n(Y)\). From \(n(A\times B)=30\) and \(n(B)=5\), we obtain \(n(A)\times5=30\), so \(n(A)=6\). Similarly, \(5\times n(C)=30\), giving \(n(C)=6\). Therefore, \(n(A\times C)=6\times6=36\), so option C is correct. The order of the sets matters in forming ordered pairs, but the cardinality is obtained by multiplication.

There are |U|=|A×B|=3×3=9 total ordered pairs. The condition x≠y excludes exactly the three diagonal pairs with equal coordinates: (1,1), (2,2), and (3,3). Using complement counting, the required number is 9−3=6. The same result can be obtained directly: for each fixed x, two of the three possible y-values are different from x, so there are 3×2=6 valid pairs. Hence option B is correct. Option D counts every pair and ignores the inequality, while option A counts only the excluded equal-coordinate pairs. Option C has no valid counting basis.

Set A has 2 elements and set B has 2 elements. In the Cartesian product A × B, each element of A is paired with every element of B, so the number of ordered pairs is 2 × 2 = 4. The pairs are (1, 3), (1, 4), (2, 3), and (2, 4). The answer 2 counts the elements of only one set; for a Cartesian product, multiply the cardinalities of both sets. Exam tip: remember the formula |A × B| = |A| × |B|.

By definition, A×A is the set of every ordered pair (a,b) in which both a and b belong to A. Since R is exactly A×A, it contains all 3×3 = 9 possible ordered pairs, including every diagonal and every reverse pair. A relation containing every pair from A×A is called the universal relation. Therefore option B is correct; it is not empty, not merely the identity relation, and it is not asymmetric.