Cartesian Products Quiz: Build and Analyze Ordered Pair Sets

An ordered pair (a,b) is in both A×B and A×C exactly when a belongs to A and the same b belongs to both B and C. This means the only possible second coordinates are those in B∩C, giving the intersection A×(B∩C).

Each element is an ordered pair (a,a) with the first and second coordinates identical and both drawn from A. This guarantees every such pair belongs to A×A, since A×A contains all possible ordered pairs whose coordinates come from A.

The size of a product equals |A|·|B| = 3·2 = 6. Every choice of a first coordinate from A can be paired independently with every choice of a second coordinate from B, producing exactly 3×2 distinct ordered pairs.

Cartesian product cardinalities multiply across each factor. When sets are combined by repeated Cartesian products, each choice in one coordinate multiplies the total number of possible choices overall, producing the product of all individual sizes.

Graphs of functions consist of ordered pairs (a,f(a)) with the first coordinate coming from the domain A, ensuring each input is paired with exactly one output. This structure distinguishes graphs from arbitrary subsets of A×B.

Pairs removed are those whose second coordinate lies in C; therefore the remaining ordered pairs are precisely those with the first coordinate in A and the second coordinate in B∖C. Removing pairs by restricting only the second slot leaves the first slot unaffected.

By definition, (A×B)×C consists of ordered pairs whose first component is itself an ordered pair drawn from A×B. Elements therefore appear in nested form ((a,b),c), reflecting the iterative construction of Cartesian products.

If A ≠ ∅ and A×B ⊆ A×C, then every b in B must also lie in C. Otherwise, some (a,b) would appear in A×B but fail to appear in A×C, contradicting the inclusion. Because A is nonempty, there is at least one a to pair with every b, forcing B ⊆ C.

Union on the right contains mixed pairs like (a,d) or (c,b), which may not lie in the union of the original rectangles A×B and C×D. Because the cross terms do not necessarily belong to either product individually, the union of the components does not equal the product of the unions.

Equality of ordered pairs requires equality of their first components and equality of their second components. No two pairs are considered equal unless both coordinates match exactly, which preserves the meaning of order in Cartesian products.

Since |A×B| = |A|·|B|, solving for the unknown cardinality reduces to dividing the product cardinality by the known factor. Thus |B| = k/m whenever the total size is k and |A| is m.

A×{b} fixes the second slot to the single element b, while the first coordinate runs freely through A. The product therefore forms a vertical “slice” in A×B containing one point above each element of A.

Nothing is guaranteed because A×B may be empty (for example, if A = ∅), causing the inclusion to hold vacuously. In such cases, the relationship tells us nothing about how A relates to C or how B relates to D, since the product contains no pairs to impose constraints.

For no (a,b) to appear in both products A×B and C×D, either the first coordinates must never overlap (A∩C = ∅) or the second coordinates must never overlap (B∩D = ∅). If either intersection is empty, then no pair can simultaneously satisfy both membership conditions.

Distributing over unions holds for Cartesian products because a pair (a,x) belongs to A×(B∪C) exactly when a is in A and x is in B or in C, which is the same condition that makes (a,x) appear in the union (A×B) ∪ (A×C). Thus both sides describe the same set of ordered pairs.

Both ∅×A and A×∅ are always empty because forming an ordered pair requires choosing one element from each factor. If either factor contributes no elements, no ordered pairs can be formed, so both products are equal and empty for all A.

Map each a in A to (a,0). This provides a one-to-one correspondence from A onto A×{0}, pairing each element with the fixed second coordinate 0. Every pair in A×{0} appears exactly once through this mapping.

A×B being finite forces both A and B to be finite, because if either factor were infinite while the other remained nonempty, the Cartesian product would automatically become infinite. Finite cardinality can only result when both sets contribute a finite number of elements.

Elements of the power-set product are ordered pairs (X,Y) where X is a subset of A and Y is a subset of B. The pair ({1},{a}) satisfies this requirement, whereas the other options fail to pair subsets in the correct manner.

A×∅ contains no ordered pairs because forming any pair requires choosing one element from each factor. Since ∅ has no elements, no second coordinate can ever be chosen, making every pair impossible regardless of how many elements A contains. Options A, B, and C all suggest a nonempty result which is incorrect.