Step 1: Cartesian Product Construction

The Cartesian product of two sets $A$ and $B$, denoted by $A \times B$, is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. [Per the definition of Cartesian product in Set Theory].

Given:

• $A = \{1, 2\}$
• $B = \{3, 4\}$

Mapping elements of $A$ to $B$:

• $(1, 3)$
• $(1, 4)$
• $(2, 3)$
• $(2, 4)$

Thus, $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

Step 2: Calculating the Number of Subsets

The total number of subsets of any set $S$ is given by the formula $2^n$, where $n$ is the number of elements (cardinality) in the set $S$. [Per the Power Set Axiom].

Here, the cardinality of $A \times B$ is $|A \times B| = 4$.

Number of subsets $= 2^4 = 16$.

Step 3: Listing the Subsets

Let $S = A \times B = \{p_1, p_2, p_3, p_4\}$, where $p_1=(1,3), p_2=(1,4), p_3=(2,3), p_4=(2,4)$. The 16 subsets are listed as follows:

Visual Representation of $A \times B$

Final Solution: $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$. The set has $16$ subsets.

Step 1: Cartesian Product Definition

The Cartesian product of two sets $X$ and $Y$, denoted by $X \times Y$, is the set of all ordered pairs $(a, b)$ such that $a \in X$ and $b \in Y$. [By Definition of Cartesian Product].

Step 2: Computing $A \times C$

Given $A = \{1, 2\}$ and $C = \{5, 6\}$.

$A \times C = \{(a, c) : a \in \{1, 2\}, c \in \{5, 6\}\}$

Performing the distribution:

• $(1, 5)$
• $(1, 6)$
• $(2, 5)$
• $(2, 6)$

Thus, $A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$.

Step 3: Computing $B \times D$

Given $B = \{1, 2, 3, 4\}$ and $D = \{5, 6, 7, 8\}$.

$B \times D = \{(b, d) : b \in \{1, 2, 3, 4\}, d \in \{5, 6, 7, 8\}\}$

The set $B \times D$ contains $4 \times 4 = 16$ elements:

Step 4: Verification of Subset Relation

A set $S$ is a subset of $T$ ($S \subseteq T$) if and only if every element of $S$ is also an element of $T$. [By Definition of Subset].

• $(1, 5) \in A \times C$ and $(1, 5) \in B \times D$
• $(1, 6) \in A \times C$ and $(1, 6) \in B \times D$
• $(2, 5) \in A \times C$ and $(2, 5) \in B \times D$
• $(2, 6) \in A \times C$ and $(2, 6) \in B \times D$

Since all four elements of $A \times C$ are present within the set $B \times D$, the condition for a subset is satisfied.

Final Solution: Since every element of $A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$ is contained within $B \times D$, it is verified that $A \times C \subseteq B \times D$.

Given Variables & Initial Setup

The set $A$ is defined as $A = \{-1, 1\}$. We are required to determine the Cartesian product of the set $A$ with itself three times, denoted as $A \times A \times A$ (also known as the set of all ordered triples $(a, b, c)$ such that $a, b, c \in A$).

Step 1: Calculating the Cartesian Product $A \times A$

By the definition of the Cartesian product of two sets [Per the definition of Cartesian Product $X \times Y = \{(x, y) : x \in X, y \in Y\}$], we first compute the product of two sets $A = \{-1, 1\}$:

Step 2: Calculating $A \times A \times A$

To find the triple Cartesian product, we map each element of the resulting set from Step 1 with each element of the original set $A = \{-1, 1\}$. Each element $(x, y) \in A \times A$ will be appended with an element $z \in A$ to form the triple $(x, y, z)$.

• For $(-1, -1) \in A \times A$:
  • $(-1, -1, -1)$
  • $(-1, -1, 1)$
• For $(-1, 1) \in A \times A$:
  • $(-1, 1, -1)$
  • $(-1, 1, 1)$
• For $(1, -1) \in A \times A$:
  • $(1, -1, -1)$
  • $(1, -1, 1)$
• For $(1, 1) \in A \times A$:
  • $(1, 1, -1)$
  • $(1, 1, 1)$

Visual Representation of the Ordered Triples

Step 3: Compilation of the Set

Consolidating the steps above, the set $A \times A \times A$ contains $2^3 = 8$ distinct elements.

Final Solution: $A \times A \times A = \{(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)\}$

Step 1: Analyzing the Set Definitions

Given the sets:

• $A = \{1, 2\}$
• $B = \{3, 4\}$
• $\phi = \emptyset$ (The empty set, which contains no elements)

Step 2: Evaluating the Intersection $(B \cap \phi)$

The intersection of any set $B$ with the empty set $\phi$ is defined by the property that no element can exist in both $B$ and the empty set simultaneously.

$B \cap \phi = \{x : x \in B \text{ and } x \in \phi\}$ [By the definition of Set Intersection]

Since $\phi$ has no elements, there is no $x$ such that $x \in \phi$. Therefore:

$B \cap \phi = \emptyset$

Step 3: Computing the Cartesian Product $A \times (B \cap \phi)$

Substitute the result from Step 2 into the expression:

$A \times (B \cap \phi) = A \times \emptyset$

The Cartesian product of any set $A$ with the empty set is defined as the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in \emptyset$. [Per the definition of Cartesian Product]

Since there are no elements in the empty set to form a pair, the resulting set must also be empty:

$A \times \emptyset = \emptyset$

Step 4: Verification of Truth Value

The original statement is: $A \times (B \cap \phi) = \phi$.

Based on the derivation in Step 3, $A \times (B \cap \phi) = \emptyset$, which is equal to $\phi$.

Final Solution: The statement is True.

Step 1: Analyzing the Definition of a Cartesian Product

The Cartesian product of two non-empty sets $A$ and $B$, denoted by $A \times B$, is defined as the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$. Symbolically, this is expressed as:

$A \times B = \{ (x, y) : x \in A \text{ and } y \in B \}$ [Per the set-theoretic definition of Cartesian products]

Step 2: Evaluating the Statement

The provided statement states: "If $A$ and $B$ are non-empty sets, then $A \times B$ is a non-empty set of ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$."

Given the formal definition established in Step 1, we observe the following:

• Since $A \neq \emptyset$, there exists at least one element $a \in A$.
• Since $B \neq \emptyset$, there exists at least one element $b \in B$.
• Consequently, the ordered pair $(a, b)$ must exist such that $(a, b) \in A \times B$.
• Therefore, $A \times B$ cannot be an empty set, making it a non-empty set containing ordered pairs $(x, y)$ satisfying the condition $x \in A$ and $y \in B$.

Step 3: Determination of Truth Value

Based on the rigorous application of set theory axioms [Zermelo-Fraenkel Set Theory principles regarding ordered pairs], the statement is logically sound and mathematically accurate.

Step 4: Visualizing the Cartesian Product

Final Solution: The statement is True. The definition of the Cartesian product $A \times B$ for non-empty sets $A$ and $B$ is definitively the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.

Step 1: Theoretical Definition of Cartesian Product

The Cartesian product of two non-empty sets $A$ and $B$, denoted as $A \times B$, is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. Formally: $A \times B = \{ (a, b) : a \in A, b \in B \}$ [Per the definition of Cartesian product in Set Theory].

Step 2: Computation of $G \times H$

Given sets $G = \{7, 8\}$ and $H = \{5, 4, 2\}$. To find $G \times H$, we pair each element of $G$ with every element of $H$:

• For $7 \in G$: $(7, 5), (7, 4), (7, 2)$
• For $8 \in G$: $(8, 5), (8, 4), (8, 2)$

Thus, $G \times H = \{ (7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2) \}$.

Step 3: Computation of $H \times G$

To find $H \times G$, we pair each element of $H$ with every element of $G$:

• For $5 \in H$: $(5, 7), (5, 8)$
• For $4 \in H$: $(4, 7), (4, 8)$
• For $2 \in H$: $(2, 7), (2, 8)$

Thus, $H \times G = \{ (5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8) \}$.

Visual Representation of Mapping (Mapping Diagram)

Comparative Analysis of Results

Note: Since ordered pairs are distinct such that $(a, b) \neq (b, a)$ unless $a=b$, it is observed that $G \times H \neq H \times G$ in the general case [Property of Non-Commutativity of Cartesian Products].

Final Solution: $G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$ and $H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

Step 1: Analyzing the Cardinality and Set A

Given that the Cartesian product $A \times A$ has $n(A \times A) = 9$ elements. [Per the property of Cartesian products, $n(A \times B) = n(A) \times n(B)$].

Since $A \times A$ represents the product of set $A$ with itself, $n(A \times A) = n(A) \times n(A) = [n(A)]^2$.

Therefore, $[n(A)]^2 = 9 \implies n(A) = 3$.

Step 2: Identifying Elements of Set A

The Cartesian product $A \times A$ consists of ordered pairs $(a, b)$ where $a \in A$ and $b \in A$. The given elements are $(-1, 0)$ and $(0, 1)$.

Extracting the components of the ordered pairs:

• From $(-1, 0)$: $-1 \in A$ and $0 \in A$.
• From $(0, 1)$: $0 \in A$ and $1 \in A$.

Combining these, we identify the distinct elements of set $A$ as $\{-1, 0, 1\}$.

Since $n(A) = 3$ and we have found three distinct elements, we conclude $A = \{-1, 0, 1\}$.

Step 3: Constructing $A \times A$

To find the total set $A \times A$, we perform the Cartesian product of $A = \{-1, 0, 1\}$ with itself by pairing every element of $A$ with every element of $A$.

Step 4: Isolating the Remaining Elements

The total set $A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}$.

Removing the given elements $(-1, 0)$ and $(0, 1)$ from the set:

Remaining elements = $\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$.

Visual Representation of $A \times A$

Final Solution: The set $A = \{-1, 0, 1\}$ and the remaining elements of $A \times A$ are $\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$.

Step 1: Identifying Given Parameters

To determine the number of elements in the Cartesian product of two sets $A$ and $B$, we define the given cardinalities:

• Number of elements in set $A$, denoted as $n(A) = 3$.
• Set $B = \{3, 4, 5\}$. Counting the distinct elements within the set, we find $n(B) = 3$.

Step 2: Applying the Cardinality Theorem for Cartesian Products

The Cartesian product of two finite sets $A$ and $B$, denoted $A \times B$, is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. [Per the Definition of Cartesian Product].

The theorem governing the number of elements in a Cartesian product states:
$n(A \times B) = n(A) \times n(B)$
[Fundamental Counting Principle applied to set theory]

Step 3: Execution of Calculation

Substitute the known cardinalities into the theorem:

$n(A \times B) = 3 \times 3$

$n(A \times B) = 9$

Step 4: Visual Representation of the Product Mapping

Final Solution: The number of elements in the set $(A \times B)$ is 9.

Step 1: Application of Equality of Ordered Pairs

According to the fundamental definition of the equality of ordered pairs, two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if their corresponding components are equal [Per Definition: $(a, b) = (c, d) \implies a = c \text{ and } b = d$].

Given the equation: $\left(\frac{x}{3} + 1, y - \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$

We equate the first components and the second components respectively:

• $\frac{x}{3} + 1 = \frac{5}{3}$
• $y - \frac{2}{3} = \frac{1}{3}$

Step 2: Solving for the Variable $x$

Starting with the first equation:

$\frac{x}{3} + 1 = \frac{5}{3}$

Subtracting $1$ from both sides [By Subtraction Property of Equality]:

$\frac{x}{3} = \frac{5}{3} - 1$

Converting the constant to a fraction with a common denominator:

$\frac{x}{3} = \frac{5}{3} - \frac{3}{3}$

$\frac{x}{3} = \frac{2}{3}$

Multiplying both sides by $3$ [By Multiplication Property of Equality]:

$x = 2$

Step 3: Solving for the Variable $y$

Starting with the second equation:

$y - \frac{2}{3} = \frac{1}{3}$

Adding $\frac{2}{3}$ to both sides [By Addition Property of Equality]:

$y = \frac{1}{3} + \frac{2}{3}$

$y = \frac{3}{3}$

$y = 1$

Step 4: Logical Verification

Substituting $x = 2$ and $y = 1$ back into the original ordered pair:

$\left(\frac{2}{3} + 1, 1 - \frac{2}{3}\right) = \left(\frac{2+3}{3}, \frac{3-2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$

The values satisfy the original equality constraint.

Final Solution: $x = 2$ and $y = 1$

Given Variables & Initial Setup

We are provided with the following information regarding sets $A$ and $B$:

• $n(A) = 3$ [Number of elements in set $A$ is 3]
• $n(B) = 2$ [Number of elements in set $B$ is 2]
• The Cartesian product $A \times B$ contains elements: $(x, 1), (y, 2), (z, 1)$
• The elements $x, y, z$ are distinct.

Step 1: Identifying Elements of Set A

By the definition of the Cartesian product, $A \times B = \{ (a, b) : a \in A \text{ and } b \in B \}$. Therefore, all first coordinates of the ordered pairs in $A \times B$ must belong to set $A$.

From the given set of ordered pairs $\{(x, 1), (y, 2), (z, 1)\}$, we extract the first components:

• $x \in A$
• $y \in A$
• $z \in A$

Since $n(A) = 3$ and we have identified three distinct elements ($x, y, z$) that belong to $A$, we can conclude that these are exactly the elements of set $A$.

Set $A = \{x, y, z\}$

Step 2: Identifying Elements of Set B

Similarly, by the definition of the Cartesian product, all second coordinates of the ordered pairs in $A \times B$ must belong to set $B$.

From the given set of ordered pairs $\{(x, 1), (y, 2), (z, 1)\}$, we extract the second components:

• $1 \in B$
• $2 \in B$
• $1 \in B$

The set of these values is $\{1, 2, 1\}$, which simplifies to $\{1, 2\}$. Since $n(B) = 2$, this set contains exactly the two required elements.

Set $B = \{1, 2\}$

Step 3: Verification of Cardinality and Set Construction

To verify, we construct the Cartesian product $A \times B$ using the sets found:

$A \times B = \{x, y, z\} \times \{1, 2\}$

$A \times B = \{(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)\}$

The provided ordered pairs $\{(x, 1), (y, 2), (z, 1)\}$ are indeed a subset of the calculated $A \times B$, confirming that our determination of sets $A$ and $B$ satisfies the given conditions.

Final Solution: $A = \{x, y, z\}$ and $B = \{1, 2\}$

Step 1: Evaluation of the Left-Hand Side (LHS)

The LHS is defined as $A \times (B \cap C)$. First, determine the intersection of sets $B$ and $C$.

• Given: $B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$
• Intersection $B \cap C = \{x : x \in B \text{ and } x \in C\}$ [By Definition of Set Intersection]
• Since there are no common elements between $\{1, 2, 3, 4\}$ and $\{5, 6\}$, $B \cap C = \emptyset$ (the empty set).

Now, calculate the Cartesian product of $A$ and $\emptyset$:

• $A \times (B \cap C) = \{1, 2\} \times \emptyset$
• By definition, the Cartesian product of any set with an empty set is an empty set [Theorem: $A \times \emptyset = \emptyset$].
• LHS = $\emptyset$

Step 2: Evaluation of the Right-Hand Side (RHS)

The RHS is defined as $(A \times B) \cap (A \times C)$.

First, find the Cartesian product $A \times B$:

• $A \times B = \{(a, b) : a \in A, b \in B\}$
• $A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\}$

Next, find the Cartesian product $A \times C$:

• $A \times C = \{(a, c) : a \in A, c \in C\}$
• $A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$

Finally, find the intersection $(A \times B) \cap (A \times C)$:

• Check for common ordered pairs between the two sets calculated above.
• There are no ordered pairs present in both sets because their second coordinates belong to disjoint sets $B$ and $C$.
• RHS = $\emptyset$

Step 3: Logical Verification

Comparing the results from Step 1 and Step 2:

Since $\text{LHS} = \text{RHS} = \emptyset$, the distributive property of Cartesian product over set intersection is verified.

Final Solution: Since $A \times (B \cap C) = \emptyset$ and $(A \times B) \cap (A \times C) = \emptyset$, the equality is verified as $\emptyset = \emptyset$.

Step 1: Theoretical Foundation of Cartesian Product

The Cartesian product $A \times B$ of two non-empty sets $A$ and $B$ is defined as the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. Formally, $A \times B = \{ (a, b) : a \in A \text{ and } b \in B \}$ [Per the definition of Cartesian Products].

Step 2: Decomposition of Ordered Pairs

Given the set $A \times B = \{ (a, x), (a, y), (b, x), (b, y) \}$, we analyze the first and second components of each ordered pair to isolate elements of $A$ and $B$.

• The first component of any ordered pair $(a, b) \in A \times B$ must belong to set $A$.
• The second component of any ordered pair $(a, b) \in A \times B$ must belong to set $B$.

Step 3: Identification of Set A

By extracting all unique elements appearing in the first positions of the given ordered pairs:

• First element of $(a, x)$ is $a$.
• First element of $(a, y)$ is $a$.
• First element of $(b, x)$ is $b$.
• First element of $(b, y)$ is $b$.

Therefore, the set of all unique first coordinates is $\{a, b\}$. Thus, $A = \{a, b\}$.

Step 4: Identification of Set B

By extracting all unique elements appearing in the second positions of the given ordered pairs:

• Second element of $(a, x)$ is $x$.
• Second element of $(a, y)$ is $y$.
• Second element of $(b, x)$ is $x$.
• Second element of $(b, y)$ is $y$.

Therefore, the set of all unique second coordinates is $\{x, y\}$. Thus, $B = \{x, y\}$.

Step 5: Verification of the Product

The resulting set $\{ (a, x), (a, y), (b, x), (b, y) \}$ matches the input provided in the problem statement, confirming the validity of the derivation [By the Axiom of Extensionality].

Final Solution: $A = \{a, b\}$ and $B = \{x, y\}$

Step 1: Identifying the Cartesian Product Definition

The Cartesian product of two non-empty sets $P$ and $Q$, denoted as $P \times Q$, is defined as the set of all ordered pairs $(p, q)$ such that $p \in P$ and $q \in Q$. Mathematically: $P \times Q = \{ (p, q) : p \in P, q \in Q \}$ [Per the definition of Cartesian Product in Set Theory].

Step 2: Analysis of Given Sets

Given sets are:

• $P = \{m, n\}$
• $Q = \{n, m\}$

The number of elements in the Cartesian product $P \times Q$ must be equal to the product of the cardinalities of the individual sets. Given $n(P) = 2$ and $n(Q) = 2$, the number of elements in $P \times Q$ is $2 \times 2 = 4$ [Per the Multiplication Principle of Sets].

Step 3: Calculating the Cartesian Product $P \times Q$

To find $P \times Q$, we take each element of $P$ and pair it with every element of $Q$:

• For $p = m$: $(m, n), (m, m)$
• For $p = n$: $(n, n), (n, m)$

Therefore, $P \times Q = \{ (m, n), (m, m), (n, n), (n, m) \}$.

Step 4: Evaluating the Truth Value

The original statement claims $P \times Q = \{(m, n), (n, m)\}$. Since our calculated set contains 4 elements and the statement suggests only 2, the statement is False.

Step 5: Rewriting the Correct Statement

The statement is false because the Cartesian product of two sets containing 2 elements each must contain $2^2 = 4$ ordered pairs.

Correct Statement: If $P = \{m, n\}$ and $Q = \{n, m\}$, then $P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$.

Final Solution: The statement is False. The correct statement is: If $P = \{m, n\}$ and $Q = \{n, m\}$, then $P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$.