Step 1: Given Sets and Definitions

The Universal set and the two subsets are defined as follows:

• $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
• $A = \{2, 4, 6, 8\}$
• $B = \{2, 3, 5, 7\}$

[Definition: The complement of a set $X$ with respect to $U$, denoted as $X'$, is the set of all elements in $U$ that are not in $X$. Formally: $X' = \{x : x \in U \text{ and } x \notin X\}$.]

Step 2: Calculating Left-Hand Side (LHS) = $(A \cap B)'$

First, we find the intersection of $A$ and $B$:

$A \cap B = \{x : x \in A \text{ and } x \in B\} = \{2\}$

Next, we determine the complement of $(A \cap B)$:

$(A \cap B)' = U \setminus \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Step 3: Calculating Right-Hand Side (RHS) = $A' \cup B'$

First, calculate the complements of individual sets $A$ and $B$:

$A' = U \setminus A = \{1, 3, 5, 7, 9\}$

$B' = U \setminus B = \{1, 4, 6, 8, 9\}$

Next, perform the union of $A'$ and $B'$:

$A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\}$

$A' \cup B' = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Step 4: Verification via Venn Diagram Representation

The Venn diagram above shows the distribution of elements. The region $(A \cap B)'$ corresponds to everything outside the central intersection overlap. By computing the union of the regions outside $A$ and $B$, we observe identical element coverage.

Step 5: Comparative Analysis

[Per De Morgan's Law for sets: The complement of the intersection of two sets is equal to the union of their complements.]

Final Solution: Since LHS = RHS = $\{1, 3, 4, 5, 6, 7, 8, 9\}$, the identity $(A \cap B)' = A' \cup B'$ is verified.

Step 1: Identifying the Universal Set and the Given Subset

The universal set $U$ is defined as the set containing all elements under consideration for the given context. The subset $B$ contains specific elements from $U$.

• Universal Set: $U = \{a, b, c, d, e, f, g, h\}$
• Subset: $B = \{d, e, f, g\}$

Step 2: Defining the Complement of a Set

Per the definition of set theory, the complement of a set $B$ with respect to a universal set $U$, denoted as $B'$ or $B^c$, is the set of all elements that are members of $U$ but are not members of $B$.

Mathematically, this is expressed as:

$B' = \{x : x \in U \text{ and } x \notin B\}$

Step 3: Executing the Set Difference Operation

To find $B'$, we remove the elements present in $B$ from the collection $U$:

$B' = U \setminus B$

$B' = \{a, b, c, \color{red}{d, e, f, g}, h\} \setminus \{d, e, f, g\}$

Following the exclusion of elements $\{d, e, f, g\}$ from $U$, we retain the remaining elements:

$B' = \{a, b, c, h\}$

Visual Representation of the Complement

Conclusion

By subtracting the subset $B$ from the universal set $U$, we identify the remaining elements that constitute the complement set.

Final Solution: $B' = \{a, b, c, h\}$

Step 1: Theoretical Foundation

To represent the complement of the union of two sets, $(A \cup B)'$, in a Venn diagram, we must satisfy the following definitions:

• Universal Set ($U$): Represented by a rectangle, containing all elements under consideration.
• Union of Sets ($A \cup B$): The collection of all elements that are in $A$, in $B$, or in both.
• Complement $(A \cup B)'$: The set of all elements in the universal set $U$ that are not in the set $(A \cup B)$. [By the definition of Complement of a Set, $(A \cup B)' = \{x \in U : x \notin (A \cup B)\}]$.

Step 2: Logical Construction

1. Draw a rectangle to denote the Universal Set $U$.
2. Draw two intersecting circles, $A$ and $B$, within the rectangle.
3. Identify the region occupied by $(A \cup B)$, which constitutes the entire interior of both circles.
4. Shade the area outside both circles but inside the rectangle. This region represents the elements that belong to neither $A$ nor $B$.

Step 3: Visual Representation

Step 4: Algebraic Summary

The operation can be broken down using De Morgan's Law [Per De Morgan's Law]:

$(A \cup B)' = A' \cap B'$

This confirms that the region $(A \cup B)'$ is the intersection of elements outside $A$ ($A'$) and elements outside $B$ ($B'$).

Final Solution: The Venn diagram for $(A \cup B)'$ is the region of the universal set $U$ excluding the union of circles $A$ and $B$.

Step 1: Identifying the Universal Set and the Given Subset

Let $U$ be the universal set of all natural numbers. By definition, the set of natural numbers is represented as:

$U = \{1, 2, 3, 4, 5, 6, 7, 8, \dots \}$ [Per the definition of the set of natural numbers, $\mathbb{N}$]

Let $A$ be the given set of odd natural numbers:

$A = \{x : x \text{ is an odd natural number}\} = \{1, 3, 5, 7, 9, \dots \}$

Step 2: Defining the Complement of a Set

The complement of a set $A$, denoted as $A'$ or $A^c$, consists of all elements in the universal set $U$ that are not members of set $A$. Mathematically, this is expressed as:

$A' = \{x \in U : x \notin A\}$ [By the definition of Set Complement]

Step 3: Calculating the Complement

To find $A'$, we subtract the set of odd natural numbers from the set of all natural numbers:

$A' = \{x \in \mathbb{N} : x \text{ is not an odd natural number}\}$

Since any natural number that is not odd must necessarily be an even natural number, we evaluate the set as follows:

$A' = \{2, 4, 6, 8, 10, \dots \}$

This corresponds to the set of all even natural numbers.

Visual Representation of the Complement

Step 4: Formal Set-Builder Notation

The resulting set $A'$ can be expressed in set-builder notation as:

$A' = \{x : x \text{ is an even natural number}\}$

Final Solution: The complement of the set of odd natural numbers is the set of all even natural numbers, represented as $\{x : x \text{ is an even natural number}\}$.

Step 1: Theoretical Foundation of Complementary Sets

By definition, the complement of a subset $A$ with respect to a universal set $U$, denoted by $A'$ or $A^c$, is the set of all elements that belong to the universal set $U$ but do not belong to the set $A$. [Per the Set Difference definition: $A^c = U \setminus A = \{x : x \in U \text{ and } x \notin A\}$].

Step 2: Identification of Given Data

• Universal Set ($U$): $\{a, b, c, d, e, f, g, h\}$
• Subset ($A$): $\{a, b, c\}$

Step 3: Systematic Logical Filtering

To determine $A^c$, we iterate through every element of $U$ and apply the exclusion criteria for set $A$:

Step 4: Visual Representation via Venn Diagram

Step 5: Final Derivation

The complement set $A^c$ comprises the elements remaining in the universal set after removing the intersection of $U$ and $A$.

$A^c = \{x : x \in \{a, b, c, d, e, f, g, h\} \text{ and } x \notin \{a, b, c\}\}$

$A^c = \{d, e, f, g, h\}$

Final Solution: The complement of set $A$ is $A^c = \{d, e, f, g, h\}$.

Step 1: Formal Definition of the Universal Set and Given Set

Given the Universal Set $U$ is the set of all natural numbers:

$U = \mathbb{N} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...\}$ [Per the standard definition of Natural Numbers in set theory].

Let $A$ be the set defined in the problem:

$A = \{x : x \text{ is a perfect cube}\}$

Listing the elements of $A$ based on $n^3$ where $n \in \mathbb{N}$:

$A = \{1^3, 2^3, 3^3, 4^3, ...\} = \{1, 8, 27, 64, ...\}$

Step 2: Defining the Complement of a Set

The complement of a set $A$, denoted by $A^c$ or $A'$, is defined as the set of all elements in the universal set $U$ that are not in $A$.

Mathematically: $A^c = \{x : x \in U \text{ and } x \notin A\}$ [Per the definition of the Complement of a Set].

Step 3: Deriving the Complement Set

To find $A^c$, we subtract the set of perfect cubes from the set of natural numbers:

$A^c = \mathbb{N} \setminus \{x : x = n^3, n \in \mathbb{N}\}$

This implies that $A^c$ contains all natural numbers except those that can be expressed as $n^3$ for some integer $n \ge 1$.

Visual Representation

Step 4: Expressing the Result in Set-Builder Notation

Using the logical condition $x \neq n^3$, we represent the complement set as follows:

$A^c = \{x : x \in \mathbb{N} \text{ and } x \neq n^3 \text{ for any } n \in \mathbb{N}\}$

Final Solution: The complement of the set $\{x : x \text{ is a perfect cube}\}$ is $\{x : x \in \mathbb{N}, x \text{ is not a perfect cube}\}$.

Step 1: Theoretical Definition of the Complement of a Set

By definition, for any set $A$ within a universal set $U$, the complement of set $A$, denoted as $A'$, is defined as the set of all elements that are present in the universal set $U$ but are not present in set $A$. Mathematically, this is expressed as:

$A' = \{x : x \in U \text{ and } x \notin A\}$ [Per the definition of Set Complement]

Step 2: Analysis of the Union Operation

The union of two sets $A$ and $B$, denoted as $A \cup B$, is defined as the set of all elements which are either in $A$, or in $B$, or in both. We are tasked with evaluating the expression $A \cup A'$.

$A \cup A' = \{x : x \in A \text{ or } x \in A'\}$ [Per the definition of Set Union]

Step 3: Logical Substitution and Simplification

Substituting the definition of $A'$ into the union expression:

• Let $x$ be an arbitrary element.
• If $x \in A$, then $x \in A \cup A'$ is true.
• If $x \notin A$, then by the definition of the complement, $x \in A'$. Consequently, $x \in A \cup A'$ is also true.

Since every element $x$ in the universal set $U$ must either be in $A$ or not in $A$ (meaning it must be in $A'$), the union of $A$ and its complement accounts for every possible element within the universal set $U$.

Step 4: Visual Representation

Step 5: Conclusion

Based on the Law of Excluded Middle and the definition of the universal set, the combination of a set and its complement covers the entirety of the universal set $U$.

Final Solution: $A \cup A' = U$

Step 1: Formal Definition of the Universal Set and Given Set

Let $U$ be the Universal set defined as the set of all natural numbers. In standard mathematical notation:

$U = \{1, 2, 3, 4, \dots\} = \mathbb{N}$

Let the given set be $A$. We are provided with $A$ in set-builder notation:

$A = \{x : 2x + 5 = 9\}$

Step 2: Algebraic Derivation of Set Elements

To find the elements of set $A$, we solve the linear equation provided [Per the Transposition Method for linear equations]:

$2x + 5 = 9$

$2x = 9 - 5$ [Subtracting 5 from both sides]

$2x = 4$

$x = \frac{4}{2}$ [Dividing both sides by 2]

$x = 2$

Since $2 \in \mathbb{N}$, the solution is valid within the Universal set $U$. Thus, the set $A$ expressed in roster form is:

$A = \{2\}$

Step 3: Calculating the Complement Set

The complement of a set $A$, denoted by $A^c$ or $A'$, is defined as the set of all elements in the Universal set $U$ that are not elements of set $A$ [Per the definition of Set Complement]:

$A^c = \{x : x \in U \text{ and } x \notin A\}$

Substituting the known values:

$A^c = \{x \in \mathbb{N} : x \neq 2\}$

This implies that $A^c$ contains all natural numbers except for 2.

Visual Representation of the Complement

Step 4: Final Formulation

The complement set $A^c$ can be written in set-builder notation as:

$A^c = \{x : x \in \mathbb{N}, x \neq 2\}$

Alternatively, by listing the elements:

$A^c = \{1, 3, 4, 5, \dots\}$

Final Solution: $\{x : x \in \mathbb{N}, x \neq 2\}$

Step 1: Definition of the Universal Set and Empty Set

In set theory, let $U$ be the Universal Set, which contains all elements under consideration for a given context. The empty set, denoted by $\phi$, is defined as the set containing no elements. [By the Axiom of the Empty Set].

Step 2: Determining the Complement of the Empty Set

The complement of any set $S$, denoted by $S'$, is defined as the set of all elements in the Universal Set $U$ that are not in $S$. Mathematically:

$S' = \{x \in U : x \notin S\}$

Applying this definition to the empty set $\phi$:

$\phi' = \{x \in U : x \notin \phi\}$

Since the empty set contains no elements, every element in the Universal Set $U$ satisfies the condition $x \notin \phi$. Therefore:

$\phi' = U$ [Per the Complement Law of Sets]

Step 3: Performing the Intersection Operation

We are required to compute $\phi' \cap A$. Substituting the result from Step 2:

$\phi' \cap A = U \cap A$

According to the properties of intersection, the intersection of the Universal Set $U$ with any subset $A$ of $U$ results in the subset $A$ itself, as all elements of $A$ are also elements of $U$.

$U \cap A = A$ [Per the Identity Law of Intersection]

Visual Representation

Step 4: Logical Deduction Summary

Final Solution: \phi' \cap A = A

Step 1: Identifying the Universal Set

In accordance with the problem statement, the Universal Set ($U$) is defined as the set of all natural numbers. Mathematically, this is represented as:

$U = \{1, 2, 3, 4, 5, 6, 7, 8, \dots\}$ [By definition of $\mathbb{N}$]

Step 2: Defining the Given Set

Let the given set be $A$. The set is defined by the condition $x + 5 = 8$. To find the elements of set $A$, we solve the linear equation:

$x + 5 = 8$

$x = 8 - 5$ [By the Subtraction Property of Equality]

$x = 3$

Thus, $A = \{3\}$.

Step 3: Calculating the Complement of the Set

The complement of a set $A$, denoted as $A'$ or $A^c$, consists of all elements that are present in the Universal Set ($U$) but are not present in set $A$.

Formula: $A' = \{x : x \in U \text{ and } x \notin A\}$ [Per the definition of the Complement of a Set in Set Theory]

Given $U = \{1, 2, 3, 4, 5, \dots\}$ and $A = \{3\}$, the complement $A'$ is derived by removing the element $3$ from the set of natural numbers:

$A' = \{1, 2, 4, 5, 6, 7, 8, \dots\}$

Step 4: Visual Representation of the Complement

Step 5: Conclusion in Set-Builder Notation

To express the result formally, we exclude the specific value $3$ from the natural number domain:

$A' = \{x : x \in \mathbb{N} \text{ and } x \neq 3\}$

Final Solution: $\{x : x \in \mathbb{N}, x \neq 3\}$ or $\{1, 2, 4, 5, 6, \dots\}$

Theoretical Foundation of Set Complementation

To determine the value of $A \cap A'$, we must refer to the foundational definition of the complement of a set within a Universal set $U$.

• Definition of Complement: For a set $A \subseteq U$, the complement $A'$ (or $A^c$) is defined as the set of all elements in $U$ that are not in $A$. Symbolically: $A' = \{x \in U : x \notin A\}$. [Per the Definition of Complementary Sets]
• Definition of Intersection: The intersection of two sets $A$ and $B$, denoted $A \cap B$, is the set of all elements that are common to both $A$ and $B$. Symbolically: $A \cap B = \{x : x \in A \text{ and } x \in B\}$. [Per the Axiom of Intersection]

Step 1: Logical Analysis of the Intersection

Consider an arbitrary element $x$ such that $x \in (A \cap A')$.

1. By the definition of intersection, $x \in A$ AND $x \in A'$.
2. From the definition of $A'$, we know that $x \in A'$ implies $x \notin A$.
3. We are now presented with the logical conjunction: $x \in A \land x \notin A$.

Step 2: Contradiction Assessment

The statement $x \in A \land x \notin A$ is a logical contradiction, as an element cannot simultaneously belong to a set and its negation. [Per the Law of Non-Contradiction in Formal Logic].

Because there is no element $x$ that satisfies both conditions simultaneously, the resulting set must contain no elements.

Visual Representation of Disjoint Sets

Conclusion

Since $A$ and $A'$ have no elements in common, their intersection is the empty set, denoted by the symbol $\emptyset$ or the empty braces $\{\}$.

Final Solution: $A \cap A' = \emptyset$

Given Variables & Initial Setup

The Universal set $U$ and subsets $A$ and $B$ are defined as follows:

• $U = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$
• $A = \{ 1, 2, 3, 4 \}$
• $B = \{ 2, 4, 6, 8 \}$

Step 1: Determining the Union of Sets A and B

The union of two sets, denoted by $A \cup B$, is the set of all elements that belong to either $A$, or $B$, or both [By the definition of Set Union].

$A \cup B = \{ x : x \in A \text{ or } x \in B \}$

$A \cup B = \{ 1, 2, 3, 4 \} \cup \{ 2, 4, 6, 8 \}$

Combining the elements while adhering to the principle that set elements are unique (no repetition):

$A \cup B = \{ 1, 2, 3, 4, 6, 8 \}$

Step 2: Calculating the Complement $(A \cup B)'$

The complement of a set $S$ with respect to the universal set $U$, denoted by $S'$, is the set of all elements in $U$ that are not in $S$ [Per the Complement Law: $S' = \{ x \in U : x \notin S \}$].

Here, $S = A \cup B = \{ 1, 2, 3, 4, 6, 8 \}$.

$(A \cup B)' = \{ x : x \in U \text{ and } x \notin (A \cup B) \}$

$(A \cup B)' = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} \setminus \{ 1, 2, 3, 4, 6, 8 \}$

Step 3: Verification via Venn Diagram Representation

Removing the elements $\{ 1, 2, 3, 4, 6, 8 \}$ from the universal set $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ leaves the set of remaining elements:

$(A \cup B)' = \{ 5, 7, 9 \}$

Final Solution: $(A \cup B)' = \{ 5, 7, 9 \}$

Given Variables & Initial Setup

Let $U$ be the Universal set and $A$ be a subset of $U$ (i.e., $A \subseteq U$). We are tasked to evaluate the intersection of the complement of the universal set ($U'$) and the set $A$.

Step 1: Definition of the Complement of the Universal Set

By the definition of the complement of a set, for any set $S \subseteq U$, its complement $S'$ is defined as $U \setminus S = \{x : x \in U \text{ and } x \notin S\}$. Applying this to the Universal set $U$:

$U' = \{x : x \in U \text{ and } x \notin U\}$

[Per the Definition of Complementary Sets]

Step 2: Logical Evaluation of $U'$

Since the universal set $U$ contains every element under consideration in the given context, there exists no element $x$ such that $x \notin U$. Therefore, the set $U'$ contains no elements.

$U' = \emptyset$ (The Empty Set or Null Set)

[Per the Law of Universal Complementation]

Step 3: Intersection with Set A

We are given the expression $U' \cap A$. Substituting the result from Step 2:

$U' \cap A = \emptyset \cap A$

By the property of the intersection with the empty set, the intersection of any set $A$ with the null set $\emptyset$ is the null set itself.

$\emptyset \cap A = \emptyset$

[Per the Null Law of Intersection]

Visual Representation of the Null Set

Summary of Properties Used

Final Solution: $U' \cap A = \emptyset$

1. Definitions & Universal Set Setup

In the context of set theory, let the universal set $U$ be the set of all natural numbers. By definition, $\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$.

The given set is defined as $A = \{x : x \in \mathbb{N} \text{ and } x \ge 7\}$.

Expressed in roster form, this set is: $A = \{7, 8, 9, 10, 11, \dots\}$.

2. Theoretical Foundation of Complementation

The complement of a set $A$, denoted by $A^c$ or $A'$, is defined as the set of all elements in the universal set $U$ that are not in $A$.

Mathematically: $A^c = \{x : x \in U \text{ and } x \notin A\}$ [By the definition of Set Complementation].

3. Derivation of the Complement

Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, \dots\}$ and $A = \{7, 8, 9, 10, \dots\}$, we perform the set subtraction $U \setminus A$.

• The elements $1, 2, 3, 4, 5, 6$ belong to $U$ but do not satisfy the condition $x \ge 7$.
• All elements $x \in U$ where $x \ge 7$ are excluded because they constitute set $A$.

Therefore, $A^c = \{x : x \in \mathbb{N} \text{ and } x < 7\}$.

4. Visual Representation of the Complement

5. Final Set Construction

The resulting set $A^c$ contains all natural numbers strictly less than 7.

In Roster Form: $A^c = \{1, 2, 3, 4, 5, 6\}$.

In Set-Builder Form: $A^c = \{x : x \in \mathbb{N}, x < 7\}$.

Final Solution: \{x : x \in \mathbb{N}, x < 7\} \text{ or } \{1, 2, 3, 4, 5, 6\}

Step 1: Formal Definition of the Universal Set and the Given Set

Let the universal set $U$ be the set of all natural numbers. Mathematically, this is expressed as:

$U = \{x : x \in \mathbb{N}\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \dots\}$

Let $A$ be the set defined by the condition that $x$ is a prime number. By definition, a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.

$A = \{x : x \text{ is a prime number}\} = \{2, 3, 5, 7, 11, 13, \dots\}$

Step 2: Defining the Complement of a Set

The complement of a set $A$ with respect to a universal set $U$, denoted as $A^c$ or $A'$, is defined as the set of all elements that belong to $U$ but do not belong to $A$ [Per the Set Theory Definition of Complements]:

$A^c = \{x : x \in U \text{ and } x \notin A\}$

Step 3: Logical Derivation of the Complement

To find $A^c$, we must identify all natural numbers that are not prime. These consist of:

• The number $1$ (which is neither prime nor composite).
• Composite numbers (natural numbers greater than $1$ that have at least one divisor other than $1$ and themselves).

Therefore, $A^c = \{x : x \in \mathbb{N}, x \neq 1, \text{ and } x \text{ is not a prime number}\}$.

Step 4: Visualizing the Set Difference

Step 5: Synthesizing the Final Set

The set $A^c$ includes all natural numbers except prime numbers. Expressed in set-builder notation:

$A^c = \{x : x \in \mathbb{N} \text{ and } x = 1 \text{ or } x \text{ is a composite number}\}$

In tabular form, noting $1$ and the sequence of composite numbers ($4, 6, 8, 9, 10, \dots$):

$A^c = \{1, 4, 6, 8, 9, 10, 12, \dots\}$

Final Solution: $\{x : x \in \mathbb{N} \text{ and } x \text{ is not a prime number}\} = \{1, 4, 6, 8, 9, 10, 12, \dots\}$

Step 1: Theoretical Foundation of De Morgan's Law

The expression $A' \cap B'$ represents the intersection of the complement of set $A$ and the complement of set $B$ within a universal set $U$. According to De Morgan’s Law of Intersection, this is equivalent to the complement of the union of the two sets:

$A' \cap B' = (A \cup B)'$ [De Morgan’s Law]

This implies that the region representing $A' \cap B'$ consists of all elements in the universal set $U$ that are neither in $A$ nor in $B$. Visually, this is the entire region inside the rectangle (Universal Set) excluding the two circles representing sets $A$ and $B$.

Step 2: Geometric Construction of the Venn Diagram

To construct the Venn diagram, we define the following:

• A rectangular boundary representing the Universal Set ($U$).
• Two overlapping circles representing set $A$ and set $B$.
• The shaded region will be the area outside both circles.

Step 3: Analytical Interpretation of Shaded Region

The Venn diagram above demonstrates that $A' \cap B'$ captures the complement of the union of sets $A$ and $B$. The shading covers all elements within the universal set $U$ except for the interior of circles $A$ and $B$.

Final Solution: The Venn diagram for $A' \cap B'$ is represented by shading the entire region within the universal set $U$ that lies outside both sets $A$ and $B$, confirming the identity $(A \cup B)' = A' \cap B'$.

Given Variables & Initial Setup

Let $U$ be the Universal set defined as the set of all natural numbers. Mathematically, this is expressed as:

$U = \{x : x \in \mathbb{N}\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, \dots\}$

Let $A$ be the set defined in the problem statement:

$A = \{x : x \in \mathbb{N} \text{ and } x \text{ is a positive multiple of } 3\}$

$A = \{3, 6, 9, 12, 15, \dots\}$

Step 1: Theoretical Definition of the Complement

By the definition of the complement of a set, for any set $A$ within a universal set $U$, the complement $A'$ (or $A^c$) is the set of all elements in $U$ that are not in $A$.

$A' = \{x : x \in U \text{ and } x \notin A\}$ [By Definition of Complement Sets]

Step 2: Analytical Identification of Elements

To determine $A'$, we extract the elements of $U$ and exclude every element that is a multiple of $3$.

• $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, \dots\}$
• $A = \{3, 6, 9, 12, \dots\}$
• $A' = U \setminus A = \{x \in \mathbb{N} : x \text{ is not a multiple of } 3\}$

Visual Representation of Set Exclusion

Step 3: Final Set Notation

The complement set $A'$ consists of all natural numbers excluding those that are multiples of 3. We can represent this using set-builder notation:

$A' = \{x : x \in \mathbb{N}, x \neq 3n \text{ for any } n \in \mathbb{N}\}$

Final Solution: The complement of the set $\{x : x \text{ is a positive multiple of } 3\}$ is $\{x : x \in \mathbb{N} \text{ and } x \text{ is not a positive multiple of } 3\}$.

Step 1: Theoretical Definition of Set Complement

The complement of a set $A$, denoted by $A'$ or $A^c$, is defined as the set of all elements in the universal set $U$ that are not members of set $A$. Mathematically, this is expressed as:

$A' = \{ x : x \in U \text{ and } x \notin A \}$

[By Definition of Set Complement]

Step 2: Identifying Given Sets

• Universal Set ($U$): $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$
• Set $A$: $\{ 1, 2, 3, 4 \}$

Step 3: Execution of Set Difference

To find $A'$, we extract all elements present in $U$ and remove any elements that appear in set $A$.

Subtracting set $A$ from $U$ leaves the remaining elements: $\{ 5, 6, 7, 8, 9 \}$.

Step 4: Visualizing the Complement in a Venn Diagram

The shaded region representing $A'$ consists of all space within the rectangular boundary of $U$ excluding the interior of the circle $A$.

Final Solution: A' = { 5, 6, 7, 8, 9 }

Step 1: Formal Definition of Sets and Universal Set

Let the universal set $U$ be the set of all natural numbers. Mathematically, this is expressed as:

$U = \mathbb{N} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \dots\}$ [Definition of Natural Numbers]

Let $A$ be the given set of perfect squares. The set $A$ is defined by the set-builder notation:

$A = \{x : x = n^2, n \in \mathbb{N}\}$

Expanding the elements of set $A$:

$A = \{1^2, 2^2, 3^2, 4^2, \dots\} = \{1, 4, 9, 16, 25, \dots\}$

Step 2: Defining the Complement

The complement of a set $A$ with respect to the universal set $U$, denoted as $A'$ or $A^c$, consists of all elements that are in $U$ but not in $A$.

$A' = \{x : x \in U \text{ and } x \notin A\}$ [Per the definition of the Complement of a Set]

Step 3: Deriving the Complement Set

To find $A'$, we subtract the set of perfect squares from the set of all natural numbers:

$A' = \{x \in \mathbb{N} : x \text{ is not a perfect square}\}$

By listing the elements, we observe the exclusion of elements $n^2$:

$A' = \{x : x \in \mathbb{N}, x \neq n^2 \text{ for any } n \in \mathbb{N}\}$

Listing the first few elements for clarity:

$A' = \{2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, \dots\}$

Step 4: Visualization

Step 5: Final Symbolic Representation

The result is the subset of natural numbers that excludes values identified as perfect squares. In logical notation:

$A' = \{x \in \mathbb{N} : x \text{ is not a perfect square}\}$

Final Solution: The complement of the set of perfect squares is $\{x : x \in \mathbb{N} \text{ and } x \text{ is not a perfect square} \}$.

Step 1: Theoretical Foundation

The expression $(A \cap B)'$ denotes the complement of the intersection of sets $A$ and $B$. In set theory, the complement of a set $X$ relative to the universal set $U$ is defined as $X' = \{x \in U \mid x \notin X\}$. Therefore, $(A \cap B)' = U \setminus (A \cap B)$. This region represents all elements in the universal set $U$ that do not belong to the common region shared by sets $A$ and $B$. [Per De Morgan’s Laws and the definition of Complement sets].

Step 2: Defining the Venn Diagram Construction

• Universal Set ($U$): Represented by a rectangle containing all elements.
• Sets $A$ and $B$: Represented by two overlapping circles within the rectangle.
• The Intersection ($A \cap B$): The central region where the two circles overlap.
• The Complement ($A \cap B$)' : The entire region within the rectangle, excluding the central intersection.

Step 3: Visual Representation

Step 4: Analytical Conclusion

To construct the required diagram:

1. Draw a rectangle to represent the Universal set $U$.
2. Draw two intersecting circles $A$ and $B$.
3. Shade the entire region inside the rectangle except for the lens-shaped area formed by the overlap of circles $A$ and $B$.
4. The unshaded central region represents the intersection $(A \cap B)$, while the shaded area represents $(A \cap B)' = A' \cup B'$. [By De Morgan's Law: $(A \cap B)' = A' \cup B'$].

Final Solution: The Venn diagram for $(A \cap B)'$ is illustrated by shading the entire region of the universal set excluding the intersection of set $A$ and set $B$.