The problem consists of several questions related to set theory. These include listing elements of sets defined by certain conditions, finding results of set operations, simplifying set expressions, expressing sets in set builder notation, proving set identities, and representing sets using Venn diagrams.

1. Problem Description

The problem consists of several questions related to set theory. These include listing elements of sets defined by certain conditions, finding results of set operations, simplifying set expressions, expressing sets in set builder notation, proving set identities, and representing sets using Venn diagrams.

2. Solution Steps

1. (a) The set contains integers greater than -2 and less than

9. Solution: $\{-1, 0, 1, 2, 3, 4, 5, 6, 7, 8\}$

(b) The set contains integers that are also members of the set $\{0, \sqrt{3}, \pi, 2i\}$.

Since integers are real numbers and $2i$ is imaginary and $\sqrt{3}, \pi$ are not integers, the only integer in the given set is

0. Solution: $\{0\}$

(c) The set contains elements of the form $x^2 + 1$, where $x$ is an element of the set $A = \{-2, -1, 0, 1, 2\}$.

For $x = -2$, $x^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5$.

For $x = -1$, $x^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2$.

For $x = 0$, $x^2 + 1 = (0)^2 + 1 = 0 + 1 = 1$.

For $x = 1$, $x^2 + 1 = (1)^2 + 1 = 1 + 1 = 2$.

For $x = 2$, $x^2 + 1 = (2)^2 + 1 = 4 + 1 = 5$.

Solution: $\{1, 2, 5\}$

(d) The set contains elements of the form $\sqrt{x + 2}$, where $x$ is an element of the set $B = \{-3, -4, 0, 1, 2\}$.

For $x = -3$, $\sqrt{x + 2} = \sqrt{-3 + 2} = \sqrt{-1}$, which is not real.

For $x = -4$, $\sqrt{x + 2} = \sqrt{-4 + 2} = \sqrt{-2}$, which is not real.

For $x = 0$, $\sqrt{x + 2} = \sqrt{0 + 2} = \sqrt{2}$.

For $x = 1$, $\sqrt{x + 2} = \sqrt{1 + 2} = \sqrt{3}$.

For $x = 2$, $\sqrt{x + 2} = \sqrt{2 + 2} = \sqrt{4} = 2$.

Solution: $\{\sqrt{2}, \sqrt{3}, 2\}$

2. Given: $A = \{0, 2, 4, 6, 8, 10\}$, $B = \{1, 3, 5, 7, 9\}$, $C = \{1, 2, 4, 5, 7, 8\}$, $D = \{1, 2, 3, 5, 7, 8, 9\}$.

(a) $(A - B) - C$:

$A - B = \{0, 2, 4, 6, 8, 10\}$.

$(A - B) - C = \{0, 2, 4, 6, 8, 10\} - \{1, 2, 4, 5, 7, 8\} = \{0, 6, 10\}$

Solution: $\{0, 6, 10\}$

(b) $(A \cap B) \cup (A \cap C)$:

$A \cap B = \emptyset$

$A \cap C = \{2, 4, 8\}$

$(A \cap B) \cup (A \cap C) = \emptyset \cup \{2, 4, 8\} = \{2, 4, 8\}$

Solution: $\{2, 4, 8\}$

(c) $D \cap (A \cap C)$:

$A \cap C = \{2, 4, 8\}$

$D \cap (A \cap C) = \{1, 2, 3, 5, 7, 8, 9\} \cap \{2, 4, 8\} = \{2, 8\}$

Solution: $\{2, 8\}$

(e) $(A \cup B) - (C \cap D)$:

$A \cup B = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

$C \cap D = \{1, 2, 5, 7, 8\}$

$(A \cup B) - (C \cap D) = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - \{1, 2, 5, 7, 8\} = \{0, 3, 4, 6, 9, 10\}$

Solution: $\{0, 3, 4, 6, 9, 10\}$

3. (a) $[A' \cup (B \cap C)]'$:

Using De Morgan's law, $[A' \cup (B \cap C)]' = (A')' \cap (B \cap C)' = A \cap (B' \cup C')$ (Using De Morgan's law again).

Solution: $A \cap (B' \cup C')$

(b) $(A - B) \cup (A \cap B)$:

$A - B = A \cap B'$.

$(A - B) \cup (A \cap B) = (A \cap B') \cup (A \cap B) = A \cap (B' \cup B) = A \cap U = A$, where U is the universal set.

Solution: $A$

(c) $(A - B)'$:

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

Solution: $A' \cup B$

(d) $(A \cup B) \cap (A \cup B')$:

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

Solution: $A$

4. (a) $A = \{1, 8, 27, 64, 125\}$

These are cubes of integers: $1 = 1^3, 8 = 2^3, 27 = 3^3, 64 = 4^3, 125 = 5^3$.

Solution: $A = \{x: x = n^3, n \in \{1, 2, 3, 4, 5\}\}$

(b) $B = \{0, 2, 4, 6, 8, 12\}$

These are even numbers. Notice that 10 is missing from the set and 12 is included. This is a bit vague because the numbers do not have a obvious rule.

Solution: $B = \{x: x = 2n, n \in \{0, 1, 2, 3, 4, 6\}\}$

(c) $C = \{1, 3, 9, 27, 81\}$

These are powers of 3: $1 = 3^0, 3 = 3^1, 9 = 3^2, 27 = 3^3, 81 = 3^4$.

Solution: $C = \{x: x = 3^n, n \in \{0, 1, 2, 3, 4\}\}$

(d) $D = \{(1, 2), (2, 4), (3, 6), (4, 8)\}$

These are ordered pairs $(x, y)$ such that $y = 2x$.

Solution: $D = \{(x, y): y = 2x, x \in \{1, 2, 3, 4\}\}$

5. Proofs:

(a) $(A \cap B)' = A' \cup B'$: This is De Morgan's Law.

(b) $(A \cup B)' = A' \cap B'$: This is De Morgan's Law.

(c) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$: Distributive Law.

(f) $(A')' = A$: Double complement law.

6. Venn diagrams can be drawn to illustrate these sets but are not given as text solutions.

7. Show that $[(A \cap B) \cup (A \cap B')]' = A'$.

$[(A \cap B) \cup (A \cap B')]' = [A \cap (B \cup B')]' = [A \cap U]' = A'$.

Since $B \cup B' = U$, and $A \cap U = A$, and $A' = A'$.

Solution: $[(A \cap B) \cup (A \cap B')]' = A'$.

3. Final Answer

1. (a) $\{-1, 0, 1, 2, 3, 4, 5, 6, 7, 8\}$

(b) $\{0\}$

(c) $\{1, 2, 5\}$

(d) $\{\sqrt{2}, \sqrt{3}, 2\}$

2. (a) $\{0, 6, 10\}$

(b) $\{2, 4, 8\}$

(c) $\{2, 8\}$

(e) $\{0, 3, 4, 6, 9, 10\}$

3. (a) $A \cap (B' \cup C')$

(b) $A$

(c) $A' \cup B$

(d) $A$

4. (a) $A = \{x: x = n^3, n \in \{1, 2, 3, 4, 5\}\}$

(b) $B = \{x: x = 2n, n \in \{0, 1, 2, 3, 4, 6\}\}$

(c) $C = \{x: x = 3^n, n \in \{0, 1, 2, 3, 4\}\}$

(d) $D = \{(x, y): y = 2x, x \in \{1, 2, 3, 4\}\}$

5. Proofs provided in solution steps.

6. Venn diagrams are not given here.

7. Proof provided in solution steps.