The problem consists of several parts related to set theory: listing elements of sets, finding set operations, simplifying set expressions using laws, describing sets in set-builder notation, proving set laws, representing sets using Venn diagrams, and proving a set identity. We will address questions 1, 2, 3 and 4 in this response.

1. Problem Description

The problem consists of several parts related to set theory: listing elements of sets, finding set operations, simplifying set expressions using laws, describing sets in set-builder notation, proving set laws, representing sets using Venn diagrams, and proving a set identity. We will address questions 1, 2, 3 and 4 in this response.

2. Solution Steps

1(a) List the elements of the set {x: x is an integer greater than -2 but less than 9}.

The integers greater than -2 are -1, 0, 1, 2, ...

The integers less than 9 are 8, 7, 6, 5, ...

Therefore, the set is {-1, 0, 1, 2, 3, 4, 5, 6, 7, 8}.

1(b) List the elements of the set {x: x is an integer} ∩ {0, √3, π, 2i }.

The set {x: x is an integer} consists of integers.

The set {0, √3, π, 2i } consists of 0, the square root of 3, pi, and 2 times the imaginary unit i.

The intersection of these two sets consists of elements that are both integers and are in {0, √3, π, 2i }.

The only integer in {0, √3, π, 2i } is

0. Therefore, the intersection is {0}.

1(c) List the elements of the set {x² + 1: x ∈ A} where A = {-2, -1, 0, 1, 2}.

We calculate $x^2 + 1$ for each element $x$ in $A$:

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

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

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

If $x = 1$, then $x^2 + 1 = (1)^2 + 1 = 1 + 1 = 2$.

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

The set is {5, 2, 1, 2, 5} = {1, 2, 5}.

1(d) List the elements of the set { √x + 2 : x ∈ B} where B = {-3, -4, 0, 1, 2}.

We calculate $\sqrt{x + 2}$ for each element $x$ in $B$:

If $x = -3$, then $\sqrt{x + 2} = \sqrt{-3 + 2} = \sqrt{-1}$, which is not a real number.

If $x = -4$, then $\sqrt{x + 2} = \sqrt{-4 + 2} = \sqrt{-2}$, which is not a real number.

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

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

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

Assuming we are looking for real numbers, the set is {√2, √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

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

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

(b) (A ∩ B) U (A ∩ C)

$A ∩ B = \{\}$ (empty set, since there are no common elements)

$A ∩ C = \{2, 4, 8\}$

$(A ∩ B) U (A ∩ C) = \{\} U \{2, 4, 8\} = \{2, 4, 8\}$.

(c) D ∩ (A ∩ C)

$A ∩ C = \{2, 4, 8\}$

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

(e) (A U B) - (C ∩ D)

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

$C ∩ D = \{1, 2, 4, 5, 7, 8\}$

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

3. Using the laws of sets, simplify the following

(a) $[A' U (B ∩ C)]'$

Using De Morgan's law: $(X U Y)' = X' ∩ Y'$

$[A' U (B ∩ C)]' = (A')' ∩ (B ∩ C)' = A ∩ (B' U C')$ (using De Morgan's law again)

(b) (A - B) U (A ∩ B)

$A - B = A ∩ B'$

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

(c) (A - B)'

$A - B = A ∩ B'$

$(A - B)' = (A ∩ B')' = A' U (B')' = A' U B$ (using De Morgan's Law)

(d) (A U B) ∩ (A U B')

$(A U B) ∩ (A U B') = A U (B ∩ B') = A U \{\} = A$

4. Describe each of the following in set builder notation:

(a) A = {1, 8, 27, 64, 125}

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

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

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

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

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

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

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

3. Final Answer

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

1(b) {0}

1(c) {1, 2, 5}

1(d) {√2, √3, 2}

2(a) {0, 6, 10}

2(b) {2, 4, 8}

2(c) {2, 8}

2(e) {0, 3, 6, 9, 10}

3(a) $A ∩ (B' U C')$

3(b) A

3(c) A' U B

3(d) A

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

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

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

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