set theory

Explanation of Rational Numbers

• Definition of Rational Numbers: A rational number is defined as any number that can be expressed as the quotient of two integers, where the denominator is not zero. This can be expressed mathematically as:
  $\text{Rational Number} = \frac{p}{q}$
  where $p$ and $q$ are integers and $q \neq 0$.

• Subsets Explanation:

  • The set of all integers, denoted as $\mathbb{Z}$, is a subset of the set of all rational numbers, denoted as $\mathbb{Q}$.

  • This means that every integer (e.g., -3, -1, 0, 1, 2) can be expressed as a rational number; for example, the integer 5 can be written as $\frac{5}{1}$.

• Conclusion:

  • Therefore, the statement "Every integer is a rational number" is true, as integers belong to the broader set of rational numbers.

Set Operations

Given sets

• Set A: $A = {1,2,3}$

• Set B: $B = {1,5,7}$

• Set C: $C = {2,3}$

Required Operations

1. a) Union of A and B

   • $A \cup B = {1,2,3} \cup {1,5,7} = {1,2,3,5,7}$

2. b) Intersection of A and B (An B)

   • $A \cap B = {1,2,3} \cap {1,5,7} = {1}$

3. c) Intersection of A and the complement of B

   • Find the complement of B with respect to universal set (assumed to be ${1,2,3,4,5,6,7}$): $B^c = {2,3,4,6}$

   • Thus, $A \cap B^c = {1,2,3} \cap {2,3,4,6} = {2,3}$

4. d) Difference of A and B

   • $A \setminus B = {1,2,3} \setminus {1,5,7} = {2,3}$

5. e) Union of A and intersection of B and C

   • First calculate $B \cap C$:

     • $B \cap C = {1,5,7} \cap {2,3} = \emptyset$

   • Therefore, $A \cup (B \cap C) = A \cup \emptyset = A = {1,2,3}$

6. f) Union of A and B minus C

   • $A \cup (B \setminus C) = {1,2,3} \cup ({1,5,7} \setminus {2,3}) = {1,2,3} \cup {1,5,7} = {1,2,3,5,7}$.

7. g) Intersection of A and (Union of B and C)

   • $A \cap (B \cup C) = {1,2,3} \cap ({1,5,7} \cup {2,3}) \rightarrow B \cup C = {1,2,3,5,7}$

   • Thus, $A \cap (B \cup C) = {1,2,3}$

8. h) Union of C and B

   • $C \cup B = {2,3} \cup {1,5,7} = {1,2,3,5,7}$

Interval Notations

Given Intervals

• I₁ = (-2, 3]

• I₂ = [1, 5)

• I₃ = (0, 4)

Required Operations

1. a) Union of intervals I₁ and I₃

   • $I₁ \cup I₃ = (-2, 3] \cup (0, 4) = (-2, 4)$

2. b) Intersection of intervals I₂ and I₃

   • $I₂ \cap I₃ = [1,5) \cap (0,4) = [1,4)$

3. c) Union of I₁ and I₂

   • $I₁ \cup I₂ = (-2, 3] \cup [1, 5) = (-2, 5)$

Absolute Value Expressions

Rewrite Expressions Without Absolute Values

1. a) 15 - 23

   • As it is, this expression does not contain an absolute value.

2. b) Involving two absolute values

   • $||-2 - (-3)|| = |1| = 1$

3. c) Involving an absolute value on a polynomial

   • $|x² + 1| = x² + 1$ (as it is always non-negative).

4. d) For a conditional expression

   • $|1 - 2x²| = 1 - 2x²$ if $2x² < 1$ (this means $x < \frac{1}{\sqrt{2}}$), otherwise, it will be the negation of that.

5. e) Polynomial condition

   • Similar structure as previous statement, depends on $|x-2|$.

Solving Inequalities Using a Sign Line

Inequalities to Solve

1. a) For 2x + 7 ≥ 3:

   • Rearranging gives: $2x <br>geq -4 \Rightarrow x <br>geq -2$

   • Solution set in interval form: $[-2, \infty)$.

2. b) For 1 < 3x + 4 ≤ 16:

   • Break into two parts:

     • $3x + 4 > 1 \Rightarrow 3x > -3 \Rightarrow x > -1$

     • $3x + 4 ≤ 16 \Rightarrow 3x ≤ 12 \Rightarrow x ≤ 4$

   • Combined solution: $(-1, 4]$.

3. c) For 0 ≤ 1 - 2x < 1:

   • Solve 1: $0 ≤ 1 - 2x \Rightarrow 2x ≤ 1 \Rightarrow x ≥ \frac{1}{2}$

   • Solve 2: 1 - 2x < 1 \Rightarrow -2x < 0 \Rightarrow x > 0

   • Combined solution set leads to: $[\frac{1}{2}, \infty)$.

4. d) For conditions like > x

   • Generally relies on direction and balance of the x-value individually.

5. e) Solving |x + 5| > 6:

   • This breaks into $x + 5 > 6$ or $x + 5 < -6$

     • Resolving both gives: $x > 1$ or $x < -11$ leading to the solution: $(-\infty, -11) \cup (1, \infty)$.

Writing Set S as a Union of Intervals

• Given Equation Set: $x² + x - 20 ≤ 0$

  • This factors to:
    $(x - 4)(x + 5) ≤ 0$

• Test Intervals:

  • Intervals: $(-\infty, -5], (-5, 4], (4, \infty)$

  • Required solution turns out to be: $[-5, 4]$.

Solving Absolute Value Equations

Problems to Solve

1. a) For |2x| = 3:

   • Two cases arise: $2x = 3 \Rightarrow x = \frac{3}{2}$ and $2x = -3 \Rightarrow x = -\frac{3}{2}$.

   • Solutions: $x = \pm \frac{3}{2}$.

2. b) For 3x + 5 = 1:

   • $3x = -4 \Rightarrow x = -\frac{4}{3}$.

3. c) For |x + 3| = |12x + 1|

   • This requires balancing both cases similarly to maintain identity.

4. d) For expressions like 12x - 3| ≤ 4:

   • Split into manageable concomitant inequalities leading to final intervals.

Conversion between Degrees and Radians

Degree to Radian Conversion Formula

• The formula used is: $Radians = Degrees \times \left( \frac{\pi}{180} \right)$

Conversion Examples

1. a) Convert 30° to Radians:

   • $30° \times \left( \frac{\pi}{180} \right) = \frac{\pi}{6}$

2. b) Convert 45° to Radians:

   • $45° \times \left( \frac{\pi}{180} \right) = \frac{\pi}{4}$

3. c) Convert 60° to Radians:

   • $60° \times \left( \frac{\pi}{180} \right) = \frac{\pi}{3}$

4. d) Convert 210° to Radians:

   • $210° \times \left( \frac{\pi}{180} \right) = \frac{7\pi}{6}$

Conversion from Radians to Degrees

Radian to Degree Conversion Formula

• The formula used is: $Degrees = Radians \times \left( \frac{180}{\pi} \right)$

Conversion Examples

1. a) Convert 4π to Degrees:

   • $4\pi \times \left( \frac{180}{\pi} \right) = 720°$

2. b) Convert 5π/2 to Degrees:

   • $\frac{5\pi}{2} \times \left( \frac{180}{\pi} \right) = 450°$

3. c) Convert 12π/5 to Degrees:

   • $\frac{12\pi}{5} \times \left( \frac{180}{\pi} \right) = 432°$

4. d) Convert 1/10 to Degrees:

   • $\left( \frac{1}{10} \right) \times \left( \frac{180}{\pi} \right) \approx 5.73°$