Let us assume there are two distinct additive identities, $e_1$ and $e_2$, where $e_1 \neq e_2$.

By the definition of an additive identity, for any element $a \in \mathbb{R}$:
$a + e_1 = a$ for all $a$
$a + e_2 = a$ for all $a$

Set $a = e_2$ in the first equation: $e_2 + e_1 = e_2$ ... (1)
Set $a = e_1$ in the second equation: $e_1 + e_2 = e_1$ ... (2)

By the Commutative Law of Addition: $e_2 + e_1 = e_1 + e_2$

From (1) and (2): $e_2 = e_1$

This contradicts our assumption that $e_1 \neq e_2$. Therefore, there is exactly one additive identity, which we denote as 0.

Using the property $|A| < a \iff -a < A < a$ (where $a > 0$):

$|3x + 2| < 1 \iff -1 < 3x + 2 < 1$

Subtract 2 from all three parts:
$-1 - 2 < 3x + 2 - 2 < 1 - 2$
$-3 < 3x < -1$

Divide all parts by 3 (positive number, so inequality signs stay the same):
$\frac{-3}{3} < \frac{3x}{3} < \frac{-1}{3}$
$-1 < x < -\frac{1}{3}$

In interval notation: $(-1, -\frac{1}{3})$

On a number line: Draw hollow circles at $-1$ and $-\frac{1}{3}$ (since the inequality is strict), and shade the line segment between them.

To convert a double inequality into absolute value form, we find the midpoint and half-width.

Midpoint: $m = \frac{-4 + (-1)}{2} = \frac{-5}{2} = -2.5$

Half-width (radius): $r = \frac{-1 - (-4)}{2} = \frac{3}{2} = 1.5$

The interval $-4 \leq x \leq -1$ means $x$ is at distance 1.5 from the center point -2.5.

Using the property $|x - m| \leq r$ (where $m$ is midpoint and $r$ is half-width):

$|x - (-2.5)| \leq 1.5$
$|x + 2.5| \leq 1.5$

To eliminate decimals, multiply both the inequality and the center by 2:
$|2x + 5| \leq 3$

Definition: The Rational Density Property states that between any two distinct real numbers, there exist infinitely many rational numbers.

Proof: Given $a < b$, we need to show that $a < \frac{a+b}{2} < b$.

Add $a$ to both sides of $a < b$:
$a + a < a + b$
$2a < a + b$

Divide by 2:
$a < \frac{a+b}{2}$ ... (1)

Add $b$ to both sides of $a < b$:
$a + b < b + b$
$a + b < 2b$

Divide by 2:
$\frac{a+b}{2} < b$ ... (2)

From (1) and (2): $a < \frac{a+b}{2} < b$

When both $a$ and $b$ are rational, their sum is rational, and dividing by 2 preserves rationality. Thus, $\frac{a+b}{2}$ is a rational number strictly between $a$ and $b$.

First, factor the quadratic expression.

$x^2 - 3x + 2 = 0$
$(x - 1)(x - 2) = 0$

So we need $(x - 1)(x - 2) < 0$.

Critical points are $x = 1$ and $x = 2$. These divide the number line into three intervals.

Construct a sign table:

| Interval | $(x-1)$ | $(x-2)$ | Product |
| :--- | :--- | :--- | :--- |
| $(-\infty, 1)$ | Negative | Negative | Positive |
| $(1, 2)$ | Positive | Negative | Negative |
| $(2, \infty)$ | Positive | Positive | Positive |

We need the product to be less than zero (negative). This occurs only in the interval $(1, 2)$.

Therefore, the solution is $(1, 2)$.

$A - B$ consists of all elements in $A$ that are NOT in $B$.

$A = [-3, 3)$ covers: all real numbers from $-3$ (included) to just before $3$.
$B = [2, 4)$ covers: all real numbers from $2$ (included) to just before $4$.

Since $B$ begins at 2, we remove everything from 2 onwards (up to 3) from set $A$.
What remains is everything from $-3$ up to (but not including) 2.

$A - B = [-3, 2)$

Graph: Draw a solid dot at $-3$ (included) and a hollow circle at $2$ (excluded), with the line segment shaded between them.

Rational Numbers: Numbers that can be expressed as $\frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \neq 0$. Their decimal representations are either terminating or repeating.
Examples:
- $\frac{1}{2} = 0.5$ (terminating decimal)
- $\frac{1}{3} = 0.333...$ (repeating decimal)

Irrational Numbers: Numbers that cannot be expressed as $\frac{p}{q}$. Their decimal representations are non-terminating and non-repeating.
Examples:
- $\sqrt{2} \approx 1.41421356...$ (non-terminating, non-repeating)
- $\pi \approx 3.14159265...$ (non-terminating, non-repeating)

Trichotomy Law: For any two real numbers $a$ and $b$, exactly one of the following is true: $a < b$, $a = b$, or $a > b$. These three relationships are mutually exclusive and exhaustive; no other relationship is possible.

Transitive Law: If $a < b$ and $b < c$, then $a < c$. This property allows us to 'chain' inequalities together. For example, if $2 < 5$ and $5 < 10$, then $2 < 10$ by transitivity.

We have a compound inequality with two inequality signs. We must perform the same operations on all three parts.

Given: $9 > 3 - 2x > 5$

Subtract 3 from all parts:
$9 - 3 > 3 - 2x - 3 > 5 - 3$
$6 > -2x > 2$

Divide all parts by $-2$. CRITICAL: Since we divide by a negative number, we MUST reverse both inequality signs:
$\frac{6}{-2} < \frac{-2x}{-2} < \frac{2}{-2}$
$-3 < x < -1$

In interval notation: $(-3, -1)$

On a number line: Place hollow circles at both $-3$ and $-1$ (strict inequalities), and shade the segment between them.

We use the fundamental property that $|a|^2 = a^2$ for any real number $a$.

Consider the left-hand side squared:
$|xy|^2 = (xy)^2 = x^2 y^2$

Using the property $|a|^2 = a^2$, we can write:
$x^2 = |x|^2$ and $y^2 = |y|^2$

Therefore:
$|xy|^2 = |x|^2 \cdot |y|^2 = (|x| \cdot |y|)^2$

Taking the non-negative square root of both sides:
$|xy| = |x| \cdot |y|$

This completes the proof.

Definition of Absolute Value: For any real number $x$, the absolute value $|x|$ is defined as:
$$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$
Geometrically, $|x|$ represents the distance of $x$ from zero on the number line. Distance is always non-negative.

Triangle Inequality Theorem: For any real numbers $x$ and $y$:
$$|x + y| \leq |x| + |y|$$

Proof:

Step 1: We know that for any real number $a$: $|a|^2 = a^2$.

Step 2: Consider the square of the left-hand side:
$$|x+y|^2 = (x+y)^2 = x^2 + 2xy + y^2$$

Step 3: Using the substitution $x^2 = |x|^2$ and $y^2 = |y|^2$:
$$|x+y|^2 = |x|^2 + 2xy + |y|^2$$

Step 4: Now we need to bound the middle term $2xy$. The key insight is that $xy \leq |x||y|$ (since the product of absolute values is always at least as large as the algebraic product).
Therefore: $2xy \leq 2|x||y|$

Step 5: Substituting this into our expression:
$$|x+y|^2 \leq |x|^2 + 2|x||y| + |y|^2$$

Step 6: The right-hand side is a perfect square:
$$|x+y|^2 \leq (|x| + |y|)^2$$

Step 7: Taking the non-negative square root of both sides (since both sides are non-negative):
$$|x+y| \leq |x| + |y|$$

This completes the proof of the Triangle Inequality.

Finding $A \cup B$ (Union - all elements in at least one set):

Set $A = [-2, 4)$ covers all real numbers from $-2$ (included, solid dot) to $4$ (excluded, hollow circle).
Set $B = (2, 5]$ covers all real numbers from $2$ (excluded, hollow circle) to $5$ (included, solid dot).

The union combines both intervals. The leftmost point is $-2$ and the rightmost point is $5$. Since $A$ includes $-2$ and $B$ includes $5$:
$$A \cup B = [-2, 5]$$

Graph: Solid dot at $-2$, solid dot at $5$, entire line shaded between them.

Finding $A \cap B$ (Intersection - elements in both sets):

For an element to be in the intersection, it must satisfy BOTH $x \in A$ AND $x \in B$.
- From $A$: $-2 \leq x < 4$
- From $B$: $2 < x \leq 5$

The overlap requires $x > 2$ (from $B$'s lower bound) and $x < 4$ (from $A$'s upper bound). Both boundaries are excluded:
$$A \cap B = (2, 4)$$

Graph: Hollow circle at $2$, hollow circle at $4$, line shaded between them.

Finding $B - A$ (Difference - elements in B but not in A):

We start with all of $B = (2, 5]$ and remove anything that belongs to $A = [-2, 4)$.
- $A$ extends from $-2$ to $4$ (not including 4)
- $B$ starts at $2$ (not including 2) and goes to $5$ (including 5)

The part of $B$ that overlaps with $A$ is $(2, 4)$. Removing this from $B$ leaves the part from $4$ to $5$:
$$B - A = [4, 5]$$

Graph: Solid dot at $4$, solid dot at $5$, line shaded between them.

To solve a rational inequality, we identify critical points and create a sign table.

Step 1: Find critical points

Critical points come from two sources:
- Numerator zeros: $x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$
- Denominator zeros: $x + 1 = 0 \Rightarrow x = -1$ (undefined point, NOT a solution)

Critical points in order: $x = -1, 0, 2$

Step 2: Create a sign table for the intervals

The critical points divide the real line into four intervals: $(-\infty, -1)$, $(-1, 0)$, $(0, 2)$, $(2, \infty)$

| Interval | $x$ | $(x-2)$ | $(x+1)$ | $\frac{x(x-2)}{x+1}$ |
| :--- | :--- | :--- | :--- | :--- |
| $(-\infty, -1)$ | $-$ | $-$ | $-$ | $\frac{(-)(-)}{{-}} = -$ |
| $(-1, 0)$ | $-$ | $-$ | $+$ | $\frac{(-)(-)}{{+}} = +$ |
| $(0, 2)$ | $+$ | $-$ | $+$ | $\frac{(+)(-)}{{+}} = -$ |
| $(2, \infty)$ | $+$ | $+$ | $+$ | $\frac{(+)(+)}{{+}} = +$ |

Step 3: Identify where the expression is $\leq 0$

We need values where $\frac{x(x-2)}{x+1} \leq 0$:
- In $(-\infty, -1)$: expression is negative ✓
- At $x = -1$: undefined (excluded)
- In $(-1, 0)$: expression is positive ✗
- At $x = 0$: expression equals 0 ✓ (can include)
- In $(0, 2)$: expression is negative ✓
- At $x = 2$: expression equals 0 ✓ (can include)
- In $(2, \infty)$: expression is positive ✗

Step 4: Combine intervals

$$\text{Solution: } (-\infty, -1) \cup [0, 2]$$

Note: The interval at $x = -1$ is never included because the function is undefined there.

We use the Triangle Inequality strategically to establish this fundamental property.

Proof:

Recall the Triangle Inequality: For any real numbers $a$ and $b$,
$$|a + b| \leq |a| + |b|$$

Step 1: Express $|x|$ cleverly as a sum using $y$:
$$|x| = |(x - y) + y|$$

Step 2: Apply the Triangle Inequality to this sum:
$$|x| = |(x-y) + y| \leq |x-y| + |y|$$

Step 3: Subtract $|y|$ from both sides:
$$|x| - |y| \leq |x-y|$$ ... (Inequality 1)

Step 4: Similarly, express $|y|$ as a sum using $x$:
$$|y| = |(y - x) + x| = |-(x-y) + x|$$

Step 5: Apply the Triangle Inequality:
$$|y| \leq |-(x-y)| + |x| = |x-y| + |x|$$
(Note: $|-(x-y)| = |x-y|$ because absolute value ignores sign)

Step 6: Subtract $|x|$ from both sides:
$$|y| - |x| \leq |x-y|$$ ... (Inequality 2)

Step 7: Combine Inequalities 1 and 2

From (1): $|x| - |y| \leq |x-y|$
From (2): $|y| - |x| \leq |x-y|$ (equivalently: $-(|x| - |y|) \leq |x-y|$)

By the definition of absolute value, $||x| - |y||$ equals either $(|x| - |y|)$ or $(|y| - |x|)$, depending on which is larger. Since both are $\leq |x-y|$:

$$||x| - |y|| \leq |x-y|$$

Rearranging:
$$|x-y| \geq ||x| - |y||$$

This completes the proof.

We use the absolute value inequality property: $|A| < a \iff -a < A < a$ (where $a > 0$).

Step 1: Apply the absolute value property

Given: $|5 - \frac{2}{x}| < 1$

This is equivalent to:
$$-1 < 5 - \frac{2}{x} < 1$$

Step 2: Subtract 5 from all parts

$$-1 - 5 < -\frac{2}{x} < 1 - 5$$
$$-6 < -\frac{2}{x} < -4$$

Step 3: Multiply all parts by $-1$ (reverse inequality signs!)

$$6 > \frac{2}{x} > 4$$

Which can be rewritten as:
$$4 < \frac{2}{x} < 6$$

Step 4: Take reciprocals (reverse inequality signs again!)

When we take reciprocals of positive numbers, the inequality signs reverse:
$$\frac{1}{4} > \frac{x}{2} > \frac{1}{6}$$

Rewriting in standard order:
$$\frac{1}{6} < \frac{x}{2} < \frac{1}{4}$$

Step 5: Multiply all parts by 2

$$\frac{2}{6} < x < \frac{2}{4}$$
$$\frac{1}{3} < x < \frac{1}{2}$$

In interval notation: $(\frac{1}{3}, \frac{1}{2})$

Graph on number line:

Place hollow circles (open endpoints) at $\frac{1}{3} \approx 0.333$ and $\frac{1}{2} = 0.5$. Shade the line segment connecting them.

Theorem: If $a$ and $b$ are distinct rational numbers with $a < b$, then there exists a rational number $c$ such that $a < c < b$.

Proof:

Since $a$ and $b$ are rational, they can be expressed as:
$$a = \frac{p_1}{q_1} \text{ and } b = \frac{p_2}{q_2}$$
where $p_1, q_1, p_2, q_2 \in \mathbb{Z}$ with $q_1 \neq 0$ and $q_2 \neq 0$.

Consider their arithmetic mean:
$$c = \frac{a + b}{2} = \frac{\frac{p_1}{q_1} + \frac{p_2}{q_2}}{2}$$

Simplifying:
$$c = \frac{p_1 q_2 + p_2 q_1}{2 q_1 q_2}$$

The numerator $p_1 q_2 + p_2 q_1$ is an integer (sum of products of integers).
The denominator $2 q_1 q_2$ is a non-zero integer (product of non-zero integers).
Therefore, $c$ is a rational number by definition.

To verify that $c$ lies strictly between $a$ and $b$:

Given $a < b$, adding $a$ to both sides: $2a < a + b$
Dividing by 2: $a < \frac{a+b}{2} = c$ ... (1)

Adding $b$ to both sides of $a < b$: $a + b < 2b$
Dividing by 2: $\frac{a+b}{2} = c < b$ ... (2)

From (1) and (2): $a < c < b$ ✓

Rational Density in the Real Number System:

By repeatedly applying this theorem, we can construct infinitely many rationals between any two rationals:
- Between $a$ and $b$, we found $c_1 = \frac{a+b}{2}$
- Between $a$ and $c_1$, we find $c_2 = \frac{a+c_1}{2}$
- Between $c_1$ and $b$, we find $c_3 = \frac{c_1+b}{2}$
- This process continues infinitely...

This means that in any arbitrarily small neighborhood on the number line, there are infinitely many rational numbers. We say that the rational numbers are dense in $\mathbb{R}$.

However, despite this density, the rationals do not 'fill' the entire real line. The irrational numbers occupy the 'gaps', and there are actually more irrational numbers than rational ones (in the sense of cardinality). Together, rationals and irrationals form the complete real number system with no 'holes' or gaps.

We prove that the left-hand side (LHS) equals the right-hand side (RHS) by applying set theory laws.

Method: Algebraic Simplification

Starting with LHS: $(A \cap B) \cup (A \cap \overline{B})$

Step 1: Factor out the common term $A$ using the Distributive Law (Reverse Direction).

Just as in algebra $ab + ac = a(b + c)$, in set theory: $(A \cap B) \cup (A \cap \overline{B}) = A \cap (B \cup \overline{B})$

Step 2: Apply the Complement Law.

We know that $B \cup \overline{B} = U$ (the Universal Set), because every element is either in $B$ or in its complement.

Therefore: $A \cap (B \cup \overline{B}) = A \cap U$

Step 3: Apply the Identity Law for Intersection.

The universal set $U$ is the identity element for intersection. Any set intersected with $U$ returns itself:
$A \cap U = A$

Step 4: Conclusion

We have shown: $(A \cap B) \cup (A \cap \overline{B}) = A$ = RHS

Therefore, the identity is proven. ✓

Intuitive Explanation:

Imagine set $A$ as a region. We partition $A$ into two non-overlapping parts:
- $(A \cap B)$: the part of $A$ that overlaps with $B$
- $(A \cap \overline{B})$: the part of $A$ that does NOT overlap with $B$ (inside the complement)

When we take the union of these two parts, we recover the entire set $A$.