Real Numbers, Order, and Absolute Value

The Real Line, Number Sets, and Order Rules

• The Real Line:

  • The set of real numbers $ℑ$ forms a continuous line where numbers represent positions.
  • Order relation: $a < b$ means that $a$ lies to the strict left of $b$ on the real line.

• Nested Number Systems:

  • The standard sets of numbers satisfy the inclusion chain: $ℕ \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$.
  • Positive Integers / Natural Numbers ($\mathbb{N}$): $ℕ = \{1, 2, 3, \dots\}$.
  • Integers ($\mathbb{Z}$): $\mathbb{Z} = \{\dots, -1, 0, 1, \dots\}$.
  • Rational Numbers ($\mathbb{Q}$): Numbers that can be expressed in the form $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$.
  • Irrational Numbers ($\mathbb{R} \setminus \mathbb{Q}$): Real numbers that cannot be expressed as a quotient of two integers.

• Set Classification Examples:

  • $-7 \in \mathbb{Z}, \mathbb{Q}, \mathbb{R}$
  • $0 \in \mathbb{Z}, \mathbb{Q}, \mathbb{R}$
  • $\frac{3}{8} \in \mathbb{Q}, \mathbb{R}$
  • $\sqrt{2} \in \mathbb{R} \setminus \mathbb{Q}, \mathbb{R}$
  • $\pi \in \mathbb{R} \setminus \mathbb{Q}, \mathbb{R}$

Decimal Representations and Irrationality Proofs

• Decimal Expansion Characteristics:

  • Every rational number possesses either a terminating decimal expansion or an eventually repeating decimal expansion.
  • Conversely, every terminating or repeating decimal expansion represents a rational number.

• Converting a Repeating Decimal to Rational Form:

  • To convert $x = 0.272727\dots$ into a rational fraction:
    • Multiply $x$ by $100$ to shift one full repeating block past the decimal point: $100x = 27.272727\dots$
    • Subtract the original equation $x = 0.272727\dots$ from $100x = 27.272727\dots$ to eliminate the infinite repeating decimal tail: $99x = 27$
    • Solve for $x$: $x = \frac{27}{99} = \frac{3}{11}$

• Proof that $\sqrt{2}$ is Irrational:
  • Proof Technique: Proof by contradiction.
  • Assume, to the contrary, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ for positive integers $p$ and $q$ that share no common factors (i.e., $\frac{p}{q}$ is in simplest form).
  • Squaring both sides gives $2 = \frac{p^2}{q^2}$, which rearranges to $p^2 = 2q^2$.
  • Since $2q^2$ is even, $p^2$ must be even. Because the square of an odd integer is always odd, $p$ itself must be an even integer.
  • Since $p$ is even, write $p = 2k$ for some positive integer $k$.
  • Substitute $p = 2k$ into $p^2 = 2q^2$: $(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$
  • Since $2k^2$ is even, $q^2$ is even, which implies that $q$ must also be an even integer.
  • This establishes that both $p$ and $q$ are even, meaning they share a common factor of $2$. This contradicts the initial assumption that $p$ and $q$ have no common factor.
  • Therefore, $\sqrt{2}$ cannot be rational; it is irrational.

Inequality Rules and Solving Inequalities

• Equivalence-Preserving Order Rules:

  • Addition: $a < b \iff a + c < b + c$
  • Positive Multiplication: For $c > 0$, $a < b \iff ac < bc$
  • Negative Multiplication: For $c < 0$, $a < b \iff ac > bc$
  • Reciprocals: For positive numbers, taking reciprocals reverses the inequality order: $0 < a < b \implies \frac{1}{a} > \frac{1}{b}$
  • Variable Multiplication Warning: Multiplying an inequality by an algebraic expression whose sign is unknown requires split-case analysis based on whether the expression is strictly positive or strictly negative.

• Worked Example: Linear Double Inequality:

  • Solve: $-3 \le 2 - 5x < 12$
  • Subtract $2$ across all parts: $-5 \le -5x < 10$
  • Divide across all parts by $-5$ (reversing all inequality directions): $1 \ge x > -2$
  • Reorder in increasing sequence: $-2 < x \le 1$

• Worked Example: Quadratic Inequality:

  • Solve: $x^2 - x - 6 \ge 0$
  • Factor into linear terms: $(x - 3)(x + 2) \ge 0$
  • Identify key roots (critical points): $x = -2$ and $x = 3$
  • Analyze interval signs:
    • For $x < -2$: $(x + 2) < 0$, $(x - 3) < 0$ $\implies$ product is positive ($+$).
    • For $-2 < x < 3$: $(x + 2) > 0$, $(x - 3) < 0$ $\implies$ product is negative ($-$).
    • For $x > 3$: $(x + 2) > 0$, $(x - 3) > 0$ $\implies$ product is positive ($+$).

*   The product equals zero at $x = -2$ and $x = 3$.
*   Solution set: $x \le -2$ or $x \ge 3$.

• Worked Example: Rational Inequality:
  • Solve: $\frac{x + 1}{x - 2} \ge 0$
  • Identify critical points and domain bounds: expression is zero at numerator root $x = -1$, and undefined at denominator root $x = 2$
  • Analyze interval signs:
    • For $x < -1$: $(x + 1) < 0$, $(x - 2) < 0$ $\implies$ quotient is positive ($+$).
    • For $-1 < x < 2$: $(x + 1) > 0$, $(x - 2) < 0$ $\implies$ quotient is negative ($-$).
    • For $x > 2$: $(x + 1) > 0$, $(x - 2) > 0$ $\implies$ quotient is positive ($+$).

*   Include $x = -1$ as it yields zero; exclude $x = 2$ because division by zero is undefined.
*   Solution set: $x \le -1$ or $x > 2$.
*   Do not multiply both sides by $(x - 2)$ directly without knowing its sign.

• Inequality Practice Questions:
  1. Classify as rational or irrational: $-\sqrt{49}$, $\sqrt{12}$, $0.6$
     • $-\sqrt{49} = -7 \in \mathbb{Q}$ (rational)
     • $\sqrt{12} = 2\sqrt{3} \notin \mathbb{Q}$ (irrational)
     • $0.6 = \frac{3}{5} \in \mathbb{Q}$ (rational)
  2. Solve $7 - 3x \le 1$:
     • $-3x \le -6 \implies x \ge 2$
  3. Solve $(x - 1)(x + 4) < 0$:
     • Product is negative between roots: $-4 < x < 1$
  4. Explain error in "$x^2 = 9 \implies x = 3$":
     • The statement ignores the negative root $x = -3$. The complete logical deduction is $x^2 = 9 \implies x = \pm 3$ (or $|x| = 3$).

Absolute Value and Distance

• Definition of Absolute Value:
  • For $x \in \mathbb{R}$, the absolute value $|x|$ represents the non-negative distance of $x$ from zero on the real line:         $|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$
  • Evaluations: $|-5| = 5$, $|0| = 0$, $|7| = 7$

• Fundamental Properties:

  • Non-negativity: $|x| \ge 0$
  • Symmetry: $|-x| = |x|$
  • Squaring: $|x|^2 = x^2$
  • Square root relationship: $\sqrt{x^2} = |x|$ (square roots always return the non-negative principal root)

• Validity Analysis of Mathematical Statements:

  • Statement (A): $\sqrt{a^2} = a$
    • Status: False for $a < 0$. Example: If $a = -1$, $\sqrt{(-1)^2} = \sqrt{1} = 1 \neq -1$. The correct identity is $\sqrt{a^2} = |a|$.
  • Statement (B): $|ab| = |a||b|$
    • Status: Always true. The magnitude of a product equals the product of the magnitudes.
  • Statement (C): $a < b \implies a^2 < b^2$
    • Status: False for negative numbers. Example: If $a = -3$ and $b = 2$, then $-3 < 2$, but $(-3)^2 = 9 > 2^2 = 4$. Order is preserved under squaring only when numbers are non-negative.

• Distance Between Numbers:

  • The distance between two numbers $x$ and $a$ on the real line is given by $|x - a|$.
  • Symmetry of distance: $|x - a| = |a - x|$.
  • Open Neighbourhoods: For a radius $r > 0$, the condition $|x - a| < r$ represents all points within distance $r$ of centre $a$:         $|x - a| < r \iff -r < x - a < r \iff a - r < x < a + r$
  • Example: $|x - 3| < 2$ describes all numbers strictly between $1$ and $5$ (centre $3$, radius $2$).

• Absolute Value Inequalities (Worked Examples):

  • Upper Bound Example: Solve $|2x - 3| \le 5$
    • Method 1 (Algebraic): $-5 \le 2x - 3 \le 5 \implies -2 \le 2x \le 8 \implies -1 \le x \le 4$
    • Method 2 (Geometric): $|2x - 3| = 2\left|x - \frac{3}{2}\right| \le 5 \implies \left|x - \frac{3}{2}\right| \le \frac{5}{2}$. Centre is $\frac{3}{2}$, radius is $\frac{5}{2}$, giving interval $[-1, 4]$.
  • Exterior Region Example: Solve $|x + 2| > 3$
    • Distance from centre $-2$ exceeds $3$ on either side:
    • $x + 2 > 3$ or $x + 2 < -3$
    • $x > 1$ or $x < -5$
    • (By contrast, $|x + 2| < 3$ yields the interior interval $-5 < x < 1$).

Triangle Inequalities and Error Analysis

• The Triangle Inequality:

  • For all real numbers $u, v$$|u + v| \le |u| + |v|$
  • Proof:
    • From the definition of absolute value: $-|u| \le u \le |u|$ and $-|v| \le v \le |v|$.
    • Adding these inequalities together yields: $-(|u| + |v|) \le u + v \le |u| + |v|$.
    • This double inequality is equivalent to $|u + v| \le |u| + |v|$.
  • Example demonstrating cancellation: $|3 + (-5)| = |-2| = 2 \le 8 = |3| + |-5|$. Cancellation of opposing signs makes the magnitude of a sum smaller than the sum of magnitudes.

• The Reverse Triangle Inequality:

  • For all real numbers $u, v$$\left| |u| - |v| \right| \le |u - v|$
  • Proof:
    • Write $u = (u - v) + v$. Apply the triangle inequality: $|u| = |(u - v) + v| \le |u - v| + |v| \implies |u| - |v| \le |u - v|$.
    • Interchanging $u$ and $v$ gives |v| - |u| \le |v - u| = |u - v| \implies -(|u| - |v|) \le |u - v|$.\n * Combining both inequalities gives \left| |u| - |v| \right| \le |u - v|.\n * **Interpretation:** Changing a number by a distance of at most d$changes its absolute magnitude by at most$d\n\n* **Controlling a Squared Quantity (Worked Example):**\n * Suppose |x - 2| < 0.1$. How close is$x^2$to$4?\n * Factor the target error expression: |x^2 - 4| = |x - 2||x + 2|\n * Bound the term |x + 2|$: Since$|x - 2| < 0.1 \implies 1.9 < x < 2.1$, adding$2$gives$3.9 < x + 2 < 4.1 \implies |x + 2| < 4.1\n * Multiply factor bounds: |x^2 - 4| < (0.1)(4.1) = 0.41\n * General Technique: Factor a difference, establish explicit numerical bounds on each individual factor, and multiply the resulting bounds.\n\n* **Absolute and Relative Error:**\n * Let a$be the true value and$\tilde{a} be the approximation.\n * \text{Absolute Error} = |\tilde{a} - a|\n * \text{Relative Error} = \frac{|\tilde{a} - a|}{|a|}$(defined for true value$a eq 0)\n * **Worked Example:**\n * If a = 200$and$\tilde{a} = 201$: absolute error is$1$; relative error is$\frac{1}{200} = 0.005 = 0.5\%\n * If a = 2$and$\tilde{a} = 3$: absolute error is$1$; relative error is$\frac{1}{2} = 0.5 = 50\%\n * Relative error is undefined when the true value a = 0\n\n\n\n# Practice Problems, Counterexamples, and Delta Bounding\n\n* **Distance and Error Exercises and Solutions:**\n 1. Solve |3x + 1| < 2\n * -2 < 3x + 1 < 2 \iff -3 < 3x < 1 \iff -1 < x < \frac{1}{3}\n 2. Solve |x - 4| \ge 2\n * x - 4 \le -2$or$x - 4 \ge 2 \iff x \le 2$or$x \ge 6\n 3. Given |a - 10| \le 0.2$and$|b - 4| \le 0.1$, bound$|(a - b) - 6|\n * Rewrite terms: (a - b) - 6 = (a - 10) - (b - 4)\n * Apply triangle inequality: |(a - b) - 6| = |(a - 10) - (b - 4)| \le |a - 10| + |b - 4| \le 0.2 + 0.1 = 0.3\n * This maximum error bound of 0.3$is attainable (e.g., set$a = 10.2$and$b = 3.9\n 4. Find all real x$for which$|x| = -x\n * By the definition of absolute value, |x| = -x$holds for all$x \le 0\n\n* **Common Mathematical Traps and Counterexamples:**\n * **Claim:** |u + v| = |u| + |v| always\n * **Counterexample:** Let u = 1, v = -1$. Then$|1 + (-1)| = 0$, but$|1| + |-1| = 2 \implies 0 eq 2\n * **Claim:** |u - v| = |u| - |v| always\n * **Counterexample:** Let u = 1, v = 3$. Then$|1 - 3| = 2$, but$|1| - |3| = -2 \implies 2 eq -2\n * **Claim:** |x| < 2$means$x < 2 only\n * **Correction:** One must also enforce the lower bound x > -2\n * **Claim:** \sqrt{x^2} = x always\n * **Correction:** The correct general identity is \sqrt{x^2} = |x|\n * **Methodological Rule:** A single counterexample disproves an "always" statement; individual examples alone cannot prove a general statement.\n\n* **Guaranteed Small Variations (Finding \delta Bounds):**\n * **Problem 1:** Choose positive \delta$such that$|x - 3| < \delta \implies |2x - 6| < 0.01\n * Note that |2x - 6| = 2|x - 3|\n * The requirement becomes 2|x - 3| < 0.01 \iff |x - 3| < 0.005\n * Choosing \delta = 0.005 guarantees the condition.\n * **Problem 2:** Choose positive \delta$such that$|x - 3| < \delta \implies |x^2 - 9| < 0.1\n * Factor the expression: |x^2 - 9| = |x - 3||x + 3|\n * Assume an initial restriction \delta \le 1$. Then$|x - 3| < 1 \implies 2 < x < 4 \implies 5 < x + 3 < 7 \implies |x + 3| < 7\n * Then |x^2 - 9| = |x - 3||x + 3| < 7\delta\n * Set 7\delta = 0.1 \implies \delta = \frac{1}{70}$. Since$\frac{1}{70} \le 1$, the choice$\delta = \frac{1}{70}$$ works.