📘 Module 1.3 AP: Absolute Value and Modular Inequalities

calculus
mathematics
course
inequalities
modulus
absolute value
Definitions of absolute value, properties, the triangle inequality, and solving inequalities with absolute value.
Author

Blog do Marcellini

Published

September 9, 2025

← Back to the Course Summary 🎓🧮 · ← Mathematics Courses · ← Mathematics Section

🎯 Previous Post: 👉 1.3 Intervals and Inequalities

🎯 Learning Objectives
  • Define the absolute value (modulus) and interpret it geometrically.
  • Present the fundamental properties of the modulus.
  • Introduce and apply the triangle inequality.
  • Solve inequalities involving absolute value.
  • Consolidate learning with examples and exercises.

1 Definition of Absolute Value

📖 Definition

The absolute value (or modulus) of a real number \(x\) is defined by: \[ |x| = \begin{cases} x, & \text{if } x \ge 0, \\ -x, & \text{if } x < 0. \end{cases} \]

2 Geometric Interpretation

  • The modulus represents the distance from \(x\) to the origin on the real line.
  • Hence, \(|x|\ge 0\) for all \(x \in \mathbb{R}\).

3 Properties of the Absolute Value

📐 Main Properties
  1. \(|x|\ge 0\) and \(|x|=0 \iff x=0\).
  2. \(|xy|=|x|\cdot|y|\).
  3. \(|\frac{x}{y}|=\frac{|x|}{|y|}\), if \(y\ne0\).
  4. \(|x+y|\le |x|+|y|\) (triangle inequality).
  5. \(||x|-|y||\le |x-y|\) (reverse triangle inequality).
  6. \(|x|=\max\{x,-x\}=\sqrt{x^2}\).

  • (1) By definition, \(|x|\) is the distance to the origin. Distances are never negative and vanish only at the origin (\(x=0\)).

  • (2) For \(x,y\in\mathbb{R}\), the sign of a product is the product of the signs, hence
    \[ |xy| = (\pm x)\cdot(\pm y) = |x|\cdot |y|. \]

  • (3) For \(y\ne0\), divide the equality from (2):
    \[ |x/y| = \frac{|x|}{|y|}. \]

  • (4) Triangle inequality. Geometrically, \(|x+y|\) is the distance from \(0\) to \(x+y\), which cannot exceed the sum of the distances from \(x\) and \(y\) to \(0\). Formally, using \(|xy|\le |x||y|\):
    \[ |x+y|^2 = (x+y)^2 = x^2 + 2xy + y^2 \le x^2 + 2|x||y| + y^2 = (|x|+|y|)^2, \] hence \(|x+y|\le |x|+|y|\).

  • (5) Reverse triangle inequality. From the triangle inequality, \(|x|=|(x-y)+y|\le |x-y|+|y|\Rightarrow |x|-|y|\le |x-y|\).
    Swapping \(x\) and \(y\) gives \(|y|-|x|\le |x-y|\).
    Combining both: \(||x|-|y||\le |x-y|\).

  • (6) Characterizations of \(|x|\).
    By cases, \(|x|=x\) if \(x\ge0\) and \(|x|=-x\) if \(x<0\).
    Thus \(|x|=\max\{x,-x\}\). Since \(|x|\ge0\) and \(|x|^2=x^2\), we also have \(|x|=\sqrt{x^2}\).

3.1 📌 Useful equivalences for absolute-value inequalities

If \(a\ge0\), the following hold:

  • \(|x|<a \iff -a<x<a\).
  • \(|x|\le a \iff -a\le x\le a\).
  • \(|x|>a \iff x<-a \ \text{or}\ x>a\).
  • \(|x|\ge a \iff x\le -a \ \text{or}\ x\ge a\).
  • \(|x-c|<r \iff c-r<x<c+r\) (open interval of radius \(r\) centered at \(c\)).

4 Worked Examples

🧮 Example 1

Solve: \[ |x| = 3 \]

By definition:
- If \(x\ge0\), then \(|x|=x=3 \Rightarrow x=3\).
- If \(x<0\), then \(|x|=-x=3 \Rightarrow x=-3\).

✅ Solution: \(\{-3,3\}\).

🧮 Example 2

Solve: \[ |x-2| < 5 \]

\(|x-2|<5 \iff -5 < x-2 < 5 \iff -3 < x < 7\).

✅ Solution: \((-3,7)\).

5 Absolute-Value Inequalities

📖 Definition

An absolute-value inequality is an inequality involving absolute value, such as \[ |f(x)| \ \{<, \le, >, \ge\}\ a, \] with \(a\ge0\). The solution uses the piecewise definition of the modulus.

5.1 🔎 Case breakdown (with \(a\ge0\))

  • Case 1 — less than
    \[ |f(x)|<a \iff -a<f(x)<a. \]

  • Case 2 — less than or equal
    \[ |f(x)|\le a \iff -a\le f(x)\le a. \]

  • Case 3 — greater than
    \[ |f(x)|>a \iff f(x)<-a \ \ \text{or}\ \ f(x)>a. \]

  • Case 4 — greater than or equal
    \[ |f(x)|\ge a \iff f(x)\le -a \ \ \text{or}\ \ f(x)\ge a. \]

🚧 Edge cases with (a=0)
  • \(|f(x)|<0\): no solution.
  • \(|f(x)|\le0\): equivalent to \(f(x)=0\).
  • \(|f(x)|>0\): equivalent to \(f(x)\ne0\).
  • \(|f(x)|\ge0\): always true (for all \(x\)).

5.2 🎯 Centered forms

For \(|x-c| \ \{<,\le,>,\ge\}\ r\) with \(r\ge0\):

  • \(|x-c|<r \iff c-r<x<c+r\) (open interval)
  • \(|x-c|\le r \iff c-r\le x\le c+r\) (closed interval)
  • \(|x-c|>r \iff x<c-r \ \text{or}\ x>c+r\) (outside the interval)
  • \(|x-c|\ge r \iff x\le c-r \ \text{or}\ x\ge c+r\)

5.3 ⚙️ Micro-examples

  • \(|x-2|<5 \iff -5<x-2<5 \iff -3<x<7\).
  • \(|2x+1|\ge3 \iff 2x+1\le-3\ \text{or}\ 2x+1\ge3 \iff x\le-2\ \text{or}\ x\ge1\).

6 Practice Exercises

📝 Exercises
  1. Solve: \(|x+1|\le4\).
  2. Solve: \(|2x-3|>5\).
  3. Solve: \(|x^2-1|<3\).
  4. Check whether \(|x+y|\le |x|+|y|\) for \(x=2, y=-5\).
  1. \(-4\le x+1\le4 \Rightarrow -5\le x\le3\).
  2. \(|2x-3|>5 \Rightarrow 2x-3>5\) or \(2x-3<-5\) ⇒ \(x>4\) or \(x<-1\).
  3. \(|x^2-1|<3 \Rightarrow -3<x^2-1<3 \Rightarrow -2<x^2<4\). Since \(x^2\ge0\), we get \(0\le x^2<4 \Rightarrow -2<x<2\).
  4. \(|2+(-5)|=|-3|=3\) and \(|2|+|{-5}|=2+5=7\). Indeed, \(3\le7\).