Absolute Value, Inequalities, and Intervals

⚡ Key Takeaways

One line: The absolute value $|x|$ measures distance from zero on the number line; inequalities express ranges of values; intervals are the standard notation for those ranges. Key result: $|x| < a \iff -a < x < a$ and $|x| > a \iff x < -a$ or $x > a$ (for $a>0$ ). Where it applies: Solving equations and inequalities, error bounds, domains of functions, distances, optimization constraints. Who/when: Notation $|x|$ introduced by Karl Weierstrass (1841); interval and inequality symbols developed from the 16th–19th centuries.

💡 The Core Insight

$|x|$ is distance, and distance is never negative. Once you read $|x - c|$ as "how far $x$ is from $c$ ," every absolute-value problem becomes a question about which points lie within (or beyond) a given distance — which is exactly what an interval describes.

DefinitionAbsolute Value

For a real number $x$ , the absolute value is defined piecewise:

1

\quad |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

Derivation hint: When $x<0$ , the quantity $-x$ is positive, so $|x|$ is always $\ge 0$ . An equivalent algebraic definition uses the square root:

2

\quad |x| = \sqrt{x^2}

Derivation hint: $\sqrt{x^2}$ returns the nonnegative root, which equals $x$ for $x\ge0$ and $-x$ for $x<0$ . Here $x$ is any real number and $|x|$ is its nonnegative magnitude (no units in pure mathematics; in applications it inherits the unit of $x$ ).

DefinitionInterval Notation

Given real numbers $a < b$ , intervals denote connected sets of real numbers:

$(a,b) = \{x : a < x < b\}$ — open, endpoints excluded
$[a,b] = \{x : a \le x \le b\}$ — closed, endpoints included
$[a,b) = \{x : a \le x < b\}$ and $(a,b]$ — half-open
$(a,\infty) = \{x : x > a\}$ , $(-\infty,b] = \{x : x \le b\}$ — unbounded

A round bracket or $\infty$ symbol never includes its endpoint; a square bracket does.

TheoremAbsolute Value Inequalities

Let $a > 0$ . Then:

3

\quad |x| < a \iff -a < x < a \iff x \in (-a, a)

Derivation hint: $|x|<a$ means the distance from $x$ to $0$ is less than $a$ , capturing all points within $a$ of the origin.

4

\quad |x| > a \iff x < -a \ \text{ or } \ x > a \iff x \in (-\infty,-a)\cup(a,\infty)

Derivation hint: distance exceeding $a$ forces $x$ beyond either endpoint; the solution splits into two rays. More generally, for a center $c$ and radius $r>0$ :

5

\quad |x - c| < r \iff c - r < x < c + r \iff x \in (c-r,\, c+r)

Derivation hint: $|x-c|$ is the distance from $x$ to $c$ ; "within $r$ " gives a symmetric interval centered at $c$ .

TheoremKey Algebraic Properties

For all real $x,y$ :

6

\quad |xy| = |x|\,|y|, \qquad \left|\frac{x}{y}\right| = \frac{|x|}{|y|}\ (y\ne0)

Derivation hint: apply $|x|=\sqrt{x^2}$ and the multiplicativity of square roots.

7

\quad |x+y| \le |x| + |y| \quad\text{(Triangle Inequality)}

Derivation hint: square both sides; the cross term $2xy \le 2|x||y|$ since $xy \le |xy|$ .

8

\quad \big||x| - |y|\big| \le |x - y| \quad\text{(Reverse Triangle Inequality)}

Derivation hint: write $|x| = |(x-y)+y| \le |x-y|+|y|$ and rearrange; repeat with roles swapped.

ProofTriangle Inequality

We prove $|x+y| \le |x|+|y|$ for all real $x,y$ . 1. Both sides are nonnegative, so the inequality holds iff it holds after squaring. 2. Left side squared: $(x+y)^2 = x^2 + 2xy + y^2$ . 3. Right side squared: $(|x|+|y|)^2 = x^2 + 2|x||y| + y^2$ (using $|x|^2=x^2$ ). 4. Subtracting, the inequality $|x+y|^2 \le (|x|+|y|)^2$ reduces to $2xy \le 2|x||y|$ , i.e. $xy \le |xy|$ . 5. But $xy \le |xy|$ is always true, since any real number is $\le$ its absolute value. 6. Therefore $|x+y|^2 \le (|x|+|y|)^2$ , and taking nonnegative square roots gives $|x+y|\le|x|+|y|$ . $\blacksquare$

∎

ProofThe "

<

" Inequality Equivalence, Eq. 3

We show $|x| < a \iff -a < x < a$ for $a>0$ . 1. ( $\Rightarrow$ ) Suppose $|x|<a$ . If $x\ge0$ then $|x|=x$ , so $x<a$ ; also $x\ge0>-a$ , giving $-a<x<a$ . If $x<0$ then $|x|=-x<a$ , so $x>-a$ ; also $x<0<a$ , giving $-a<x<a$ . 2. ( $\Leftarrow$ ) Suppose $-a<x<a$ . If $x\ge0$ , then $|x|=x<a$ . If $x<0$ , then $|x|=-x$ , and from $-a<x$ we get $-x<a$ , so $|x|<a$ . 3. Both directions hold, establishing the equivalence. $\blacksquare$

∎

mathematician · Karl Weierstrass | 1815–1897

The German analyst Karl Weierstrass introduced the notation $|x|$ for absolute value in 1841 while building the rigorous "epsilon-delta" foundations of calculus. His framework recast continuity and limits in the language $|f(x)-L|<\varepsilon$ whenever $|x-c|<\delta$ — making absolute-value inequalities the literal grammar of modern analysis. Weierstrass's insistence on precise inequality-based definitions ended centuries of vague appeals to "infinitely small" quantities.

🧠 Intuition

Picture the number line as a ruler with $0$ at the center. $|x|$ asks: how many units from the start am I? — and direction is irrelevant. So $|x|=5$ has two answers, $x=5$ and $x=-5$ , because two points sit five units from zero. Now slide the center: $|x-3|$ measures distance from $3$ . The inequality $|x-3|<2$ asks for all points within 2 units of 3, which is the interval $(1,5)$ — center $3$ , reaching out $2$ in each direction. The " $>$ " case flips this: $|x-3|>2$ wants points farther than 2 units from 3, leaving everything outside $(1,5)$ , i.e. $(-\infty,1)\cup(5,\infty)$ . The one habit that prevents nearly all errors: never solve absolute-value problems by guessing — convert to distance or to a double inequality first.

Figure

Number-line picture of

|x-3|<2

vs

|x-3|>2

ExampleSolving a "less-than" inequality

Given: Solve $|2x - 1| \le 7$ .

Steps: 1. The " $\le$ " form means: $-7 \le 2x - 1 \le 7$ (apply Eq. 3 with $\le$ ). 2. Add $1$ to all three parts: $-6 \le 2x \le 8$ . 3. Divide by $2$ : $-3 \le x \le 4$ . 4. Endpoints are included because the inequality is non-strict.

\boxed{x \in [-3,\, 4]}

Check: At $x=4$ : $|2(4)-1|=|7|=7\le7$ ✓. At $x=-3$ : $|-7|=7\le7$ ✓.

ExampleSolving a "greater-than" inequality with a tolerance interpretation

Given: A machined rod must have length $L$ within $0.05$ cm of the target $12.00$ cm. (a) Express the acceptable lengths as an absolute-value inequality and an interval. (b) A second rod fails inspection because $|L-12.00| > 0.05$ ; describe its lengths.

Steps: 1. "Within $0.05$ of $12.00$ ": $|L - 12.00| \le 0.05$ (Eq. 5 with $\le$ ). 2. Expand: $12.00 - 0.05 \le L \le 12.00 + 0.05$ , so $11.95 \le L \le 12.05$ .

\boxed{L \in [11.95,\ 12.05]\ \text{cm (acceptable)}}

3. For the failing rod, $|L-12.00|>0.05$ gives $L<11.95$ or $L>12.05$ :

\boxed{L \in (-\infty,\,11.95)\cup(12.05,\,\infty)\ \text{cm, restricted to }L>0}

Check: $L=12.00$ gives $|0|=0\le0.05$ ✓ (acceptable). $L=12.10$ gives $|0.10|>0.05$ ✓ (rejected).

⚠️ Common mistake

Splitting the wrong way. $|x| < a$ becomes a single double-inequality $-a<x<a$ (an "and"), while $|x|>a$ becomes two separate inequalities joined by "or." Students frequently write $-a > x > a$ for the "greater" case, which is impossible (no number is both less than $-a$ and greater than $a$ ). The " $>$ " case is never a single interval — it is always a union.

Negative right-hand side. If $a<0$ : the inequality $|x|<a$ has no solution (distance can't be negative), while $|x|>a$ is satisfied by all real numbers. Never blindly apply Eqs. 3–4 when $a\le0$ .

Forgetting to isolate first. In $3|x-2|+1 \le 10$ , you must isolate the absolute value ( $|x-2|\le3$ ) before splitting.

Counterexample

Claim (false): " $|x+y| = |x| + |y|$ always."

Counterexample: Take $x=5$ , $y=-3$ . Then $|x+y| = |2| = 2$ , but $|x|+|y| = 5+3 = 8$ . So $2 \ne 8$ . Equality in the triangle inequality holds only when $x$ and $y$ have the same sign (or one is zero); otherwise cancellation makes the left side strictly smaller.

⚠️ Common mistake

Squaring inequalities carelessly. From $|x|<a$ you may square to $x^2<a^2$ only because $a>0$ . But $x < 5$ does not imply $x^2 < 25$ (try $x=-10$ ). Squaring is safe only when both sides are known nonnegative — exactly the situation absolute values guarantee.

connection

Absolute value and intervals thread through nearly every later topic. The definition $|x|=\sqrt{x^2}$ links to Square Roots and Radicals, and the piecewise definition is a first encounter with Piecewise Functions. Inequality solution sets feed directly into Domain and Range of Functions, since domains are typically described by inequalities. The distance interpretation generalizes to the Distance Formula in the plane and ultimately to Metric Spaces in advanced analysis. Most importantly, the inequality $|f(x)-L|<\varepsilon$ is the backbone of Limits and Continuity, where Weierstrass's notation becomes indispensable.

🔬 Where this appears

⏳ History

- 1557 — Robert Recorde introduces the equals sign "

=

" in The Whetstone of Witte, the ancestor of relational symbols.

- 1631 — Thomas Harriot's posthumous work popularizes the inequality signs "

<

" and "

>

".

- 1734 — Pierre Bouguer introduces "

\le

" and "

\ge

" (the symbols with the bar underneath).

- 1821 — Augustin-Louis Cauchy uses inequality-based reasoning extensively in Cours d'analyse, foreshadowing rigorous limits.

- 1841 — Karl Weierstrass introduces the notation

|x|

for absolute value (then called "numerischer Wert" / numerical value).

- 1854 — The triangle inequality is formalized in the study of metric/normed quantities (geometry context far older, dating to Euclid's Elements, Book I, Prop. 20).

- 1872 — Weierstrass and Heine fully cement the

\varepsilon

–

\delta

definition, making

|f(x)-L|<\varepsilon

standard.

- 1906 — Maurice Fréchet generalizes distance to metric spaces, abstracting

|x-y|

into a general distance function

d(x,y)

.

- 20th century — Interval notation with round/square brackets becomes the international standard (with the French convention

]a,b[

for open intervals still in use today).

open-problem

The absolute value is not differentiable at $x=0$ — its graph has a sharp corner there, so $|x|$ has no well-defined slope at the origin. This single non-smooth point launches deep questions: optimization with absolute-value (L1) penalties is non-differentiable, requiring subgradient methods rather than ordinary calculus, and these underlie modern techniques like LASSO regression and compressed sensing. Generalizing $|x|$ to a norm $\|\mathbf{v}\|$ on vectors raises the question of which norms (L1, L2, L∞, ...) best capture "size" — a choice with no universal answer and active consequences in machine learning, where the geometry of the chosen norm shapes which solutions an algorithm prefers. Whether sparse (L1-favoring) representations are always recoverable is governed by subtle conditions still refined in current research.

Absolute Value, Inequalities, and Intervals

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also