Absolute value and its properties – Operations on the Set of Real Numbers (Senior 4 Mathematics) | Ganzaa

Operations on the Set of Real Numbers

Lessons

Senior 4 · Mathematics

Operations on the Set of Real Numbers

5 lessons

Lesson 1 of 5

Absolute value and its properties

25 min

How far, not which way

A workshop cuts steel pins that should be $8.00\,\mathrm{mm}$ thick. No machine is perfect, so the drawing allows an error of $0.05\,\mathrm{mm}$ either way. A pin $7.97\,\mathrm{mm}$ thick is $0.03\,\mathrm{mm}$ too thin; one $8.06\,\mathrm{mm}$ thick is $0.06\,\mathrm{mm}$ too thick. The inspector does not care about the direction of the error, only its size. The number that measures a size without a sign is the absolute value, and the inspector's rule is one line:

\text{accept a pin of thickness } d \iff |d - 8.00| \le 0.05 .

In S1 you met $∣ x ∣$ as "the number without its sign". At A-level you need more: a definition that works for letters as well as numbers, its link with distance, its algebraic properties, and a method for equations that contain it.

Definition

Absolute value: for a real number $x$ ,

|x| = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x \lt 0. \end{cases}

Notation: $∣ x ∣$ is read "the absolute value of $x$ " or "the modulus of $x$ ".

Examples: $∣ 9 ∣ = 9$ , $∣ 0 ∣ = 0$ , $∣ - 4.5 ∣ = - (- 4.5) = 4.5$ , and $|3 - \pi| = -(3 - \pi) = \pi - 3$ because $3 - \pi \lt 0$ .

Warning

$- x$ is not always negative. When $x \lt 0$ , the number $- x$ is positive: if $x = - 6$ , then $- x = 6$ . That is why the second line of the definition gives a positive result. A common error is to write $∣ x - 5 ∣ = x + 5$ "to make it positive"; the correct rule depends on the sign of $x - 5$ : it is $x - 5$ when $x \ge 5$ and $5 - x$ when $x \lt 5$ .

Absolute value is distance

On the number line, $∣ x ∣$ is the distance from the point $x$ to 0, and for any two real numbers $a$ and $b$

|a - b| = \text{the distance between } a \text{ and } b .

Distance does not depend on the order, so $∣ a - b ∣ = ∣ b - a ∣$ . In the diagram, $- 3$ and $4$ are $∣ 4 - (- 3) ∣ = ∣ 7 ∣ = 7$ units apart, and $- 3$ is $3$ units from 0.

A second link you will use in every unit that follows: a square root is never negative, so

\sqrt{x^2} = |x| \quad \text{for every real } x .

For example, $\sqrt{(-7)^2} = \sqrt{49} = 7 = |-7|$ , not $- 7$ .

Worked example 1: removing the bars

Write without absolute value signs: (a) $|\sqrt{2} - 1.5|$ (b) $∣ x - 5 ∣$ given that $x \lt 5$ (c) $\sqrt{(1 - x)^2}$ given that $x \gt 1$ .

Solution

$\sqrt{2} \approx 1.414$ , so $\sqrt{2} - 1.5 \lt 0$ and $|\sqrt{2} - 1.5| = 1.5 - \sqrt{2}$ .
$x \lt 5$ means $x - 5 \lt 0$ , so $∣ x - 5 ∣ = - (x - 5) = 5 - x$ .
$\sqrt{(1 - x)^2} = |1 - x|$ . Since $x \gt 1$ , $1 - x \lt 0$ , so the answer is $- (1 - x) = x - 1$ .

In each part the method is the same: decide the sign of the expression inside the bars, then apply the definition.

The properties of absolute value

For all real numbers $x$ and $y$ :

Property	Statement	Why it holds
Non-negative	$\|x\| \ge 0$ , and $\|x\| = 0 \iff x = 0$	A distance is never negative, and only 0 is at distance 0 from 0.
Symmetry	$∣ - x ∣ = ∣ x ∣$	$x$ and $- x$ are the same distance from 0, on opposite sides.
Product	$\|xy\| = \|x\|\,\|y\|$	$\|xy\| = \sqrt{x^2y^2} = \sqrt{x^2}\,\sqrt{y^2} = \|x\|\,\|y\|$ .
Quotient	$\left\|\frac{x}{y}\right\| = \frac{\|x\|}{\|y\|}$ , $y \ne 0$	Apply the product rule to $\frac{x}{y} \cdot y = x$ .
Square	$∣ x ∣^{2} = x^{2}$	Both equal $x \cdot x$ or $(- x) (- x)$ .
Bounds	$-\|x\| \le x \le \|x\|$	$x$ is either $∣ x ∣$ or $- ∣ x ∣$ .
Triangle inequality	$\|x + y\| \le \|x\| + \|y\|$	Proved below.

There is no sum rule: $∣ x + y ∣$ and $∣ x ∣ + ∣ y ∣$ are different in general. Try $x = 3$ , $y = - 5$ : $∣ x + y ∣ = 2$ but $∣ x ∣ + ∣ y ∣ = 8$ .

Important

Theorem (triangle inequality). For all real $x$ and $y$ ,

|x + y| \le |x| + |y|,

with equality exactly when $xy \ge 0$ (that is, when $x$ and $y$ do not have opposite signs).

Proof. Both sides are non-negative, so it is enough to compare their squares:

(|x| + |y|)^2 - |x + y|^2 = x^2 + 2|x|\,|y| + y^2 - (x^2 + 2xy + y^2) = 2\big(|xy| - xy\big) \ge 0,

since $xy \le |xy|$ by the bounds property. The difference is 0 exactly when $x y = ∣ x y ∣$ , that is, when $xy \ge 0$ . $\blacksquare$

The name comes from geometry: one side of a triangle is never longer than the other two together.

Worked example 2: using the properties

(a) Simplify $\left|\frac{-6x^2}{3}\right|$ . (b) Given $∣ a ∣ = 4$ and $∣ b ∣ = 7$ , find the greatest and the least possible values of $∣ a + b ∣$ .

Solution

$\left|\frac{-6x^2}{3}\right| = |-2x^2| = |-2|\,|x^2| = 2x^2$ , because $x^2 \ge 0$ .
By the triangle inequality, $|a + b| \le 4 + 7 = 11$ , with equality when $a$ and $b$ have the same sign ( $a = 4, b = 7$ ). For the least value, $|b| = |(a + b) - a| \le |a + b| + |a|$ , so $|a + b| \ge 7 - 4 = 3$ , reached when the signs are opposite ( $a = 4, b = - 7$ ). So $3 \le |a + b| \le 11$ .

Equations with absolute values

Read $∣ x - 2 ∣ = 5$ as "the distance from $x$ to 2 is 5". There are two such points, one on each side of 2:

In general, for a real number $r$ :

if $r \gt 0$ : $|A| = r \iff A = r$ or $A = - r$ ;
if $r = 0$ : $|A| = 0 \iff A = 0$ ;
if $r \lt 0$ : $∣ A ∣ = r$ has no solution.

Two more forms appear in exams. When both sides are absolute values, $|A| = |B| \iff A = B$ or $A = - B$ . When the right-hand side contains the unknown, $∣ A ∣ = B$ needs the condition $B \ge 0$ , and every candidate must be checked against it.

Worked example 3: two absolute values

Solve $∣ x + 3 ∣ = ∣ 2 x - 1 ∣$ .

Solution

Either $x + 3 = 2 x - 1$ , giving $x = 4$ ; or $x + 3 = - (2 x - 1)$ , giving $3 x = - 2$ and $x = -\frac{2}{3}$ .

Check: $x = 4$ : $∣ 7 ∣ = 7$ and $∣ 7 ∣ = 7$ . ✓ $x = -\frac{2}{3}$ : $\left|\frac{7}{3}\right| = \frac{7}{3}$ and $\left|-\frac{4}{3} - 1\right| = \frac{7}{3}$ . ✓

Solution set: $S = \left\{ -\frac{2}{3},\ 4 \right\}$ .

Worked example 4: the unknown on both sides

Solve $∣ x - 1 ∣ = 2 x + 4$ .

Solution

Condition: an absolute value is never negative, so we need $2x + 4 \ge 0$ , that is, $x \ge -2$ .

Case 1: $x - 1 = 2 x + 4$ gives $x = - 5$ . This breaks the condition $x \ge -2$ , so reject it. (Indeed $∣ - 6 ∣ = 6$ but $2 (- 5) + 4 = - 6$ .)

Case 2: $x - 1 = - (2 x + 4)$ gives $3 x = - 3$ , so $x = - 1$ . It meets the condition. Check: $∣ - 2 ∣ = 2$ and $2 (- 1) + 4 = 2$ . ✓

Solution set: $S = {- 1}$ .

Writing an expression in pieces

To remove the bars from an expression with several absolute values, split the number line at the points where each expression inside is zero, then use the definition on each interval. For $f (x) = ∣ x - 1 ∣ + ∣ x + 2 ∣$ the split points are $- 2$ and $1$ :

Interval	$∣ x - 1 ∣$	$∣ x + 2 ∣$	$f (x)$
$x \lt -2$	$1 - x$	$- x - 2$	$- 2 x - 1$
$-2 \le x \le 1$	$1 - x$	$x + 2$	$3$
$x \gt 1$	$x - 1$	$x + 2$	$2 x + 1$

The flat part has a meaning: $f (x)$ is the distance from $x$ to 1 plus the distance from $x$ to $- 2$ . Any point between $- 2$ and $1$ has total distance exactly 3, the length of the gap; any point outside has more. So the least value of $f$ is 3. (On a straight road, a water point placed anywhere between two homes gives the same total walking distance to both.)

Tip

Read every absolute value as a distance before you calculate: $∣ x + 3 ∣$ is "the distance from $x$ to $- 3$ ". An equation such as $∣ x + 3 ∣ = 4$ then solves itself on a number line sketch ( $x = 1$ or $x = - 7$ ), and a sketch catches answers that have the wrong sign.

GreeneMath: Absolute Value and the Distance Between Two Points on a Number Line. Watch for why the distance between a and b is |a − b| whichever number you write first.

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Practice

Try each one before opening the answers. They get harder as you go.

Evaluate: (a) $∣ - 12 ∣ - ∣ 5 ∣$ (b) $|3 - 8| \times |-2|$ (c) $|2 - \sqrt{5}|$ .
Write without absolute value signs: (a) $∣ x - 4 ∣$ given $x \lt 4$ (b) $∣ x^{2} + 1 ∣$ (c) $\sqrt{(x + 3)^2}$ given $x \le -3$ .
Find the distance on the number line between $- 7.5$ and $2.3$ .
A pin passes inspection when $|d - 12.00| \le 0.04$ , with $d$ in mm. Which of the pins $11.95\,\mathrm{mm}$ , $12.03\,\mathrm{mm}$ and $12.05\,\mathrm{mm}$ pass?
Solve $∣ 3 x + 1 ∣ = 8$ .
Solve $∣ x - 2 ∣ = ∣ 3 x + 6 ∣$ .
Solve $∣ 2 x - 5 ∣ = x - 1$ , stating the condition first.
Use the triangle inequality to prove that $|a| - |b| \le |a - b|$ for all real $a$ and $b$ .

Show answersHide answers

(a) $12 - 5 = 7$ (b) $5 \times 2 = 10$ (c) $\sqrt{5} \approx 2.236 \gt 2$ , so $\sqrt{5} - 2$ .
(a) $4 - x$ (b) $x^{2} + 1$ , since $x^2 + 1 \gt 0$ for every $x$ (c) $∣ x + 3 ∣ = - x - 3$ , since $x + 3 \le 0$ .
$∣ - 7.5 - 2.3 ∣ = ∣ - 9.8 ∣ = 9.8$ .
$|{-0.05}| = 0.05 \gt 0.04$ : fails. $|0.03| = 0.03 \le 0.04$ : passes. $|0.05| = 0.05 \gt 0.04$ : fails. Only the $12.03\,\mathrm{mm}$ pin passes.
$3 x + 1 = 8$ gives $x = \frac{7}{3}$ ; $3 x + 1 = - 8$ gives $x = - 3$ . $S = \left\{-3, \frac{7}{3}\right\}$ .
$x - 2 = 3 x + 6$ gives $x = - 4$ ; $x - 2 = - (3 x + 6)$ gives $x = - 1$ . Checks: $∣ - 6 ∣ = ∣ - 6 ∣$ and $∣ - 3 ∣ = ∣ 3 ∣$ . $S = \{-4, -1\}$ .
Condition $x - 1 \ge 0$ , so $x \ge 1$ . $2 x - 5 = x - 1$ gives $x = 4$ ✓; $2 x - 5 = - (x - 1)$ gives $x = 2$ ✓. Checks: $∣ 3 ∣ = 3$ , $∣ - 1 ∣ = 1$ . $S = \{2, 4\}$ .
$|a| = |(a - b) + b| \le |a - b| + |b|$ by the triangle inequality; subtract $∣ b ∣$ from both sides.

Quick check

Given that x < 0, which expression is equal to |x|?

Key points

$∣ x ∣ = x$ for $x \ge 0$ and $∣ x ∣ = - x$ for $x \lt 0$ ; also $\sqrt{x^2} = |x|$ .
$∣ a - b ∣$ is the distance between $a$ and $b$ .
$|xy| = |x|\,|y|$ and $|x + y| \le |x| + |y|$ , with equality when $xy \ge 0$ .
$∣ A ∣ = B$ needs the condition $B \ge 0$ ; check every candidate.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Photo: Messschieber (vernier caliper), ArtMechanic, Wikimedia Commons, CC BY-SA 3.0.
Video: Absolute Value and the Distance Between Two Points on a Number Line, GreeneMath.com, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1How far, not which way

2Absolute value is distance

3The properties of absolute value

4Equations with absolute values

5Writing an expression in pieces

How far, not which way

Absolute value is distance

The properties of absolute value

Equations with absolute values

Writing an expression in pieces