. This page covers what the bars mean, the rules they obey, how to remove them, and where they appear in graphs and calculus. For solving, see absolute value equations and absolute value inequalities.

Key Terms

Absolute Value— the distance of a number from zero on the number line

Absolute Value Function— the function

f(x) = |x|

and its V-shaped graph

Principal Root— the non-negative root, which is why

\sqrt{x^2} = |x|

Absolute Value Inequality— an inequality whose solutions are read as distances

See All Algebra Definitions →

What Absolute Value Measures

The absolute value of a real number

x

, written

|x|

, is the distance from

x

0

on the number line.

Distance has no direction, so a number and its opposite have the same absolute value. Both

-3

and

3

lie three units from zero, so

|-3| = 3

and

|3| = 3

. The bars keep the size of a number and discard its sign.

Example

|-12| = 12

\; |7.5| = 7.5

\; |0| = 0

\; \left|-\frac{2}{3}\right| = \frac{2}{3}

The bars are also grouping symbols, like parentheses: the expression inside is simplified first, and only then is the absolute value taken.

Pitfall: Splitting the bars over a difference

Writing

|3 - 8| = |3| - |8| = -5

produces a negative absolute value, which is impossible. Simplify inside first:

|3 - 8| = |-5| = 5

Example

|3 - 8| = |-5| = 5

\;

|2 \cdot (-4) + 1| = |-7| = 7

\;

|4| - |-9| = 4 - 9 = -5

, because here the bars close before the subtraction.

A minus sign outside the bars is not touched by them:

-|-4| = -(4) = -4

. The absolute value itself is never negative, but its opposite can be.

Absolute value as distance from zero

The amber segment runs from −3 to 0, the blue one from 0 to 3. Both are three units long, so |−3| = |3| = 3. The sign says which side of zero a number is on; the absolute value says only how far.

The Piecewise Definition

Distance is the idea; the working definition is a rule with two cases.

The Piecewise Definition

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

For

x \geq 0

, the number already equals its distance from zero and is left unchanged. For

x < 0

, the rule returns

-x

, the opposite of

x

Pitfall: Reading −x as a negative number

Seeing

|x| = -x

and concluding that

|x|

is negative. In this case

x

itself is negative, so

-x

is positive:

|-6| = -(-6) = 6

The second case looks as if it produced a negative number, but it does the reverse. When

x

is negative,

-x

is positive: for

x = -6

-x = -(-6) = 6

. The minus sign in

-x

means "the opposite of

x

", not "a negative number".

Example

For

x = 4

: the first case applies,

|4| = 4

. For

x = -2.5

: the second case applies,

|-2.5| = -(-2.5) = 2.5

The two cases agree at

x = 0

, where both give

0

, so the boundary point could go in either one. Putting it in the first case,

x \geq 0

, is the usual convention.

Absolute Value as Distance Between Numbers

Subtracting two numbers gives a signed gap: positive or negative depending on the order. The bars remove the sign, so the order no longer matters.

Absolute Value as Distance Between Numbers

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

Example

The distance between

-2

and

5

|5 - (-2)| = |7| = 7

, and in the other order

|-2 - 5| = |-7| = 7

The definition of

|x|

is the special case

b = 0

|x| = |x - 0|

, the distance from

x

0

This reading turns statements about absolute value into statements about position:

$|x - 3| = 2$ asks for the numbers at distance $2$ from $3$ : they are $1$ and $5$ .
$|x - 3| < 2$ asks for the numbers closer than $2$ to $3$ : the open interval $(1, 5)$ .
$|x + 4|$ is the distance from $x$ to $-4$ , because $x + 4 = x - (-4)$ .

The same picture is used to solve absolute value equations and absolute value inequalities without algebra.

Absolute value as distance between numbers

Top: the gap from −2 to 5 is |5 − (−2)| = 7, whichever number is subtracted from which. Bottom: |x − 3| < 2 collects every number less than 2 away from 3, the open interval from 1 to 5. The hollow ends are 1 and 5 themselves, which are exactly 2 away.

Properties of Absolute Value

For all real numbers

a

and

b

, the bars obey the rules below.

RULE

Absolute value passes through multiplication and division:

|ab| = |a|\,|b|

and

\left|\frac{a}{b}\right| = \frac{|a|}{|b|}

for

b \neq 0

. It does not pass through addition or subtraction.

The product rule holds because the size of a product is the product of the sizes, and the sign is decided separately. For sums there is no such rule:

|3 + (-5)| = 2

, while

|3| + |-5| = 8

. The best that can be said about a sum is the triangle inequality.

Property	Rule	Example
Never negative	\|a\| ≥ 0	\|−5\| = 5
Zero only at zero	\|a\| = 0 exactly when a = 0	\|0\| = 0
Opposites	\|−a\| = \|a\|	\|−7\| = \|7\| = 7
Product	\|ab\| = \|a\| · \|b\|	\|(−3)(4)\| = 12 = 3 · 4
Quotient	\|a / b\| = \|a\| / \|b\|, b ≠ 0	\|−8 / 2\| = 4 = 8 / 2
Powers	\|aⁿ\| = \|a\|ⁿ	\|(−2)³\| = 8 = 2³
Square	\|a\|² = a²	\|−5\|² = 25 = (−5)²
Square root	√(a²) = \|a\|	√((−5)²) = 5
Bounds	−\|a\| ≤ a ≤ \|a\|	−5 ≤ −5 ≤ 5
Triangle inequality	\|a + b\| ≤ \|a\| + \|b\|	\|3 + (−5)\| = 2 ≤ 8
Reverse triangle inequality	\| \|a\| − \|b\| \| ≤ \|a − b\|	\| 3 − 5 \| = 2 ≤ \|3 − (−5)\| = 8

Absolute value splits over products and quotients, never over sums and differences.

Square Roots and Squares

The square root symbol always means the principal root, the non-negative one. So

\sqrt{x^2}

is the non-negative number whose square is

x^2

, and that number is

|x|

, not

x

Square Roots and Squares

\sqrt{x^2} = |x|

Pitfall: Writing √(x²) = x

For

x = -4

this gives

\sqrt{(-4)^2} = -4

, a negative square root. In fact

\sqrt{16} = 4 = |-4|

Example

For

x = -4

\sqrt{(-4)^2} = \sqrt{16} = 4 = |-4|

. For

x = 4

\sqrt{16} = 4 = |4|

. The answer is

|x|

in both cases.

The same holds for every even root,

\sqrt[4]{x^4} = |x|

, but not for odd roots:

\sqrt[3]{(-2)^3} = -2

, so

\sqrt[3]{x^3} = x

. The properties of roots page covers the general rule.

Squaring goes the other way and removes bars:

|x|^2 = x^2

, because squaring discards the sign just as the bars do. This is why

|A| = |B|

holds exactly when

A^2 = B^2

, which gives a second way to solve equations with absolute value on both sides.

The Triangle Inequality

The absolute value of a sum is at most the sum of the absolute values.

The Triangle Inequality

|a + b| \leq |a| + |b|

When

a

and

b

have the same sign, nothing cancels and the two sides are equal:

|2 + 5| = 7 = |2| + |5|

. When the signs differ, part of one number cancels part of the other, and the left side is smaller:

|2 + (-5)| = 3

, while

|2| + |-5| = 7

RULE

|a + b| = |a| + |b|

exactly when

ab \geq 0

, that is, when

a

and

b

do not have opposite signs.

On the number line: start at

0

, take a step of

a

, then a step of

b

. The walk covers

|a| + |b|

in total, but if the second step turns back, the end point

a + b

is closer to the start than that.

Replacing

b

b - a

and rearranging gives the reverse form,

\big||a| - |b|\big| \leq |a - b|

: two numbers are at least as far apart as their distances from zero differ.

The same inequality holds for complex numbers, where

a

b

and

a + b

form the sides of an actual triangle. The proof and the reverse form are on the modulus of complex numbers page.

Removing the Bars

To rewrite an expression without bars, split at the point where the inside equals zero. On the side where the inside

A

is non-negative, replace

|A|

A

; on the side where it is negative, replace

|A|

-A

Example

|x - 2|

is zero at

x = 2

. For

x \geq 2

|x - 2| = x - 2

. For

x < 2

|x - 2| = -(x - 2) = 2 - x

Pitfall: Negating only the first term

For

x < 5

, writing

|x - 5| = -x - 5

. At

x = 2

this gives

-7

, but

|2 - 5| = 3

. The whole inside changes sign:

-(x - 5) = 5 - x

With several absolute values, each one has its own split point. Together the split points cut the number line into intervals, and on each interval every inside expression keeps one sign.

Example

Simplify

f(x) = |x - 2| + |x + 1|

. The split points are

x = -1

and

x = 2

For

x < -1

, both insides are negative:

f(x) = -(x - 2) - (x + 1) = -2x + 1

For

-1 \leq x < 2

, only

x + 1

is non-negative:

f(x) = -(x - 2) + (x + 1) = 3

For

x \geq 2

, both are non-negative:

f(x) = (x - 2) + (x + 1) = 2x - 1

Check one point in each interval:

f(-2) = 4 + 1 = 5 = -2(-2) + 1

\; f(0) = 2 + 1 = 3

\; f(3) = 1 + 4 = 5 = 2(3) - 1

The result is a piecewise function. Every expression built from absolute values of linear terms becomes one this way.

Removing the bars piece by piece

The dashed lines are the split points, where x + 1 and x − 2 change sign. Each interval gets its own formula. Between −1 and 2 the graph is flat at 3: there the two distances, from x to 2 and from x to −1, always add up to the distance between −1 and 2.

The Graph of the Absolute Value Function

The graph of

y = |x|

follows the two cases of the definition: the line

y = x

for

x \geq 0

and the line

y = -x

for

x < 0

. The two half-lines meet at the origin and form a V.

The vertex $(0, 0)$ is the lowest point; the range is $[0, \infty)$ .
The graph is symmetric about the $y$ -axis, because $|-x| = |x|$ : the function is even.
The slope is $-1$ on the left and $1$ on the right.

Shifting and stretching move the V without changing its shape. The graph of

y = a|x - h| + k

has its vertex at

(h, k)

, opens upward when

a > 0

and downward when

a < 0

, and has slopes

-a

and

a

on the two sides.

Example

y = 2|x - 1| - 3

has its vertex at

(1, -3)

, opens upward, and has slopes

-2

and

2

. It crosses the

x

-axis where

|x - 1| = \frac{3}{2}

: at

x = -\frac{1}{2}

and

x = \frac{5}{2}

The absolute value function section of the function families page places this graph among the other basic shapes.

The V shape

The graph keeps the part of y = x on the right and the part of y = −x on the left; the faded dashed continuations are the halves the bars throw away. The two half-lines meet at the vertex, where the slope jumps from −1 to 1.

Absolute Value in Equations and Inequalities

Equations and inequalities with absolute value all reduce to a few forms. In each,

A

stands for the expression inside the bars and

k

for a constant.

The two signs of

k

behave differently. An absolute value is never negative, so

|A| = k

and

|A| < k

have no solutions when

k < 0

, and

|A| > k

holds for every

x

.

The full methods, with expressions on both sides and quadratic insides, are on the absolute value equations and absolute value inequalities pages. To check an answer, use the absolute value equation calculator.

Form	Equivalent statement	Example	Solution
\|A\| = k, k > 0	A = k or A = −k	\|x − 3\| = 2	x = 1 or x = 5
\|A\| < k, k > 0	−k < A < k	\|x − 3\| < 2	1 < x < 5
\|A\| > k, k > 0	A < −k or A > k	\|x − 3\| > 2	x < 1 or x > 5
\|A\| = 0	A = 0	\|x − 3\| = 0	x = 3
\|A\| = k or \|A\| < k, k < 0	impossible	\|x − 3\| = −2	no solution
\|A\| > k, k < 0	always true	\|x − 3\| > −2	every real x

Absolute Value in Calculus

The function

|x|

is the standard example of a function that is continuous but not differentiable.

Continuity. The graph has no gaps or jumps:

|x|

is continuous at every real number.

The corner at 0. To the left of

0

the slope is

-1

; to the right it is

1

. The two one-sided derivatives disagree, so

|x|

has no derivative at

0

. The corners section of the differentiability page treats this case.

The derivative elsewhere. For

x \neq 0

Absolute Value in Calculus

\frac{d}{dx}|x| = \frac{x}{|x|} = \begin{cases} -1 & \text{if } x < 0 \\ 1 & \text{if } x > 0 \end{cases}

One-sided limits. The quotient

\frac{|x|}{x}

approaches

-1

from the left and

1

from the right, so its two-sided limit at

0

does not exist. See one-sided limits.

Antiderivatives.

\int |x|\,dx = \frac{x|x|}{2} + C

. The bars also appear in

\int \frac{1}{x}\,dx = \ln|x| + C

: the logarithm accepts only positive inputs, while

\frac{1}{x}

is defined for negative

x

as well, and the bars cover both sides. This formula is listed with the basic antiderivatives.

Area. An integral counts area below the axis as negative; the integral of

|f(x)|

counts all of it as positive. For example,

\int_{-1}^{2} |x|\,dx = \frac{1}{2} + 2 = \frac{5}{2}

, while

\int_{-1}^{2} x\,dx = \frac{3}{2}

Absolute Value Beyond the Real Numbers

The idea of size without direction carries over to other kinds of numbers.

For a complex number

z = a + bi

, the absolute value, also called the modulus, is

|z| = \sqrt{a^2 + b^2}

: the distance from

z

to the origin in the complex plane. When

b = 0

it reduces to the ordinary absolute value,

\sqrt{a^2} = |a|

. See modulus of complex numbers.

For a vector, the same role is played by its length, written

\|v\|

.

In every setting, four rules hold: the size is never negative, it is zero only for zero, it multiplies over products, and it obeys the triangle inequality. These four rules are what make a quantity a measure of size.

Common Pitfalls

Splitting the bars over a sum or difference

The bars are distributed over the terms inside, as if they were a factor.

Wrong working:

|3 - 8| = |3| - |8| = 3 - 8 = -5

, and

|x + 2| = |x| + 2

.

Right working: simplify inside first,

|3 - 8| = |-5| = 5

. The expression

|x + 2|

cannot be split; it is

x + 2

for

x \geq -2

and

-x - 2

for

x < -2

.

Why it happens: the product rule

|ab| = |a|\,|b|

is true, and it is carried over to sums, where it fails.

How to check: an absolute value can never be negative, so

-5

is wrong on sight. For the expression, test a negative value: at

x = -5

|-5 + 2| = 3

, but

|-5| + 2 = 7

.

Reading −x as a negative number

The case

|x| = -x

is read as "the absolute value is negative".

Wrong working: "For

x < 0

|x| = -x

, so on this side the absolute value is negative."

Right working: on this side

x

is negative, so

-x

is positive. For

x = -6

|x| = -(-6) = 6

.

Why it happens: a minus sign in front of a letter is read as a negative number, when it means the opposite of whatever the letter stands for.

How to check: substitute a number from the case. Any negative

x

gives a positive

-x

.

Writing √(x²) = x

The square root and the square are treated as canceling.

Wrong working:

\sqrt{x^2} = x

, so

\sqrt{(-4)^2} = -4

.

Right working:

\sqrt{(-4)^2} = \sqrt{16} = 4

. In general

\sqrt{x^2} = |x|

.

Why it happens: for non-negative

x

the square root and the square do undo each other, and the rule is remembered without its condition.

How to check: a square root symbol never gives a negative result. If the answer can be negative, bars are missing.

Negating only the first term

When the inside is negative, the minus sign is applied to the first term only.

Wrong working: for

x < 5

|x - 5| = -x - 5

.

Right working: the whole inside changes sign,

|x - 5| = -(x - 5) = 5 - x

.

Why it happens: the parentheses around

x - 5

are left out when

-A

is written down.

How to check: test a value from the interval. At

x = 2

|2 - 5| = 3

and

5 - 2 = 3

, while

-2 - 5 = -7

Summary: Absolute Value at a Glance

The definition, the rules, the graph and the calculus facts on this page reduce to the table below.

Idea	Statement	Example
Meaning	\|x\| is the distance from x to 0	\|−3\| = \|3\| = 3
Definition	\|x\| = x for x ≥ 0, \|x\| = −x for x < 0	\|−6\| = −(−6) = 6
Distance between numbers	\|a − b\|	distance from −2 to 5 is 7
What splits over the bars	products and quotients; never sums or differences	\|−2 · 3\| = 6, \|3 − 8\| ≠ 3 − 8
Square roots	√(x²) = \|x\|	√((−4)²) = 4
Triangle inequality	\|a + b\| ≤ \|a\| + \|b\|, equal exactly when ab ≥ 0	\|2 + (−5)\| = 3 < 7
Removing the bars	split where the inside is 0; use A or −A on each side	\|x − 2\| = 2 − x for x < 2
Graph	y = a\|x − h\| + k is a V with vertex (h, k)	y = 2\|x − 1\| − 3: vertex (1, −3)
Calculus	continuous everywhere, corner at 0; ∫ 1/x dx = ln\|x\| + C	slopes −1 and 1 meet at 0

Absolute Value FAQ

Can an absolute value be negative?

No. A distance is never negative, so |x| ≥ 0 for every real x, and |x| = 0 only when x = 0. An expression such as −|4| = −4 is negative, but that is the opposite of an absolute value, not an absolute value.Read more →

Is the square root of x squared equal to x?

Only when x ≥ 0. In general √(x²) = |x|, because the square root symbol always gives the non-negative root. For x = −4, √((−4)²) = √16 = 4, not −4.Read more →

Is |a + b| equal to |a| + |b|?

Not in general. The triangle inequality says |a + b| ≤ |a| + |b|, with equality exactly when a and b do not have opposite signs. For example, |2 + (−5)| = 3, while |2| + |−5| = 7. Absolute value does split over products: |ab| = |a| · |b|.Read more →

Why is |x| not differentiable at 0?

The graph of |x| has a corner at 0: the slope is −1 just to the left and 1 just to the right. The two one-sided derivatives disagree, so no single derivative exists there, even though |x| is continuous at 0.Read more →

Why is $|x + 4|$ the distance to $-4$ and not to $4$ ?

Because the distance reading needs a difference. The distance between

x

and

b

|x - b|

, so

x + 4

has to be rewritten first as

x - (-4)

, and the point it measures from is

-4

. That makes

|x + 4| = 1

a question about the numbers one unit from

-4

: they are

-5

and

-3

, and both check, since

|-1| = |1| = 1

.Read more →

Is $-|x|$ the same as $|-x|$ ?

No. In

|-x|

the minus sign is inside the bars, and a number and its opposite lie the same distance from zero, so

|-x| = |x|

and

|-4| = 4

. In

-|x|

the minus sign is outside, so it acts after the bars:

-|-4| = -(4) = -4

. The first is never negative; the second is never positive.Read more →

What is the antiderivative of $|x|$ ?

\int |x|\,dx = \frac{x|x|}{2} + C

, one formula for both sides of

0

. Differentiating checks it: for

x > 0

the function is

\frac{x^2}{2}

, with derivative

x

, and for

x < 0

it is

-\frac{x^2}{2}

, with derivative

-x

, and both equal

|x|

. Unlike

|x|

itself,

\frac{x|x|}{2}

has no corner at

0

; its slope there is

0

.Read more →

Absolute Value

Distance Without Direction

Key Terms

What Absolute Value Measures

The Piecewise Definition

Absolute Value as Distance Between Numbers

Properties of Absolute Value

Square Roots and Squares

The Triangle Inequality

Removing the Bars

The Graph of the Absolute Value Function

Absolute Value in Equations and Inequalities

Absolute Value in Calculus

Absolute Value Beyond the Real Numbers

Common Pitfalls

Summary: Absolute Value at a Glance

Absolute Value FAQ

Can an absolute value be negative?

Is the square root of x squared equal to x?

Is |a + b| equal to |a| + |b|?

Why is |x| not differentiable at 0?

Why is ∣x+4∣|x + 4|∣x+4∣ the distance to −4-4−4 and not to 444?

Is −∣x∣-|x|−∣x∣ the same as ∣−x∣|-x|∣−x∣?

What is the antiderivative of ∣x∣|x|∣x∣?

Why is $|x + 4|$ the distance to $-4$ and not to $4$ ?

Is $-|x|$ the same as $|-x|$ ?

What is the antiderivative of $|x|$ ?