Absolute Value

Defining Absolute Value

The order properties developed in “Transitivity of Inequalities” let us combine comparisons into useful chains. A related idea is to assign a nonnegative quantity to every real number, whether that number is positive, negative, or zero. This quantity is its absolute value. In this tutorial, we define absolute value algebraically and establish properties that let us work with it using order and field operations.

The definition depends on the sign of the input. A nonnegative input is left unchanged; a negative input is replaced by its additive inverse. The latter is positive, so the result is nonnegative in both cases.

Definition (Absolute value). For $x\in\mathbb{R}$, the absolute value of $x$, denoted $|x|$, is defined by

|x|= \begin{cases} x,&x\geq0,\\ -x,&x<0. \end{cases}

The cases include every real number: by trichotomy, either $x\geq0$ or $x<0$, and these alternatives do not overlap. In particular, zero belongs to the first case, so $|0|=0$. The definition is sometimes described by saying that absolute value removes a number’s sign. It does not mean that the input itself is always nonnegative; it means that the value produced by the definition is nonnegative.

Worked Example: Evaluating Absolute Values

For $x=-\frac{7}{3}$, the input is negative, so the second branch of the definition applies:

\left|-\frac{7}{3}\right| =-\left(-\frac{7}{3}\right) =\frac{7}{3}.

For $y=\frac{5}{4}$, the input is nonnegative, so the first branch applies:

\left|\frac{5}{4}\right|=\frac{5}{4}.

These computations use different branches, but both outputs are nonnegative. The first does not say that a negative input equals its absolute value; rather, its absolute value is the additive inverse of that input.

Basic Order Properties

The definition immediately gives several facts that will be used repeatedly. In particular, $|x|$ bounds $x$ from above, and $-|x|$ bounds it from below. We prove these facts together, also identifying exactly when the absolute value is zero.

Theorem (Basic properties of absolute value). For every $x\in\mathbb{R}$, $|x|\geq0$, $|x|=0$ if and only if $x=0$, and $$ -|x|\leq x\leq |x|. $$ Moreover, $|-x|=|x|$.

Proof. First suppose $x\geq0$. Then $|x|=x$, so $|x|\geq0$, and $x\leq|x|$ holds with equality. Also, $-|x|=-x\leq x$, because $0\leq x+x$, or equivalently because $x\geq0$ implies $-x\leq x$.

Now suppose $x<0$. Then $|x|=-x$, and $-x>0$, so again $|x|\geq0$. The upper bound $x\leq|x|$ follows from $x<0<-x$, and the lower bound is $-|x|=x$, an equality. Thus the two bounds hold in either case.

If $x=0$, the definition gives $|x|=0$. Conversely, if $|x|=0$, then the bounds just proved give $0\leq x\leq0$, so trichotomy implies $x=0$.

Finally, consider $|-x|$. If $x\geq0$, then $-x\leq0$, and the definition gives $|-x|=-(-x)=x=|x|$. If $x<0$, then $-x>0$, so $|-x|=-x=|x|$. These cases prove the identity for every real $x$. $\square$

The bounds in the theorem are useful even when the sign of $x$ is unknown. They say that both $x$ and its negative are no greater than $|x|$, while $|x|$ itself is never negative. A common mistake is to use $|x|=x$ without checking that $x\geq0$. For a negative $x$, the correct formula is $|x|=-x$.

Worked Example: Bounding a Number of Unknown Sign

Suppose $|u|=\frac{9}{5}$. The basic bounds give

-\frac{9}{5}=-|u|\leq u\leq |u|=\frac{9}{5}.

Thus $u$ must lie between $-\frac{9}{5}$ and $\frac{9}{5}$. The equation alone does not tell us which value $u$ is: both $u=\frac{9}{5}$ and $u=-\frac{9}{5}$ satisfy it. Indeed, the definition gives $\left|\frac{9}{5}\right|=\frac{9}{5}$ and $\left|-\frac{9}{5}\right|=\frac{9}{5}$.

Turning Absolute-Value Bounds into Inequalities

A bound involving $|x|$ can be rewritten as two ordinary inequalities. This is one of the most useful ways to apply the definition, especially when solving inequalities. The condition that the bound $a$ be nonnegative is essential: a nonnegative absolute value cannot be less than a negative number.

Theorem (Absolute-value bound criterion). If $a\geq0$, then $$ |x|\leq a \quad\Longleftrightarrow\quad -a\leq x\leq a. $$

Proof. Suppose first that $|x|\leq a$. The basic properties give $-|x|\leq x\leq|x|$. Since $|x|\leq a$, order reversal under negation gives $-a\leq-|x|$. Chaining these comparisons yields $-a\leq-|x|\leq x\leq|x|\leq a$, so $-a\leq x\leq a$.

Conversely, suppose $-a\leq x\leq a$. We show $|x|\leq a$ by considering the sign of $x$. If $x\geq0$, then $|x|=x\leq a$. If $x<0$, then $|x|=-x$, and $-a\leq x$ implies $-x\leq a$ by order reversal under negation. Hence $|x|\leq a$ in either case. This proves both directions. $\square$

This equivalence preserves the endpoint information: $|x|\leq a$ becomes a lower and an upper bound, both non-strict. To solve an inequality involving a more complicated expression, first apply the criterion to that whole expression, then use the order rules for addition and multiplication.

Worked Example: Solving an Absolute-Value Inequality

Solve $|3t-2|\leq7$. Since $7\geq0$, the absolute-value bound criterion gives an equivalent pair of bounds:

|3t-2|\leq7 \quad\Longleftrightarrow\quad -7\leq3t-2\leq7.

Add $2$ throughout, which preserves order under translation:

-5\leq3t\leq9.

Since $3>0$, multiplication by $1/3$ preserves the order, so the solution is

-\frac{5}{3}\leq t\leq3.

The endpoints satisfy the original inequality: for $t=-\frac{5}{3}$, $3t-2=-7$, and $|-7|=7$; for $t=3$, $3t-2=7$, and $|7|=7$. Every value between the endpoints satisfies the equivalent double inequality, so no additional restriction is needed.

Absolute Value and Algebraic Operations

Absolute value also interacts predictably with multiplication. The product rule below can be verified directly from the definition by checking the signs of the factors. We include zero in the cases so that no assumption that either factor is nonzero is needed.

Theorem (Absolute value of a product). For all $x,y\in\mathbb{R}$, $$ |xy|=|x||y|. $$

Proof. If $x=0$ or $y=0$, then $xy=0$, and both sides are zero. Assume now that both are nonzero. Each is either positive or negative.

If both are positive, then $xy>0$, so $|xy|=xy=|x||y|$. If $x>0$ and $y<0$, then $xy<0$, so $|xy|=-xy=x(-y)=|x||y|$. If $x<0$ and $y>0$, then $xy<0$, and $|xy|=-xy=(-x)y=|x||y|$. If both are negative, then $xy>0$, so $|xy|=xy=(-x)(-y)=|x||y|$. These four sign cases exhaust the possibilities for nonzero $x$ and $y$, and together with the zero cases prove the result. $\square$

Worked Example: Applying the Product Rule

Consider the product $\left(-\frac{8}{3}\right)\left(\frac{9}{4}\right)$. Direct multiplication gives

\left(-\frac{8}{3}\right)\left(\frac{9}{4}\right) =-\frac{72}{12}=-6, \qquad \left|\left(-\frac{8}{3}\right)\left(\frac{9}{4}\right)\right|=6.

The product rule gives the same result by taking absolute values first:

\left|-\frac{8}{3}\right|\left|\frac{9}{4}\right| =\frac{8}{3}\cdot\frac{9}{4} =\frac{72}{12}=6.

The calculation illustrates why the product rule is useful: the sign of the product is handled by absolute value, while the magnitudes multiply as usual.

The Triangle Inequality

The absolute value of a sum is bounded above by the sum of the absolute values. This statement is called the triangle inequality. It does not say that the absolute value of a sum always equals the sum of the absolute values; cancellation between terms can make the left side smaller.

Theorem (Triangle inequality). For all $x,y\in\mathbb{R}$, $$ |x+y|\leq|x|+|y|. $$

Proof. By the basic properties of absolute value, $x\leq|x|$ and $y\leq|y|$. Adding these inequalities gives $x+y\leq|x|+|y|$. The lower bounds $-|x|\leq x$ and $-|y|\leq y$ also give, upon addition, $-|x|-|y|\leq x+y$. Since $|x|+|y|\geq0$, these two comparisons state that \[ -\bigl(|x|+|y|\bigr)\leq x+y\leq|x|+|y|. \] The absolute-value bound criterion, with $a=|x|+|y|$, now yields $|x+y|\leq|x|+|y|$. $\square$

The same result can be applied repeatedly to a sum of more than two terms. For example, first apply it to $x+y$ and $z$, and then apply it to $x$ and $y$; this gives $|x+y+z|\leq|x+y|+|z|\leq|x|+|y|+|z|$. At each step, the bound is non-strict, and the argument does not require the terms to have the same sign.

Worked Example: Bounding a Sum with Opposite Signs

Let $x=\frac{13}{6}$ and $y=-\frac{5}{2}$. The sum is

x+y=\frac{13}{6}-\frac{15}{6}=-\frac{2}{6}=-\frac{1}{3}, \qquad |x+y|=\frac{1}{3}.

The triangle inequality bounds this value by the sum of the absolute values:

|x|+|y|=\frac{13}{6}+\frac{5}{2} =\frac{13}{6}+\frac{15}{6} =\frac{28}{6}=\frac{14}{3}, \qquad \frac{1}{3}\leq\frac{14}{3}.

Here the terms have opposite signs, and their sum is smaller in absolute value than either the sum of their absolute values or the larger individual magnitude. In contrast, equality can occur when the terms have the same sign; for instance, $|2+5|=7=|2|+|5|$. Thus the triangle inequality is an upper bound, not an identity in every case.

Equations and a Common Pitfall

For an equation $|u|=a$, the sign of $a$ must be checked first. If $a<0$, there are no solutions because absolute values are nonnegative. If $a=0$, the basic theorem says the only possibility is $u=0$. If $a>0$, the definition shows that $u$ can be either $a$ or $-a$. For inequalities, the bound criterion gives a different but related translation: $|u|\leq a$ corresponds to the entire interval from $-a$ to $a$, provided $a\geq0$.

Worked Example: Solving an Absolute-Value Equation

Solve $|2s+1|=9$. The expression $2s+1$ must equal $9$ or $-9$. In the first case,

2s+1=9 \quad\Longrightarrow\quad 2s=8 \quad\Longrightarrow\quad s=4.

In the second case,

2s+1=-9 \quad\Longrightarrow\quad 2s=-10 \quad\Longrightarrow\quad s=-5.

Both candidates check in the original equation: when $s=4$, $|2s+1|=|9|=9$; when $s=-5$, $|2s+1|=|-9|=9$. Therefore the solutions are $s=4$ and $s=-5$.

A frequent error is to write $|x+y|=|x|+|y|$ without checking the terms. For $x=4$ and $y=-4$, the left side is $|0|=0$, while the right side is $4+4=8$. The valid general statement is the triangle inequality, $|x+y|\leq|x|+|y|$. Another error is to replace $|x|\leq a$ by a double inequality when $a<0$; in that case the original inequality has no solutions, whereas the usual criterion requires $a\geq0$.

Key takeaway. Absolute value is defined by the sign of its input and is always nonnegative. Its basic bounds are $-|x|\leq x\leq|x|$; for $a\geq0$, $|x|\leq a$ is equivalent to $-a\leq x\leq a$. Absolute values preserve products and satisfy the triangle inequality for sums.

Check Your Understanding

Use the definition and the proved properties of absolute value to answer the following questions.

Evaluate $\left|-\frac{11}{6}\right|$, and state which branch of the definition applies.
If $|v|\leq4$, what double inequality must $v$ satisfy?
Why can $|w|\leq-2$ have no real solutions?
Use the product rule to evaluate $|(-5)(-3)|$.
Give real numbers $p$ and $q$ for which $|p+q|<|p|+|q|$, and verify the inequality.
Solve $|r|=6$, and check both proposed values in the equation.

What You'll Learn