The modulus function

The mod­u­lus of a num­ber mea­sures its size with­out car­ing about its sign: the mod­u­lus of $-5$ and of $5$ are both $5$. You can think of $|x|$ as the dis­tance of $x$ from $0$ on the num­ber line. The mod­u­lus func­tion is the source of a large fam­ily of equa­tions and inequal­i­ties, and this les­son builds the foun­da­tion: the def­i­n­i­tion, the graph, and the prop­er­ties you will use in every mod­u­lus prob­lem after­wards.

The mod­u­lus of a real num­ber#

The mod­u­lus (or absolute value) of a real num­ber $x$ is writ­ten $|x|$ and read "mod $x$". It is defined by

$$|x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$$

For a num­ber that is zero or pos­i­tive, the mod­u­lus is the num­ber itself. For a neg­a­tive num­ber, the mod­u­lus is the neg­a­tive of the num­ber, which is pos­i­tive.

Stu­dents some­times think "$-x$ is always neg­a­tive", so $|x| = -x$ looks wrong. It is not: when $x$ itself is neg­a­tive, $-x$ is pos­i­tive. For $x = -3$, $-x = 3$.

The mod­u­lus func­tion#

The func­tion $f(x) = |x|$, $x \in \mathbb{R}$, is called the mod­u­lus func­tion.

$$\text{Domain} = \mathbb{R}, \qquad \text{Range} = [0, \infty)$$

Every real num­ber has a mod­u­lus, and the mod­u­lus is never neg­a­tive; every non-neg­a­tive num­ber is the mod­u­lus of some­thing (itself).

The graph of $y = |x|$#

For $x \geq 0$ the graph is the line $y = x$; for $x < 0$ it is the line $y = -x$. Together they make a V shape with its cor­ner at the ori­gin. The right arm goes through $(1, 1)$, $(2, 2)$, $(3, 3)$, and the left arm through $(-1, 1)$, $(-2, 2)$, $(-3, 3)$. The graph is sym­met­ric about the $y$-axis and never goes below the $x$-axis.

Prop­er­ties of the mod­u­lus#

In these prop­er­ties $x$ and $y$ are any real num­bers, and "for all" is writ­ten $\forall$.

1. $|x| \geq 0$, $\forall x \in \mathbb{R}$.
2. $|x| = 0 \implies x = 0$.
3. $|x| = x \implies x \geq 0$.
4. $|x| > x \implies x < 0$.
5. $|x| < x$ is true for no real $x$: the solu­tion set is $\phi$.
6. $|-x| = |x|$.
7. $|x^n| = |x|^n$.
8. $|xy| = |x||y|$.
9. $\displaystyle \left|\frac{x}{y}\right| = \frac{|x|}{|y|}$, for $y \neq 0$.
10. (a) $\big||x| - |y|\big| \leq |x + y| \leq |x| + |y|$ and (b) $\big||x| - |y|\big| \leq |x - y| \leq |x| + |y|$.
11. $|x + y| = |x| + |y|$ when $xy \geq 0$.
12. $|x + y| < |x| + |y|$ when $xy < 0$.

Prop­er­ties 3 and 5 are used a great deal, and prop­erty 10 is the famous tri­an­gle inequal­ity.

A com­mon mis­take. Prop­erty 6 says $|-x| = |x|$. It does not say $|-x| = x$. Test with $x = -2$: the left side is $|-(-2)| = |2| = 2$, while $x = -2$. The two sides dif­fer, so $|-x| = x$ is false in gen­eral.

Prop­erty 11 explained. When $x$ and $y$ have the same sign (or one is zero), adding them makes the size grow, so $|x + y|$ is the full $|x| + |y|$. For exam­ple, with $x = 0$ and $y = -2$ we have $xy = 0$, and $|x + y| = |0 - 2| = 2$ while $|x| + |y| = 0 + 2 = 2$. When $x$ and $y$ have oppo­site signs, they partly can­cel, so $|x + y|$ is smaller: with $x = 3$, $y = -5$, $|x + y| = 2$ but $|x| + |y| = 8$. That is prop­erty 12.

Mod­u­lus inequal­i­ties#

Think of $|x|$ as the dis­tance from $0$. Let $a > 0$.

13. $|x| < a \implies -a < x < a$, that is, $x \in (-a, a)$.

    $|x| \leq a \implies -a \leq x \leq a$, that is, $x \in [-a, a]$.

    $|x| > a \implies x < -a$ or $x > a$, that is, $x \in (-\infty, -a) \cup (a, \infty)$.

    $|x| \geq a \implies x \leq -a$ or $x \geq a$, that is, $x \in (-\infty, -a] \cup [a, \infty)$.

A dis­tance less than $a$ means $x$ lies between $-a$ and $a$; a dis­tance greater than $a$ means $x$ lies out­side. The same short rule as for qua­drat­ics: less, lies between.

Exam­ple 1#

Solve $|x| < 2$.

$-2 < x < 2$, that is, $x \in (-2, 2)$.

Exam­ple 2#

Solve $|x - 3| < 2$.

Treat $x - 3$ as the quan­tity inside the mod­u­lus:

$$\begin{aligned} -2 &< x - 3 < 2 \\ -2 + 3 &< x < 2 + 3 \\ 1 &< x < 5 \end{aligned}$$

So $x \in (1, 5)$. In words, $x$ is less than $2$ units away from $3$.

Exam­ple 3#

Solve $|2x - 3| \geq 2$.

Greater than or equal, so the quan­tity lies out­side $[-2, 2]$:

$$2x - 3 \leq -2 \quad \text{or} \quad 2x - 3 \geq 2$$

$$2x \leq 1 \quad \text{or} \quad 2x \geq 5$$

$$\displaystyle x \leq \frac{1}{2} \quad \text{or} \quad x \geq \frac{5}{2}$$

So $\displaystyle x \in \left(-\infty, \frac{1}{2}\right] \cup \left[\frac{5}{2}, \infty\right) = \mathbb{R} - \left(\frac{1}{2}, \frac{5}{2}\right)$. On a num­ber line, shade every­thing to the left of $\displaystyle \frac{1}{2}$ and to the right of $\displaystyle \frac{5}{2}$, with filled dots at both ends.

Prac­tice#

1. Find $|-9|$.
2. Find $|5 - 8|$.
3. Find $|-4| \times |3|$ and $|(-4) \times 3|$, and say which prop­erty they show.
4. Is $|x| > x$ true for $x = -3$?
5. For which real num­bers is $|x| = x$?
6. Check prop­erty 11 or 12 for $x = 3$, $y = 5$ and for $x = 3$, $y = -5$.
7. Solve $|x| \leq 4$.
8. Solve $|x + 1| < 3$.
9. Solve $|x - 5| > 2$.
10. Solve $|3x + 1| \geq 5$.
11. Solve $|2x - 1| < 0$.

Answers#