Inequalities Involving the Modulus Function

The modulus (or absolute value) of a real number $x$, written $|x|$, is defined piecewise as

Geometrically, $|x|$ is the distance of the point $x$ from the origin $0$ on the number line, always non-negative: $|x| \ge 0$ for every real $x$, with $|x| = 0$ only when $x = 0$. More generally, $|x - a|$ is the distance between the points $x$ and $a$.

Rule A — $|x| < a$, with $a > 0$.

Proof. Case $x \ge 0$: here $|x| = x$, so $|x| < a$ becomes $x < a$; combined with $x \ge 0 > -a$, this gives $-a < x < a$. Case $x < 0$: here $|x| = -x$, so $|x| < a$ becomes $-x < a$, i.e. $x > -a$; combined with $x < 0 < a$, this again gives $-a < x < a$. Both cases give exactly the interval $-a < x < a$. $\blacksquare$

Rule B — $|x| > a$, with $a > 0$.

Proof. Case $x \ge 0$: $|x| = x$, so $|x| > a$ becomes $x > a$. Case $x < 0$: $|x| = -x$, so $|x| > a$ becomes $-x > a$, i.e. $x < -a$. Combining the two cases gives $x < -a$ or $x > a$ — two separate, disjoint rays, never a single interval, because $|x| > a$ says $x$ is far from $0$ in either direction. $\blacksquare$ (The slack versions $|x| \le a \iff -a \le x \le a$ and $|x| \ge a \iff x \le -a$ or $x \ge a$ follow the same way, replacing every strict sign with a slack one.)

Shifted modulus inequalities. Replacing $x$ by any linear expression in Rules A and B — valid, since the proof never used any special property of $x$ itself — gives the forms used constantly in practice:

This matches the distance reading directly: $|x - c| < a$ says "$x$ is within a distance $a$ of $c$," i.e. $x$ lies strictly between $c-a$ and $c+a$.

Worked illustration. Solve $|x - 3| < 4$. Here $c = 3$, $a = 4$, so directly $c - a < x < c + a$, i.e. $-1 < x < 7$. Solution set: $(-1, 7)$. …