Modulus Inequalities

In short

The absolute value |x| measures distance from zero. Modulus inequalities like |x| < a translate to -a < x < a (a band around zero), while |x| > a translates to x < -a or x > a (outside the band). For compound expressions, |f(x)| < |g(x)| is equivalent to f(x)^2 < g(x)^2, which avoids case-splitting. The triangle inequality |a + b| \leq |a| + |b| is the master inequality that governs all absolute-value reasoning.

A cricket pitch is 22 yards long. The umpire stands at the centre and wants every fielder within 15 yards of him. If the umpire is at position 0 on a number line running along the pitch, a fielder at position x satisfies the constraint when |x| \leq 15. What range of positions works? That is a modulus inequality — an inequality involving absolute value.

The absolute value |x| strips the sign and returns the magnitude: |3| = 3, |-3| = 3, |0| = 0. Geometrically, |x| is the distance from x to the origin on the number line. This distance interpretation is the key to every modulus inequality.

The two basic types

Every modulus inequality falls into one of two patterns.

Type 1: |x| < a (distance less than a). The points whose distance from 0 is less than a form the open interval (-a, a). Equivalently:

|x| < a \iff -a < x < a

This works only when a > 0. If a \leq 0, no real number has a non-negative absolute value less than a non-positive number, so the solution set is empty.

Type 2: |x| > a (distance greater than a). The points whose distance from 0 exceeds a lie outside the interval [-a, a]:

|x| > a \iff x < -a \text{ or } x > a

$|x| < 3$ selects the band within distance $3$ of the origin. $|x| > 3$ selects everything outside that band. Every modulus inequality is a variation on one of these two patterns.

These two patterns are the foundation. Every modulus inequality — no matter how complicated — eventually reduces to one of these.

Shifting the centre: |x - c| < a

The expression |x - c| measures the distance from x to the point c. So |x - c| < a means "the distance from x to c is less than a," which gives:

|x - c| < a \iff c - a < x < c + a

The solution is the interval (c - a, c + a), centred at c with radius a.

Take |x - 7| \leq 2. This says: x is within distance 2 of 7. The solution is [5, 9].

Take |2x - 5| < 3. First isolate the modulus: |2x - 5| < 3 means -3 < 2x - 5 < 3. Add 5: 2 < 2x < 8. Divide by 2: 1 < x < 4. The solution is (1, 4).

$|x - 7| \leq 2$ captures all points within distance $2$ of $7$: the interval $[5, 9]$.

Solving |f(x)| < |g(x)|: the squaring trick

When both sides of an inequality have absolute values, case-splitting gets messy — you would need four cases (both positive, both negative, mixed). There is a cleaner approach.

Since both sides are non-negative, you can square without changing the direction of the inequality:

|f(x)| < |g(x)| \iff f(x)^2 < g(x)^2 \iff f(x)^2 - g(x)^2 < 0 \iff (f - g)(f + g) < 0

The last expression is a polynomial inequality, and you can solve it with the sign scheme.

Take |x - 1| < |2x + 3|. Square both sides: (x-1)^2 < (2x+3)^2. Expand: x^2 - 2x + 1 < 4x^2 + 12x + 9. Rearrange: 0 < 3x^2 + 14x + 8. Factor: 3x^2 + 14x + 8 = (3x + 2)(x + 4).

The roots are x = -\frac{2}{3} and x = -4. The leading coefficient is positive, so the quadratic is positive outside the roots: x \in (-\infty, -4) \cup (-\frac{2}{3}, \infty).

Alternatively, using the (f-g)(f+g) < 0 form directly: f - g = (x-1) - (2x+3) = -x - 4 and f + g = (x-1) + (2x+3) = 3x + 2. So (-x-4)(3x+2) < 0, which is (x+4)(3x+2) > 0 (multiplying by -1 flips the inequality). Roots at x = -4 and x = -\frac{2}{3}, positive outside: (-\infty, -4) \cup (-\frac{2}{3}, \infty). Same answer.

The dashed V is $y = |x-1|$ and the solid V is $y = |2x+3|$. The inequality $|x-1| < |2x+3|$ holds where the dashed graph lies below the solid graph — in the shaded regions.

The triangle inequality

For any real numbers a and b:

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

Equality holds if and only if a and b have the same sign (or at least one is zero).

Why is this true? Consider a = 5 and b = -3. Then |a + b| = |2| = 2, while |a| + |b| = 5 + 3 = 8. The sum a + b benefits from cancellation (the positive 5 and negative -3 partially cancel), so |a + b| is smaller than the sum of the individual magnitudes.

When a and b have the same sign — say a = 5, b = 3 — there is no cancellation: |a + b| = 8 = |a| + |b|. Equality.

The name "triangle inequality" comes from geometry: in any triangle, the length of one side is at most the sum of the other two sides. On the number line, it says that the distance from 0 to a + b is at most the distance from 0 to a plus the distance from 0 to b. You might overshoot if a and b point in the same direction, but you can never exceed the total.

When $a$ and $b$ have opposite signs, partial cancellation makes $|a+b|$ strictly less than $|a| + |b|$. When they have the same sign, there is no cancellation and equality holds.

A useful companion: the reverse triangle inequality:

|a - b| \geq \big||a| - |b|\big|

This says the distance between two points is at least the difference of their distances from the origin.

Max and min using modulus

Absolute value gives clean formulas for the maximum and minimum of two real numbers:

\max(a, b) = \frac{a + b + |a - b|}{2}, \qquad \min(a, b) = \frac{a + b - |a - b|}{2}

Check with a = 7, b = 3: |a - b| = 4. Then \max = \frac{10 + 4}{2} = 7 and \min = \frac{10 - 4}{2} = 3. Correct.

Check with a = 2, b = 9: |a - b| = 7. Then \max = \frac{11 + 7}{2} = 9 and \min = \frac{11 - 7}{2} = 2. Correct.

Why do these work? The term \frac{a+b}{2} is the midpoint of a and b. The term \frac{|a-b|}{2} is half the gap between them. Adding half the gap to the midpoint gives the larger number; subtracting gives the smaller. This is a useful identity in optimization problems where you want to express max or min algebraically.

Interactive: visualise |x - c| < r

Drag the centre point c and the radius marker to see how the solution interval changes. The highlighted region is always symmetric around c.

The dark red point is the centre $c$; the lighter point sets the radius $r$. The solution to $|x - c| < r$ is the interval $(c - r, c + r)$, shown as the highlighted segment.

Two worked examples

Example 1: Solve $|2x - 3| \leq 5$

Step 1. Apply the Type 1 pattern. |2x - 3| \leq 5 translates to -5 \leq 2x - 3 \leq 5.

Why: |E| \leq a means -a \leq E \leq a. This is the definition of "distance from E to zero is at most a."

Step 2. Add 3 to all three parts: -2 \leq 2x \leq 8.

Why: isolating x requires removing the constant term inside the absolute value, step by step.

Step 3. Divide by 2: -1 \leq x \leq 4.

Why: dividing by a positive number preserves the inequality direction.

Step 4. Write the solution in interval notation: x \in [-1, 4].

Step 5. Verify the endpoints. At x = -1: |2(-1) - 3| = |-5| = 5 \leq 5. At x = 4: |2(4) - 3| = |5| = 5 \leq 5. Both satisfy the inequality. At x = -2 (outside): |2(-2) - 3| = |-7| = 7 > 5. Does not satisfy. Consistent.

Result: x \in [-1, 4].

The solution is the interval $[-1, 4]$, centred at $\frac{3}{2}$ with radius $\frac{5}{2}$. The absolute value inequality carves out a symmetric band around the point where the expression inside equals zero.

The centre of the interval is \frac{-1 + 4}{2} = \frac{3}{2}, which is the value of x that makes 2x - 3 = 0. The radius is \frac{5}{2}, which is \frac{a}{\text{coefficient of } x} = \frac{5}{2}. This pattern — centre at the zero of the inner expression, radius scaled by the coefficient — holds for all linear modulus inequalities.

Example 2: Solve $|x + 1| > |3x - 5|$

Step 1. Square both sides. Since both sides are non-negative, this preserves the inequality direction: (x + 1)^2 > (3x - 5)^2.

Why: for non-negative quantities A and B, A > B if and only if A^2 > B^2. Squaring avoids four separate cases.

Step 2. Expand: x^2 + 2x + 1 > 9x^2 - 30x + 25.

Step 3. Bring everything to one side: 0 > 8x^2 - 32x + 24, or equivalently 8x^2 - 32x + 24 < 0. Divide by 8: x^2 - 4x + 3 < 0.

Why: dividing by the positive number 8 does not change the inequality direction.

Step 4. Factor: (x - 1)(x - 3) < 0. The roots are x = 1 and x = 3. The quadratic opens upward (positive leading coefficient), so it is negative between the roots.

x \in (1, 3)

Step 5. Verify. At x = 2: |3| > |1|, i.e., 3 > 1. True. At x = 0: |1| > |-5|, i.e., 1 > 5. False. At x = 4: |5| > |7|, i.e., 5 > 7. False. Consistent.

Result: x \in (1, 3).

The solid V is $|x+1|$ and the dashed V is $|3x - 5|$. Between the intersection points $x = 1$ and $x = 3$, the solid graph lies above the dashed one — confirming the solution $(1, 3)$.

The squaring method converted a modulus inequality into a plain quadratic inequality in three lines. The graphical picture confirms: between x = 1 and x = 3, the shallower V (|x+1|) sits above the steeper V (|3x - 5|).

Common confusions

"|x| < -2 has the solution -2 < x < 2." The right-hand side is negative. An absolute value is always \geq 0, so |x| < -2 has no solution. Always check whether the bound a is positive before applying the pattern.
"|x| > a gives -a < x < a." This is backwards. |x| > a means x is far from zero, not close to zero. The solution is x < -a or x > a — two rays, not a single interval.
"I can remove the absolute value by just dropping the bars." Only if the expression inside is guaranteed non-negative. If x - 3 could be negative, then |x - 3| \neq x - 3 for those values. Either use the two-case definition or the squaring trick.
"Squaring both sides always works." It works when both sides are non-negative, which is guaranteed when both sides are absolute values. If one side is not an absolute value — say |x| > x - 2 — squaring may introduce false solutions. In that case, use case analysis or the definition directly.
"The triangle inequality is just |a + b| = |a| + |b|." Equality holds only when a and b have the same sign. In general, |a + b| is strictly less than |a| + |b| because of cancellation. The inequality, not the equality, is the default.

Going deeper

If you can handle |E| < a, |E| > a, and |f| < |g| via squaring, and you know the triangle inequality with its equality condition, you have the complete toolkit for modulus inequalities. What follows extends the ideas.

Nested absolute values

Expressions like ||x| - 3| < 2 arise occasionally. Apply the definition from outside in: ||x| - 3| < 2 means -2 < |x| - 3 < 2, so 1 < |x| < 5. Now |x| > 1 gives x < -1 or x > 1, and |x| < 5 gives -5 < x < 5. Intersecting: x \in (-5, -1) \cup (1, 5).

Triangle inequality with n terms

The triangle inequality extends by induction:

|a_1 + a_2 + \cdots + a_n| \leq |a_1| + |a_2| + \cdots + |a_n|

Equality holds when all the a_i have the same sign (or are zero). This generalisation is used in analysis and in bounding sums — you will meet it when studying convergence of series.

Connection to complex numbers

For a complex number z = a + bi, the modulus |z| = \sqrt{a^2 + b^2} generalises absolute value from the real line to the complex plane. The triangle inequality extends: |z_1 + z_2| \leq |z_1| + |z_2|, with equality when z_1 and z_2 point in the same direction in the complex plane. The modulus of a complex number article explores this further.

Where this leads next

Absolute Value Inequalities — the prerequisite article covering the definition and basic properties of absolute value.
Linear and Polynomial Inequalities — the sign scheme method that modulus inequalities often reduce to after squaring or case-splitting.
Rational Inequalities — another class of inequalities where sign analysis is the key technique.
Advanced Inequalities — AM-GM, Cauchy-Schwarz, and other powerful inequalities that build on the triangle inequality.
Modulus of a Complex Number — where absolute value extends from the number line to the complex plane.