What is Absolute Value?
The absolute value of a number, written |x|, is its distance from zero on the number line — always a non-negative result. The formal definition is:
- |x| = x if x ≥ 0
- |x| = −x if x < 0
So |7| = 7 (already positive), |−7| = 7 (strip the negative sign), and |0| = 0 (zero is zero).
The concept of absolute value is fundamental to algebra, calculus, and analysis. It appears wherever you need the magnitude of a quantity without regard to its direction or sign: distance between points, error in a measurement, deviation from a mean, and the magnitude of a complex number.
The notation |x| was introduced by Karl Weierstrass in 1841. Before that, mathematicians used phrases like “the modulus” (still common in British usage: the modulus of a number). For complex numbers z = a + bi, the modulus |z| = √(a² + b²) generalizes the same idea to two dimensions.
Definition:
|x| = x (if x ≥ 0) or −x (if x < 0)
Distance interpretation:
|x − a| = distance from x to a on the number line
Solving |ax + b| = c (c > 0):
Case 1: ax + b = c → x = (c − b) / a
Case 2: ax + b = −c → x = (−c − b) / a
Solving |ax + b| < c:
−c < ax + b < c → (−c − b)/a < x < (c − b)/a (flip inequalities if a < 0)
Solving |ax + b| > c:
ax + b > c OR ax + b < −c → x > (c − b)/a OR x < (−c − b)/a
How to Use
- Single Number mode — type any real number and instantly see |x|, the sign of x, and the negation −x.
- Expression mode — type any numeric expression using |…| bars, such as
|-4| + |2 - 9| or |3*5 - 20|. The calculator evaluates the whole expression. - Inequality mode — enter coefficients a and b for |ax + b|, choose an operator (<, ≤, >, ≥), and enter the right-hand side c. The solution is shown in interval notation with step-by-step working.
- Read the steps — the working panel shows how the sign is determined or how the compound inequality is split and solved.
- Interpret the interval — for inequalities, a bounded interval like (1, 7) means all x strictly between 1 and 7; a union like (−∞, 1) ∪ (7, +∞) means all x outside the interval.
Example Calculations
Example 1 — Absolute Value of a Negative Number
Find |−13.5|
1
−13.5 is negative, so |−13.5| = −(−13.5) = 13.5
|−13.5| = 13.5. The absolute value strips the sign, measuring the distance from 0.
Try this example →Example 2 — Absolute Value Inequality: |x − 4| < 3
Solve |x − 4| < 3 (a = 1, b = −4, c = 3)
1
|x − 4| < 3 means the distance from x to 4 is less than 3: −3 < x − 4 < 3
2
Add 4 throughout: 1 < x < 7
Solution: x ∈ (1, 7) — all numbers within distance 3 of 4.
Try this example →Example 3 — Absolute Value Inequality: |2x + 1| ≥ 5
Solve |2x + 1| ≥ 5 (a = 2, b = 1, op = ≥, c = 5)
1
Split into two cases: 2x + 1 ≥ 5 → 2x ≥ 4 → x ≥ 2
2
Second case: 2x + 1 ≤ −5 → 2x ≤ −6 → x ≤ −3
Solution: x ∈ (−∞, −3] ∪ [2, +∞)
Try this example →❓ Frequently Asked Questions
What is the absolute value of a number?+
The absolute value |x| is the distance from x to zero on the number line — always non-negative. Formally, |x| = x when x ≥ 0 and |x| = −x when x < 0. Examples: |8| = 8, |−8| = 8, |0| = 0. It gives the magnitude of a number without regard to its sign.
How do you solve an absolute value equation?+
For |ax + b| = c: if c < 0, there is no solution (absolute value is always ≥ 0). If c = 0, solve ax + b = 0. If c > 0, split into two equations: ax + b = c and ax + b = −c, giving two solutions x = (c − b)/a and x = (−c − b)/a. Always verify by substituting back into |ax + b| = c.
What is the difference between |x| < c and |x| > c?+
|x| < c means x is within distance c of zero: −c < x < c, an interval (a bounded set). |x| > c means x is more than distance c from zero: x < −c or x > c, a union of two rays (an unbounded set). Memorize: “less than → single interval (AND), greater than → union of two rays (OR).”
Can the absolute value ever be negative?+
No. |x| ≥ 0 for every real number x. This is a fundamental property — absolute value measures distance, and distance is never negative. A consequence: the equation |x + 3| = −5 has no solution, and the inequality |x| > −2 is true for all real x.
What is the triangle inequality for absolute value?+
The triangle inequality: |x + y| ≤ |x| + |y|. This says the absolute value of a sum is at most the sum of the absolute values. It is one of the most important inequalities in mathematics, appearing throughout analysis, linear algebra (as the norm inequality), and metric geometry. Equality holds when x and y have the same sign.
What are the key properties of absolute value?+
Six essential properties: (1) |x| ≥ 0 for all x. (2) |x| = 0 ⇔ x = 0. (3) |−x| = |x|. (4) |xy| = |x||y|. (5) |x/y| = |x|/|y| (y ≠ 0). (6) |x + y| ≤ |x| + |y| (triangle inequality). These properties make absolute value a norm on the real number line, satisfying the axioms of a metric space.
What does interval notation mean in an absolute value solution?+
Interval notation describes solution sets concisely. Parentheses ( ) mean the endpoint is excluded (strict inequality); brackets [ ] mean the endpoint is included. Example: (1, 7) means 1 < x < 7 (endpoints excluded). [−3, 5] means −3 ≤ x ≤ 5 (endpoints included). The union ∪ joins two separate intervals, as in (−∞, −3] ∪ [2, +∞).
How is absolute value used in statistics?+
The mean absolute deviation (MAD) is the average of |x⊂i; − mean| across all data points — a measure of spread that is more robust to outliers than standard deviation. The L1 norm (sum of absolute deviations) is used in median regression (least absolute deviations) and LASSO regularization. Absolute percentage error (APE) = |actual − forecast| / |actual| is standard in forecasting evaluation.
What is the absolute value of a complex number?+
For a complex number z = a + bi, the absolute value (or modulus) is |z| = √(a² + b²). This is the distance from z to the origin in the complex plane. Example: |3 + 4i| = √(9 + 16) = √25 = 5. Properties carry over: |z⊂1;z⊂2;| = |z⊂1;||z⊂2;| and |z⊂1; + z⊂2;| ≤ |z⊂1;| + |z⊂2;|.
How do you graph y = |x| and transformations of it?+
y = |x| is V-shaped: slope +1 for x ≥ 0, slope −1 for x < 0, vertex at (0, 0). Transformations: y = |x − h| + k shifts the vertex to (h, k). y = a|x − h| + k changes the slopes to ±a (steeper if |a| > 1, flatter if |a| < 1, reflected if a < 0). y = |2x − 4| has vertex at x = 2 (where 2x − 4 = 0) and slopes ±2.
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