If $x$ is a real number ($x \in \mathbb{R}$), its modulus is denoted by $|x|$ and is defined piecewise.
The formal definition is:
$|2| = 2$
$|0| = 0$
$|-2| = -(-2) = 2$
The modulus function satisfies several important properties:
Zero Property: $|x|=0$ if and only if $x=0$.
Positive Case: $|x|=x$, if $x \ge 0$.
Negative Case: $|x|=-x$, if $x < 0$.
Power: $|x^n| = |x|^n$.
Reflection: $|-x| = |x|$.
A crucial property, especially in advanced mathematics, is the Triangle Inequality:
This inequality shows that the distance of the sum of two numbers is less than or equal to the sum of their individual distances from zero.
Equality holds (i.e., $|x+y| = |x| + |y|$) when $xy \ge 0$ (when $x$ and $y$ have the same sign or one is zero).
Example: $|2+3| = |5| = 5$, and $|2| + |3| = 2 + 3 = 5$.
Example: $|-2+(-3)| = |-5| = 5$, and $|-2| + |-3| = 2 + 3 = 5$.
Strict inequality holds (i.e., $|x+y| < |x| + |y|$) when $xy < 0$ (when $x$ and $y$ have opposite signs).
Example: $|-1+2| = |1| = 1$, and $|-1| + |2| = 1 + 2 = 3$. Here $1 < 3$.
The modulus helps define intervals when solving inequalities:
If the modulus of $x$ is less than a positive number $a$ (i.e., $a>0$), then $x$ must lie between $-a$ and $a$.
This means $x$ is in the open interval $(-a, a)$.
If the modulus of $x$ is less than or equal to a positive number $a$, then $x$ must lie in the closed interval $[-a, a]$.
The graph of the basic modulus function, $y = |x|$, is a distinctive V-shape.
| Input ($x$) | Output ($|x|$) |
| :—: | :—: |
| $0$ | $0$ |
| $1$ | $1$ |
| $-1$ | $1$ |
| $2$ | $2$ |
| $-2$ | $2$ |
Vertex: The minimum point is at the origin $(0, 0)$.
Right Side ($x \ge 0$): The graph follows the line $y = x$ (slope is 1).
Left Side ($x < 0$): The graph follows the line $y = -x$ (slope is $-1$).
This V-shape perfectly illustrates the function’s property: any non-zero input, whether positive or negative, results in a positive output.
The general rule for solving equations involving the modulus is to remove the absolute value signs by considering both the positive and negative possibilities of the expression inside.
| Rule | Statement | Explanation |
| Equation 1 | $ | x |
| Equation 2 | $ | x |
When the absolute value is less than a positive number, the solution is a single, bounded interval.
If $a$ is a positive number, then:
The solution set is the open interval $x \in (-a, a)$.
If $a$ is a positive number, then:
The solution set is the closed interval $x \in [-a, a]$.
with solid dots at -a and a]
When the absolute value is greater than a positive number, the solution is two separate, unbounded intervals connected by “or.”
If $a$ is a positive number, then:
The solution set is the union of two open intervals: $x \in (-\infty, -a) \cup (a, \infty)$.
If $a$ is a positive number, then:
The solution set is the union of two intervals, where the endpoints are included: $x \in (-\infty, -a] \cup [a, \infty)$.
When dealing with more complex expressions like $|2x-3|$, we use the definition of modulus to break the problem into cases based on the sign of the inner expression.
To define $|2x-3|$, we find the critical point where $2x-3 = 0$, which is $x = \frac{3}{2}$.