Absolute Value Inequalities
The absolute value rule makes it possible to rewrite an inequality containing a modulus as an equivalent inequality without absolute value symbols.
- An inequality of the form \[ |A| < B \quad \text{with } B > 0 \] is equivalent to \[ -B < A < B \] In other words, the expression \( A \) must lie between \( -B \) and \( +B \).
- An inequality of the form \[ |A| > B \quad \text{with } B > 0 \] is equivalent to \[ A < -B \quad \text{or} \quad A > B \] That is, \( A \) must lie outside the interval \( (-B, B) \).
The condition $ B > 0 $ is essential because the absolute value $ |A| $ is always greater than or equal to zero.
A Practical Example
Example 1
Solve the inequality
\[ |x - 2| < 3 \]
Apply the rule
\[ |A| < B \iff -B < A < B \]
Here, \( A = x - 2 \) and \( B = 3 \). Therefore:
\[ -3 < x - 2 < 3 \]
Add 2 to each part of the inequality:
\[ -3 +2 < x - 2 +2 < 3 +2 \]
\[ -1 < x < 5 \]
Thus, the inequality is satisfied for all values of \( x \) between -1 and 5.
Example 2
Consider the inequality
\[ |x + 1| > 2 \]
Apply the rule
\[ |A| > B \iff A < -B \ \text{or} \ A > B \]
In this case, \( A = x + 1 \) and \( B = 2 \).
\[ x + 1 < -2 \quad \text{or} \quad x + 1 > 2 \]
Solve each branch:
- \( x + 1 < -2 \Rightarrow x < -3 \)
- \( x + 1 > 2 \Rightarrow x > 1 \)
Therefore, the solutions are:
\[ x < -3 \quad \text{or} \quad x > 1 \]
Equivalently, the solution set consists of all values of \( x \) outside the interval \( (-3, 1) \).
The Proof
To justify the absolute value rule for inequalities, begin with the definition of absolute value:
\[ |A| = \begin{cases} A \quad \text{if } A \ge 0 \\ \\ -A \quad \text{if } A < 0 \end{cases} \]
Let \( A \) and \( B \) be real numbers and assume \( B > 0 \).
1) Proof of \( |A| < B \iff -B < A < B \)
Because this is a double implication $ \iff $, we must prove both the forward implication (⇒) and the reverse implication (⇐).
Forward implication (⇒)
Assume \( |A| < B \) and consider the two cases \( A \ge 0 \) and \( A < 0 \).
- Case 1: \( A \ge 0 \)
If \( A \ge 0 \), then \( |A| = A \). Hence \[ |A| < B \Rightarrow A < B \] Since \( A \ge 0 \) and \( B > 0 \), it follows that \( A > -B \). Therefore, \[ -B < A < B \] - Case 2: \( A < 0 \)
If \( A < 0 \), then \( |A| = -A \). Hence \[ |A| < B \Rightarrow -A < B \] Multiply by \(-1\) and reverse the inequality: \[ A > -B \] Because \( A < 0 \) and \( B > 0 \), we also have \( A < B \). Thus, \[ -B < A < B \]
In both cases we obtain
\[ |A| < B \Rightarrow -B < A < B \]
Reverse implication (⇐)
Assume \( -B < A < B \) and again consider the cases \( A \ge 0 \) and \( A < 0 \).
- Case 1: \( A \ge 0 \)
Then \( |A| = A \). From \( A < B \), we immediately obtain \[ |A| < B \] - Case 2: \( A < 0 \)
Then \( |A| = -A \). From \( -B < A \), multiply by \(-1\): \( B > -A \), that is \( -A < B \). Hence \[ |A| < B \]
In conclusion,
$$ -B < A < B \Rightarrow |A| < B $$
Combining the two implications:
\[ |A| < B \iff -B < A < B \quad (B > 0) \]
The rule is proven.
2) Proof of \( |A| > B \iff A < -B \ \text{or} \ A > B \)
We now prove the statement in both directions.
Assume throughout that $ B > 0 $.
Forward implication (⇒)
Suppose \( |A| > B \).
- Case 1: \( A \ge 0 \)
Then \( |A| = A \), so \[ |A| > B \Rightarrow A > B \] - Case 2: \( A < 0 \)
Then \( |A| = -A \), so \[ |A| > B \Rightarrow -A > B \] Multiply by \(-1\) and reverse the inequality: \[ A < -B \]
Therefore,
\[ |A| > B \Rightarrow A < -B \ \text{or} \ A > B \]
Reverse implication (⇐)
Consider the alternatives \( A > B \) and \( A < -B \).
- Case 1: \( A > B \)
Since \( B > 0 \), we have \( A > 0 \). Thus \( |A| = A \), and $$ A > B \Rightarrow |A| > B. $$ - Case 2: \( A < -B \)
Since \( B > 0 \), we have \( A < 0 \). Thus \( |A| = -A \). From $$ A < -B $$ multiply by \(-1\): $$ -A > B. $$ Hence $$ |A| > B. $$
In both cases:
\[ A > B \ \text{or} \ A < -B \Rightarrow |A| > B \]
Combining forward and reverse implications:
\[ |A| > B \iff A < -B \ \text{or} \ A > B \]
As required.
And so on.
