We are tasked with proving that for $a \geq 0$, the statement $|x| \leq a$ is true if and only if $-a \leq x \leq a$.

Part 1: Proving $|x| \leq a \implies -a \leq x \leq a$

Step 1: Definition of Absolute Value

By definition, the absolute value of $x$ is given as:

\begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$$ So, $$|x| \leq a$$ means that the magnitude of $$x$$ is no larger than $$a$$. #### Step 2: Consider Two Cases Based on $$x$$ 1. **Case 1**: If $$x \geq 0$$, then $$|x| = x$$. The condition $$|x| \leq a$$ becomes: $$x \leq a$$ Since $$x \geq 0$$, combining these gives: $$0 \leq x \leq a$$ 2. **Case 2**: If $$x < 0$$, then $$|x| = -x$$. The condition $$|x| \leq a$$ becomes: $$-x \leq a \implies x \geq -a$$ Since $$x < 0$$, combining these gives: $$-a \leq x < 0$$ #### Step 3: Combine Both Cases From both cases, we conclude that $$|x| \leq a$$ implies: $$-a \leq x \leq a$$ ### Part 2: Proving $$-a \leq x \leq a \implies |x| \leq a$$ Assume that $$-a \leq x \leq a$$. #### Step 1: Definition of Absolute Value We want to show that $$|x| \leq a$$. Again, by the definition of absolute value: $$|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$$ #### Step 2: Consider Two Cases Based on $$x$$ 1. **Case 1**: If $$x \geq 0$$, then $$|x| = x$$. Since $$-a \leq x \leq a$$, we have $$x \leq a$$, so: $$|x| = x \leq a$$ 2. **Case 2**: If $$x < 0$$, then $$|x| = -x$$. Since $$-a \leq x \leq a$$, we have $$x \geq -a$$, which implies: $$-x \leq a \implies |x| = -x \leq a$$ #### Step 3: Conclusion In both cases, we find that $$|x| \leq a$$. Thus, $$-a \leq x \leq a$$ implies $$|x| \leq a$$. ### Final Conclusion: Since both directions have been proven, we conclude that: $$|x| \leq a \iff -a \leq x \leq a$$ for $$a \geq 0$$. Would you like more details or have any further questions? ### Related Questions: 1. How does the definition of absolute value affect inequalities? 2. Can the inequality $$|x| \leq a$$ have negative solutions? 3. What happens when $$a < 0$$ in this context? 4. How would this proof change for $$|x| < a$$? 5. Can similar reasoning be applied to functions of $$x$$, not just $$x$$ itself? ### Tip: When solving absolute value inequalities, it's helpful to always break the problem into cases based on the definition of the absolute value function.