We are tasked with proving that if $a \geq 0$, then the statement $|x| \leq a$ is equivalent to $-a \leq x \leq a$.

Proof:

1. Implication $|x| \leq a \implies -a \leq x \leq a$:

Start by assuming that $|x| \leq a$, where $a \geq 0$. The absolute value $|x|$ is defined as:

\begin{cases} x, & \text{if } x \geq 0, \\ -x, & \text{if } x < 0. \end{cases}$$ The inequality $$|x| \leq a$$ means that the magnitude (or distance) of $$x$$ from zero is less than or equal to $$a$$. Thus, we have two cases to consider: - If $$x \geq 0$$, then $$|x| = x$$. Therefore, $$x \leq a$$. - If $$x < 0$$, then $$|x| = -x$$. Therefore, $$-x \leq a$$, which implies $$x \geq -a$$. Combining both cases, we conclude that $$-a \leq x \leq a$$. #### 2. **Implication $$-a \leq x \leq a \implies |x| \leq a$$:** Now, assume that $$-a \leq x \leq a$$. We will show that this implies $$|x| \leq a$$. - If $$x \geq 0$$, then $$|x| = x$$, and since $$x \leq a$$, we have $$|x| \leq a$$. - If $$x < 0$$, then $$|x| = -x$$, and since $$x \geq -a$$, this means $$-x \leq a$$, or $$|x| \leq a$$. Thus, in both cases, we conclude that $$|x| \leq a$$. ### Conclusion: Since both directions of the implication are true, we have shown that: $$|x| \leq a \iff -a \leq x \leq a.$$ This completes the proof. --- Would you like further clarification on any step? Here are 5 related questions to deepen your understanding: 1. How is the absolute value function defined for real numbers? 2. What is the geometric interpretation of $$|x| \leq a$$? 3. How can you visualize $$|x| \leq a$$ on a number line? 4. What happens if $$a < 0$$? Does the proof still hold? 5. Can you generalize this result for other inequalities involving absolute value? **Tip:** Always remember that inequalities involving absolute values are usually two-sided and can be broken down into simpler inequalities.