Answer
See explanation
Work Step by Step
**Proof:**
Let \( r \) be any real number and \( c \) be any positive real number (\( c > 0 \)). Suppose that
\[
|r| \le c.
\]
We want to show that \(-c \le r \le c\).
By the definition of absolute value:
- **Case 1:** If \( r \ge 0 \), then \(|r| = r\).
The inequality \(|r| \le c\) becomes
\[
r \le c.
\]
And since \( r \ge 0 \) and \( c > 0 \), it is clear that
\[
-c \le r.
\]
Hence, in this case, we have
\[
-c \le r \le c.
\]
- **Case 2:** If \( r < 0 \), then \(|r| = -r\).
The inequality \(|r| \le c\) becomes
\[
-r \le c.
\]
Multiplying both sides of this inequality by \(-1\) (and reversing the inequality sign, since \(-1\) is negative) gives:
\[
r \ge -c.
\]
Additionally, since \( r < 0 \) and \( c > 0 \), it follows that
\[
r \le c.
\]
Therefore, in this case too, we have
\[
-c \le r \le c.
\]
Since both cases lead to the conclusion that
\[
-c \le r \le c,
\]
we have shown that for all real numbers \( r \) and \( c \) with \( c > 0 \), if \(|r| \le c\) then \(-c \le r \le c\).
This completes the proof.
