# Proving |a|≤ b iff -b≤|a|≤ b

Hi, I'm working through "Passage to Abstract Mathematics," Watkins and Meyer, over break. I think I have some idea on how to approach this proof, but I'm not very confident with my work thus far.

## Homework Statement

Let $b \geq 0$. Prove that $|a| \leq b \Leftrightarrow -b \leq a \leq b$.

## Homework Equations

The definition for the absolute value function provided in the book is
$|x| = \begin{cases} x, & \mbox{if } x \geq 0 \\ -x, & \mbox{if } x < 0 \end{cases}$

I have a relevant side question. Prior to this section, the book covers some symbolic logic and logical equivalence. Am I to understand this definition as an "and" or an "or" definition? Meaning,
$|x| := (a \: \mbox{for} \: a \geq 0) \wedge (-a \: \mbox{for} \: a < 0)$, or
$|x| := (a \: \mbox{for} \: a \geq 0) \vee (-a \: \mbox{for} \: a < 0)$

## The Attempt at a Solution

(1) Left to right: Assume $|a| \leq b$.

$|a|$ is defined to be
$|a| = \begin{cases} a, & \mbox{if } a \geq 0 \\ -a, & \mbox{if } a < 0 \end{cases}$.

If $a \geq 0$, we have
$|a| = a \leq b$.

If $a < 0$, we have
$|a| = -a \leq b$, or equivalently,
$a \geq -b$.

From here, I conclude that $-b \leq a \leq b$, but I'm not sure if this would be accepted if this were actually assigned. I have a feeling that I should also show what happens when b = 0 and b > 0, but I'm not sure it's necessary.

(2) Right to left: Assume $-b \leq a \leq b$.

Suppose $b = 0$.
$-0 \leq a \leq 0 \Rightarrow a = 0 \Rightarrow |a| = b$​

Suppose $b > 0$.
1. $a = 0 \Rightarrow |a| = 0 \Rightarrow |a| < b$
2. $a > 0 \Rightarrow |a| = a > 0 \Rightarrow 0 < |a| < b \Rightarrow |a| < b$
3. $a < 0 \Rightarrow |a| = -a > 0 \Rightarrow |a| < b$

Therefore, for $b \geq 0$, $|a| \leq b$.

Thanks for any input!