Thanks for your help!Proving the Equivalence of |a|≤ b and -b≤|a|≤ b

[SithsNGiggles]

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!

 

[kru_]

For the first part, you might want to show the step between your use of the definition and your conclusion. For example:

b ≥ 0 => -b ≤ 0 => -b ≤ 0 ≤ b. Now assume |a| ≤ b, if a ≥ 0 then |a| = a and we have 0 ≤ |a| = a ≤ b => -b ≤ 0 ≤ a ≤ b => -b ≤ a ≤ b, which is what was desired.

Same for a < 0.

 

[SithsNGiggles]

Oh, I see. Thanks. But other than that, everything else is fine?