Proving |a|≤ b iff -b≤|a|≤ b

  • #1
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 [itex]b \geq 0[/itex]. Prove that [itex]|a| \leq b \Leftrightarrow -b \leq a \leq b[/itex].

Homework Equations


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

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,
[itex]|x| := (a \: \mbox{for} \: a \geq 0) \wedge (-a \: \mbox{for} \: a < 0)[/itex], or
[itex]|x| := (a \: \mbox{for} \: a \geq 0) \vee (-a \: \mbox{for} \: a < 0)[/itex]

The Attempt at a Solution


(1) Left to right: Assume [itex]|a| \leq b[/itex].

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

If [itex]a \geq 0[/itex], we have
[itex]|a| = a \leq b[/itex].

If [itex]a < 0[/itex], we have
[itex]|a| = -a \leq b[/itex], or equivalently,
[itex] a \geq -b[/itex].

From here, I conclude that [itex]-b \leq a \leq b[/itex], 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 [itex]-b \leq a \leq b[/itex].

Suppose [itex]b = 0[/itex].
[itex]-0 \leq a \leq 0 \Rightarrow a = 0 \Rightarrow |a| = b[/itex]​

Suppose [itex]b > 0[/itex].
  1. [itex]a = 0 \Rightarrow |a| = 0 \Rightarrow |a| < b[/itex]
  2. [itex]a > 0 \Rightarrow |a| = a > 0 \Rightarrow 0 < |a| < b \Rightarrow |a| < b[/itex]
  3. [itex]a < 0 \Rightarrow |a| = -a > 0 \Rightarrow |a| < b[/itex]

Therefore, for [itex]b \geq 0[/itex], [itex]|a| \leq b[/itex].

Thanks for any input!
 

Answers and Replies

  • #2
85
0
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.
 
  • #3
Oh, I see. Thanks. But other than that, everything else is fine?
 

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