Proving Absolute Value Inequality: |a| ≤ b → -b ≤ a ≤ b (b≥0)

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Homework Help Overview

The discussion revolves around proving the absolute value inequality |a| ≤ b leading to the conclusion that -b ≤ a ≤ b, under the condition that b ≥ 0. Participants are examining the logical connections between the statements and the necessity of case analysis in the proof.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the proof can be established without considering cases for a, questioning if the implications of the absolute value can be directly applied. There is discussion about the meaning of the inequalities and whether the sign of a affects the reasoning.

Discussion Status

The conversation is ongoing, with some participants affirming the validity of the proof under certain assumptions, while others suggest that a case-based approach may be necessary for clarity. There is no explicit consensus, but productive dialogue is occurring regarding the implications of the absolute value and the conditions of the proof.

Contextual Notes

Participants are considering the implications of the absolute value inequality and whether the proof requires case analysis based on the value of a. The discussion also touches on the assumption that b is non-negative.

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Prove the following: if [tex]|a| \leq b[/tex] then [tex]-b \leq a \leq b[/tex] (where [tex]b \geq 0[/tex]).

So [tex]a \leq b[/tex] and [tex]-a \leq b[/tex]. Then [tex]-b \leq a[/tex] so that [tex]-b \leq a \leq b[/tex].

Suppose that [tex]-b \leq a \leq b[/tex]. Then [tex]a \leq b[/tex] and [tex]-a \leq b[/tex] so that [tex]|a| \leq b[/tex].

Is this a correct proof? You don't have to consider cases (e.g. [tex]a <0, \ a = 0, \ a > 0[/tex])?
 
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It's fine if you already know |a|<=b implies a<=b and -a<=b. If you don't know that you might have to use cases to prove that.
 
So if you know that already, then it is ok to deduce from [tex]-b \leq a \leq b[/tex] that [tex]a \leq b[/tex] and [tex]-a \leq b[/tex]? The sign does not matter?
 
That's definitely ok. That's what -b<=a<=b MEANS. I'm talking about saying |a|<=b implies -a<=b and a<=b.
 
So if we have [tex]-b \leq a \leq b[/tex] and [tex]a = -5[/tex] then we can say that [tex]-a \leq b[/tex]?

Similarly, if [tex]a = 5[/tex] then [tex]-a \leq b[/tex]?

It doesn't matter what the sign is?

Thanks
 
I'm saying I think you should PROVE |a|<=b implies -a<=b and a<=b by splitting it into cases.
 
but if you already know that then post #5 doesn't depend on the sign of [tex]a[/tex]?

thanks
 
If you already know that |a|<=b implies -a<=b and a<=b then this whole thread should have been over at post 2. I already said the proof is fine in that case.
 

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