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

[tronter]

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

So a \leq b and -a \leq b. Then -b \leq a so that -b \leq a \leq b.

Suppose that -b \leq a \leq b. Then a \leq b and -a \leq b so that |a| \leq b.

Is this a correct proof? You don't have to consider cases (e.g. a <0, \ a = 0, \ a > 0)?

 

[Dick]

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.

 

[tronter]

So if you know that already, then it is ok to deduce from -b \leq a \leq b that a \leq b and -a \leq b? The sign does not matter?

 

[Dick]

That's definitely ok. That's what -b<=a<=b MEANS. I'm talking about saying |a|<=b implies -a<=b and a<=b.

 

[tronter]

So if we have -b \leq a \leq b and a = -5 then we can say that -a \leq b?

Similarly, if a = 5 then -a \leq b?

It doesn't matter what the sign is?

Thanks

 

[Dick]

I'm saying I think you should PROVE |a|<=b implies -a<=b and a<=b by splitting it into cases.

 

[tronter]

but if you already know that then post #5 doesn't depend on the sign of a?

thanks

 

[Dick]

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.