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.