Homework Help: Absolute value

1. Oct 6, 2008

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. $$a <0, \ a = 0, \ a > 0$$)?

2. Oct 6, 2008

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.

3. Oct 6, 2008

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?

4. Oct 6, 2008

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.

5. Oct 6, 2008

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

6. Oct 6, 2008

Dick

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

7. Oct 6, 2008

tronter

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

thanks

8. Oct 6, 2008

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.