# 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.