# Prove -|x|</x</|x|

1. Homework Statement
Prove -|x|</x</|x|

2. Homework Equations

3. The Attempt at a Solution

We are trying to prove -|x|</x</|x| (less than or equal to)

This means that we need to show that x>/ -|x| and x</ |x|.

Let x be a real number.

Assume |x|=x if x>0 and -x if x<0

I'm not sure were to go from here, as I cannot use actual values to prove it. Could someone please show me what to do?

Thank you very much

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Case 1:
$$-|x| \leq x$$

$$-x \leq -x \leq x$$

So it's obvious that case 1 holds..

Can you see where to go form there?

Thank you very much

Could you please explain to me how you got that? Would I do the same thing for |x|?

Thank you

It's just basic a absolute value identity, and yes you would^^

Thank you

After I do that, would that complete the proof?

|x|>x

x>x or -x>x

In order to prove that |x|</a if and only if -a</x</a where a>/0, would I do the same thing as in the previous problem?

Thank you

Last edited:
HallsofIvy
Homework Helper
The simple way to do it is to use the definition: |x|= x if $x\ge 0$, -x is x< 0.

If x$\ge$ 0, then $-|x|\le x\le |x|$ becomes $-x\le x\le x$. Isn't that obvious?

If x> 0, then $-|x|\le x\le |x|$ becomes [tex]x\le x\le -x[/itex]. Isn't that obvious?

Thank you very much

In order to prove that |x|</a if and only if -a</x</a where a>/0, would I do the same thing as in the previous problem?

Thank you

Th

If x> 0, then $-|x|\le x\le |x|$ becomes [tex]x\le x\le -x[/itex]. Isn't that obvious?
there is a typo here, it should read i guess x<0. I just wanted to point it out.

Thank you very much

Regards