Proving -|x|</x</|x| Homework Statement

[chocolatelover]

Homework Statement

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

Homework Equations

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

 

[Feldoh]

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?

 

[chocolatelover]

Thank you very much

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

Thank you

 

[Feldoh]

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

 

[chocolatelover]

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

 

[HallsofIvy]

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 x\le x\le -x[/itex]. Isn&#039;t that obvious?

 

[chocolatelover]

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

 

[sutupidmath]

Th

If x> 0, then -|x|\le x\le |x| becomes x\le x\le -x[/itex]. Isn&#039;t that obvious?

<br /> <br /> there is a typo here, it should read i guess x&lt;0. I just wanted to point it out.

 

[chocolatelover]

Thank you very much

Regards