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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
 

Answers and Replies

1,341
3
Case 1:
[tex]-|x| \leq x[/tex]

[tex]-x \leq -x \leq x[/tex]

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
 
1,341
3
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
Science Advisor
Homework Helper
41,732
893
The simple way to do it is to use the definition: |x|= x if [itex]x\ge 0[/itex], -x is x< 0.

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

If x> 0, then [itex]-|x|\le x\le |x|[/itex] 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
 
1,631
4
Th

If x> 0, then [itex]-|x|\le x\le |x|[/itex] 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
 

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