# Proving ||z|-|w|| <= |z-w|

• Nov 24th 2010, 03:24 PM
Pedro
Proving ||z|-|w|| <= |z-w|
hi, I stumbled upon a simple demonstration but i canīt solve it:

||z|-|w|| <= |z-w|

Can someone help me?

Pedro
• Nov 24th 2010, 03:29 PM
Prove It
Quote:

Originally Posted by Pedro
hi, I stumbled upon a simple demonstration but i canīt solve it:

||z|-|w|| <= |z-w|

Can someone help me?

Pedro

This is the reverse triangle inequality.
• Nov 24th 2010, 04:25 PM
Pedro
ok, thank you!
• Nov 25th 2010, 08:37 AM
Plato
This proof relies on this fact: $a \geqslant 0\;\& \, - a \leqslant b \leqslant a\, \Leftrightarrow \,\left| b \right| \leqslant \left| a \right|$.

So $\left| u \right| \leqslant \left| {u - w} \right| + \left| w \right|\, \Rightarrow \,\left| u \right| - \left| w \right| \leqslant \left| {u - w} \right|$

Likewise $\left| w \right| - \left| u \right| \leqslant \left| {w - u} \right| = \left| {u - w} \right|$

Thus we now have $- \left( {\left| {u - w} \right|} \right) \leqslant \left( {\left| u \right| - \left| w \right|} \right) \leqslant \left( {\left| {u - w} \right|} \right)$

Can you use the fact to finish?
• Nov 27th 2010, 04:07 PM
rcs
i understood what is discussed above... but i would like know how to explain and admit that
-|a| ≤ a ≤ |a|

and

-|b| ≤ b ≤ |b|

can anybody help me to prove this?
all i know is that a negative is of course less than a positive number
• Nov 27th 2010, 04:35 PM
Plato

You said that you understand that $a \geqslant 0\;\& \, - a \leqslant b \leqslant a\, \Leftrightarrow \,\left| b \right| \leqslant \left| a \right| .$

If you do, can we say that $|a|\le ||a||~?$

If so then is it not true that $-|a|\le a \le |a|~?$
• Nov 27th 2010, 05:33 PM
Prove It
The fact is, that any nonnegative number is equal to its absolute value. So if $\displaystyle a \geq 0$, then $\displaystyle a = |a|$.

However, any negative number is less than its absolute value (since the absolute value is always nonnegative). So if $\displaystyle a < 0$, then $\displaystyle a < |a|$.

So that means for any $\displaystyle a$ that $\displaystyle a \leq |a|$.

A similar argument works the other way to show $\displaystyle -|a| \leq a$.

So that means $\displaystyle -|a| \leq a \leq |a|$.
• Nov 27th 2010, 11:16 PM
rcs
thanks you! so same in the -|b| ≤ b ≤ |b|.