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So the question asks to express $\, a - |(a-|a|)|$ without using absolute value signs and instead with in multiple cases.

I feel as though I'm on the right track but something just isn't clicking in my head on how to properly do this.

My attempt so far has gone like this.

$a - |(a - |a|)|$

Two cases:

$|a - |a|| < 0 \quad or \quad |a-|a|| >0 $

which then each simplify to

$-a + |a| \quad and \quad a-|a|$

so the whole expression becomes

$a + a - |a| \quad and \quad a -a + |a|$

from here both of these have two cases

$a < 0 \quad then \quad a + a + a =3a$

$a > 0 \quad then \quad a + a - a =a$

$a < 0 \quad then \quad a - a - a =-a$

$a > 0 \quad then \quad a - a + a =a$

The answer given is that: $ a \;\;if\;\; a>0;\quad 3a\;\;if\;\; a<0$

I see how the first interval makes sense, but I don't how I can get $3a$ to appear in the 3rd case since the $a -a$ seems to make it necessarily either $-a$ or $a$.

Thank you in advance for any help.

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Note that we always have $|a-|a|| \ge 0$.

If $a \ge 0$, $\, a - |(a-|a|)|=a-|a-a|=a-0=a$

If $a <0$, $a-|a-|a||=a-|a-(-a)|=a-|2a|=a-(-2a)=a+2a=3a$ where in the second case I have used the property that if $a<0$ then $2a<0$.

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  • $\begingroup$ Thank you! This is much clearer than my attempt and I see what I messed up on $\endgroup$ – peaches Apr 4 '18 at 22:38
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The key inequality is that $$ -|a|\leq a\le|a| $$ where $a\in\mathbb{R}$. In particular $$ a-|a|\leq 0\implies|a-|a||=-(a-|a|)=|a|-a $$ so $$ a - |(a-|a|)|=a-|a|+a=2a-|a|=\begin{cases} a&\text{if}\quad a\ge0\\ 3a&\text{if}\quad a<0 \end{cases} $$

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  • $\begingroup$ Thanks for the help! I see where I went wrong now $\endgroup$ – peaches Apr 4 '18 at 22:40

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