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.


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 $2a<0$.

| cite | improve this answer | |
  • $\begingroup$ 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

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 $$ a - |(a-|a|)|=a-|a|+a=2a-|a|=\begin{cases} a&\text{if}\quad a\ge0\\ 3a&\text{if}\quad a<0 \end{cases} $$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.