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Let $\text{abs}(a)$ denote the absolute value of $a$. Is it true that $\text{abs}(a)\geq{-a}$? I suppose that $\text{abs}(a)>{-a}$, but my math book says the other way. Please help me to understand is it a misprint in my book, or my misunderstanding. Thank you in advance.

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    $\begingroup$ If $a\leq 0$ then $\mathrm{abs}(a)=-a$. $\endgroup$ – Thomas Andrews Jan 2 '15 at 13:22
  • $\begingroup$ For non-real complex numbers $a$ one cannot even compare $a$ and $\operatorname{abs}(a)=|a|$. $\endgroup$ – Marc van Leeuwen Jan 2 '15 at 18:27
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Consider the example of $a=0$. Then $\operatorname{abs}(a) = -a$.

Or consider the example of $a = -1$. Then $\operatorname{abs}(a) = -a = 1$. Similarly, $\operatorname{abs}(a) = -a$ whenever $a<0$.

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  • $\begingroup$ Oh poor zero always forget him ;) Thank you! $\endgroup$ – dimaastronom Jan 2 '15 at 13:07
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    $\begingroup$ Don't forget zero! Always think of him first! $\endgroup$ – MJD Jan 2 '15 at 13:13
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    $\begingroup$ Actually, it is for all $a\leq 0$ that $\mathrm{abs}(a)=-a$. @dimaastronom $\endgroup$ – Thomas Andrews Jan 2 '15 at 13:23
  • $\begingroup$ Dont understand with -1. Abs(-1)=1 $\endgroup$ – dimaastronom Jan 2 '15 at 18:33
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    $\begingroup$ If $a=-1$, both $\operatorname{abs}(a)$ and $-a$ are equal to $1$. $\endgroup$ – MJD Jan 2 '15 at 19:38
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yes, it's correct - if $a\leq 0$, then $|a|=-a$, and the inequality $|a|\geq -a$ holds.

if $a>0$, then $-a<0$, and so $|a|>0>-a$.

either way, the inequality $|a|\geq -a$ holds.

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We have

$$\operatorname{abs}(a)=\max(a,-a)=\left\{\begin{array}{cl}a\;&\text{if}\; a\ge0\\-a\;&\text{otherwise}\end{array}\right.$$

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The $abs$ function is defined by:

$\forall{x}\in\mathbb{R},\,abs(x)=|x|=\left\{ \begin{array}{lr} x & : x\ge0\\ -x & : x <0 \end{array} \right.$

So $\forall x\in\mathbb{R},\,|x|\ge0$

Let $a\in\mathbb{R}$.

If $a\ge0$ then $|a|=a$ and so $a\le|a|$

If $a<0$ then $|a|=-a>0>a$ and so $a\le|a|$

Now, if $a\ge0$ then $-a\le0\le|a|$

If $a<0$ then $-a=|a|$ and so $-a\le|a|$.

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