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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8If $a\leq 0$ then $\mathrm{abs}(a)=-a$. – Thomas Andrews Jan 02 '15 at 13:22
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For non-real complex numbers $a$ one cannot even compare $a$ and $\operatorname{abs}(a)=|a|$. – Marc van Leeuwen Jan 02 '15 at 18:27
4 Answers
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
$$\operatorname{abs}(a)=\max(a,-a)=\left\{\begin{array}{cl}a\;&\text{if}\; a\ge0\\-a\;&\text{otherwise}\end{array}\right.$$
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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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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12Actually, it is for all $a\leq 0$ that $\mathrm{abs}(a)=-a$. @dimaastronom – Thomas Andrews Jan 02 '15 at 13:23
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Yeah its obvious sorry for the stupid question, did everything on the phone in subway, could not think clear ;) – dimaastronom Jan 02 '15 at 20:25
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