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.

  • 8
    $\begingroup$ 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

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 $a<0$.

  • $\begingroup$ Oh poor zero always forget him ;) Thank you! $\endgroup$ – dimaastronom Jan 2 '15 at 13:07
  • 4
    $\begingroup$ Don't forget zero! Always think of him first! $\endgroup$ – MJD Jan 2 '15 at 13:13
  • 12
    $\begingroup$ 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
  • 1
    $\begingroup$ If $a=-1$, both $\operatorname{abs}(a)$ and $-a$ are equal to $1$. $\endgroup$ – MJD Jan 2 '15 at 19:38

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.


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 $-a\le|a|$.


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.