Property of absolute value

Question

How to prove for $x\in\Bbb R$, $-|x|\leq x\leq |x|$?I could prove it for the cases $x=0$ and for $x>0$.

$$|x| = \left\{ \begin{array}{ll} 0 & \mbox{if } x =0 \\ x & \mbox{if } x > 0 \\ -x & \mbox{if } x < 0 \end{array} \right.$$

Case 1:When $x=0$

$-|x|=x=|x|$ hence the inequality holds

Case 2:When $x>0$

We know that for all $x\in\mathbb R$:

$$\begin{align} &|x|≥x && (i)\\ \Rightarrow&-|x|≤-x && (ii) \end{align}$$

and when $x>0$ this implies $-x<0$

so $$–x<0<x\qquad (iii)$$

From (i),(ii) and (iii) We get

$$-|x|\leq-x<0<x\leq|x|\implies -|x|≤x≤|x|$$ hence the inequality holds

but I am not able to prove for the case $x<0$.

Answer 1

You are way overcomplicating the case when $x>0$. Remember, from the definition of absolute values, what is the absolute value of a positive number?

Hint: Take a look at some examples of positive numbers. What are $|1|, |2|, |1000|, |\pi|, |e|$?

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For the negative numbers, the task should be even simpler. Remember that one of the fundamental properties of the absolute value is that it is always non-negative. So, from that, if $x<0$, can you see why $x<|x|$ (remember, again, $|x|$ is non-negative!)

Answer 2

By definition of absolute value we can write,

$$\begin{align}&x≥0\\ \iff &-x≤0≤x\\ \iff &|x|=x≥x≥-x \end{align}$$

and

$$\begin{align}&x<0\\ \iff&-x>0>x\\ \iff& |x|=-x≥-x>x \end{align} $$

This immediately implies,

$$|x|≥x ~~\text{and}~~ |x|≥-x$$

This means,

$$-|x|≤x≤|x|.$$