# Prove that $|x|\leq c\iff -c\leq x\leq c$

I want to show $|x|\le c$ is equivalent to $-c\le x\le c$. But I've taken this for granted so long I'm not actually sure where to start. Can someone give me some hints (not the full solution).

• Note that $|x|=\max(-x,x),$ for all real $x$. – Git Gud Dec 28 '14 at 16:00
• A similar question has been asked here, though it gives complete solutions, not hints. – Milo Brandt Dec 28 '14 at 16:05

Since $x\leqslant|x|$ and $-x\leqslant|x|$, if $|x|\leqslant c$ then $x\leqslant c$ and $-x\leqslant c$, hence $-c\leqslant x\leqslant c$.

If $-c\leqslant x\leqslant c$, then $x\leqslant c$ and $-x\leqslant c$, hence $|x|\leqslant c$ as $|x|=\max\{x,-x\}$.

Considering cases where $x<0$, $x>0$, etc. just makes arguments like these more confusing, in my opinion.

Split into cases. Show that it's true for $x = 0$, for $x > 0$ , and for $x < 0$. In each case, you have a simple expression for $|x|$, which is what you need to get going.

(The $x = 0$ case can be lumped in with $x > 0$ if you like.)

$$|x| \leq c\\ +x \leq c\text{ if }x > 0\\ -x \leq c\text{ if }x < 0\\ x \leq c\text{ for }x > 0\\ x \geq -c\text{ for }x < 0$$ hence $c \geq x \geq -c$

Since absolute values are non negative, we have $$0\le|x|\le c.$$ By squaring both sides $$x^2\le c^2.$$ $$(x-c)(x+c)\le0$$ Solving above inequality, we get $$-c\le x\le c.$$