# Prove the $x\leq \lvert x \rvert$ (absolute value)

I want to prove $x\leq \lvert x \rvert$.

We have two cases:

If $x\geq 0$: We have $x \leq \lvert x \rvert$ (or should it be $x \leq \lvert x \rvert=x$ ?)

If $x<0$: We have $-x<- \lvert x \rvert \iff x> \lvert x \rvert$ (Or maybe $x<- \lvert x \rvert \iff -x> \lvert x \rvert$ ?)

So I have $x\leq \lvert x \rvert$ and $x>\lvert x\rvert$, but now what?

When I check the two cases, should I only change the sign for the $x$ in absolute value-sign or also the normal $x$ (case 2)?

Thanks!

• In problems like these, graphing the two sides may make it easier to see what's going on. Your $x < 0$ analysis is exactly backwards. – Brian Tung Mar 14 '17 at 15:30
• First case : Yes $x \le |x|=x$. – Mauro ALLEGRANZA Mar 14 '17 at 15:34
• Second case : if $x < 0$ then $x < 0 \le |x|$, because $|x|$ is always non-negative. And obviously, if $x < |x|$ then also $x \le |x|$. – Mauro ALLEGRANZA Mar 14 '17 at 15:36

For $x\ge 0$ the equality $x=|x|$ holds. For $x<0$ we have $|x|=-x$, so $x<0$ and $|x|\ge 0$, hence trivially $x\le |x|$.

It's simple ...just remember the definition of absolute value function $$|x| = \begin{cases}{x} & x\geq0 \\ {-x} & x<0 \end{cases}$$

SO ...For $x \geq0$ The inequality is trivial.. $x\le x=|x|$ which is true

For $x<0$ which means $-x>0$

So,the inequality is as follows

$$x<0< -x=|x|$$

• In the last inequality you have $x<0< -x=\lvert x \rvert$ shouln't it be $x<x< -x=\lvert x \rvert$? Why the zero? – JDoeDoe Mar 14 '17 at 16:48
• @JDoeDoe....The first case justifies $x \geq 0$ the second case justifies $x<0$...what is wrong? – user35508 Mar 14 '17 at 16:50
• Think I got it. But is $x<0< -x=\lvert x \rvert$ the same as $x<\lvert x \rvert$? And the first case then justify $x\leq \lvert x \rvert$? – JDoeDoe Mar 14 '17 at 16:56
• Yes the last inequality in your comment is true only $if$ $x<0$ – user35508 Mar 14 '17 at 16:58
• You mean my second inequality? (I edited my comment) – JDoeDoe Mar 14 '17 at 17:00