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)?


  • $\begingroup$ 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. $\endgroup$ – Brian Tung Mar 14 '17 at 15:30
  • $\begingroup$ First case : Yes $x \le |x|=x$. $\endgroup$ – Mauro ALLEGRANZA Mar 14 '17 at 15:34
  • $\begingroup$ Second case : if $x < 0$ then $x < 0 \le |x|$, because $|x|$ is always non-negative. And obviously, if $x < |x|$ then also $x \le |x|$. $\endgroup$ – 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

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

  • $\begingroup$ 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? $\endgroup$ – JDoeDoe Mar 14 '17 at 16:48
  • $\begingroup$ @JDoeDoe....The first case justifies $x \geq 0$ the second case justifies $x<0$...what is wrong? $\endgroup$ – user35508 Mar 14 '17 at 16:50
  • $\begingroup$ 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 $x\leq \lvert x \rvert$? $\endgroup$ – JDoeDoe Mar 14 '17 at 16:56
  • $\begingroup$ Yes the last inequality in your comment is true only $if$ $x<0$ $\endgroup$ – user35508 Mar 14 '17 at 16:58
  • $\begingroup$ You mean my second inequality? (I edited my comment) $\endgroup$ – JDoeDoe Mar 14 '17 at 17:00

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.