Is the property $|x| \geq x$ true? I was thinking that $|x| \geq x$ must always hold because if $x$ is a negative number like $-1$, then $|-1|\geq -1$ and if $x$ is positive like $1$, then $|1|\geq 1$, too.
However, I have not seen it in any textbook, so I am not sure whether I can surely affirm this. Does it have any counterexample?
 A: Of course, yes. If $x$ is positive or zero, then ${|x|=x}$. If $x$ is negative, then ${|x| = -x > 0}$, so clearly ${|x| > x}$. Hence for any real $x$, ${|x|\geq x}$. This is a proof.
Obviously this only makes sense for reals. We cannot in general compare complex numbers in the same way we can real numbers.
But we can say that
$|x|\ge\Re(x)\quad$ and $\quad|x|\ge\Im(x)\quad$ for any $\;x\in\mathbb C\;,$
where $\;\Re(x)\;$ is the real part of $\,x\,$ and $\;\Im(x)\;$ is the imaginary part of $\,x\,.$
A: The definition of $\ \vert x \vert\ $ is the following:
$\vert x \vert = \begin{cases}
 x&\text{if}\, x\geq 0\\
      -x&\text{if}\ x<0
\end{cases}
$
So we see that, when $\ x\geq0,\ \vert x \vert = x \geq x.$
And when $\ x<0,\ \vert x \vert = -x > 0 > x,\ $ and so $\ \vert x \vert\geq x\ $ here too.
So yes, for all (real) values of $\ x,\ \vert x \vert \geq x.$
A: Yes, it is true. If$ x$ is positive or zero, then $|x|=x. $If x is negative, then $|x|=−x>0$,  $|x|>x$. So  $x$, $|x|≥x$. You can also draw a picture to think about this
