kind of a simple question , I know, but just in case I'm wrong

is writing an absolute value necessary when said absolute value is squared?

for example something like this:

$(\lvert x-3\rvert)^2 = (x-3)^2$

since absolute value functions can be written as

$\lvert x \rvert = \sqrt{(x)^2}$

I thought that maybe $ (\lvert x \rvert)^2 = (\sqrt{(x)^2})^2 $ and the square root cancels out ,making $(x)^2$


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    $\begingroup$ For real $x$, it is absolutely fine to write $|x|^2=x^2$. For complex x, these two have different meaning. $\endgroup$ – Math Lover Aug 16 '17 at 18:46
  • $\begingroup$ No, it's not wrong. $\endgroup$ – John Wayland Bales Aug 16 '17 at 18:46
  • $\begingroup$ okay, thanks. I'm not dealing with complex numbers or vectors, so it works then $\endgroup$ – Dahen Aug 16 '17 at 18:47
  • $\begingroup$ It is a basic result that on $\mathbf R$, $\;\lvert x\rvert^2=x^2$. Of course this is wrong on $\mathbf C$. $\endgroup$ – Bernard Aug 16 '17 at 18:47
  • $\begingroup$ For vectors in $\mathbb R^n$ writing $|x|^2 = x \cdot x = x^2$ can be motivated by geometric algebra: en.wikipedia.org/wiki/Geometric_algebra $\endgroup$ – md2perpe Aug 16 '17 at 19:24

In general, it is not the case that $|A|^2 = A^2$ (for example, if $A$ is a complex number, or a vector, or some other kind of object where some notions of absolute values and squaring make sense). However, if $A$ is a real number, then you are fine.

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    $\begingroup$ ok , thanks. real numbers is what I'm dealing with right now so I'll just write it like that $\endgroup$ – Dahen Aug 16 '17 at 18:48
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    $\begingroup$ However for a vector you have an analog: $\lVert A\rVert^2=A\cdot A$. $\endgroup$ – Bernard Aug 16 '17 at 18:48

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