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The title seems like a simple statement, but my lack of skill with absolute values prevents me from being too creative with the intuition. I can't seem to find an answer to this online, so I assume this result follows from a few elementary absolute value arithmetic laws.

I can reason intuitively that the sign of the quotient can never be negative because of the square and the absolute value. I have trouble proving it mathematically.

Scene
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  • $x^2=|x|^2$ ... Equivalently, $(x-|x|)(x+|x|)=0$. – Blue Oct 13 '22 at 04:33
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    If $x \neq 0$, then yes it's true since you can just write $|x|^2 = x^2$ and cancel out the $|x|$. – Accelerator Oct 13 '22 at 04:37
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    In a slightly different approach, use $|x|=x\cdot \operatorname{sgn}(x)$ and $\operatorname{sgn}(x)=\dfrac{1}{\operatorname{sgn}(x)}$. – insipidintegrator Oct 13 '22 at 05:07
  • @Accelerator Ah, this was that trivial step I could not think of. Thanks! If you'd like to submit the comment as an answer, I'd be happy to accept it :) – Scene Oct 13 '22 at 06:59

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Let $x \in \mathbb{R}$. By definition, $|x|=\begin{cases} x, & \text{if }x \geqslant 0\\ -x, & \text{if }x<0 \end{cases}$

Let $x\neq0$.

If $x>0, \frac{x^2}{|x|}=\frac{x^2}{x}=x=|x|$;

If $x<0,\frac{x^2}{|x|}=\frac{x^2}{-x}=-x=|x|$

Either way, it's true: $\forall x\neq0, \frac{x^2}{|x|}=|x|$.

Stéphane Jaouen
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