Equivalent form of $\frac{\text{sgn}(x)}{|x|}$

Question

Wikipedia says that any real number $x$ can be written as $|x| = \text{sgn}(x)x$ where $\text{sgn}(x)$ is the sign function. Rearranging, this means that $\frac{\text{sgn}(x)}{|x|} = \frac{1}{x}$ for $x\neq 0$.

On the other hand, WolframAlpha says that $\frac{\text{sgn}(x)}{|x|} = \frac{1}{x^*}$ where $x^*$ is the conjugate of $x$ and that $\frac{\text{sgn}(x)}{|x|}$ is only equal to $\frac{1}{x}$ for $x>0$.

Is this an inconsistency? Why does WolframAlpha omit the $x<0$ case?

Wolfram Alpha says for $x>0$, $\dfrac{\text{sgn}(x)}{|x|}=\dfrac1x$; but Wolfram Alpha does not say only for $x>0$ is that true — J. W. Tanner, May 21 '20 at 22:13
Alpha considers complex $x$. For real $x$ we have $x = x^{\ast}$, so that's in agreement with wikipedia. — Daniel Fischer, May 21 '20 at 22:13

score 1 · Accepted Answer · answered May 21 '20 at 22:13

1

That WA is talking about conjugates should be a clear hint that you are thinking about $x$ as a real number and WA is working with complex $x$. If you give WA the hint that $x$ is real, it give you what you expect: WA: sgn(x)/|x| for x in reals

answered May 21 '20 at 22:13

Eric Towers

67,037

Equivalent form of $\frac{\text{sgn}(x)}{|x|}$

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