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Prove-that-Sgn-0-0-




Question Number 207832 by Davidtim last updated on 28/May/24
Prove that Sgn(0)=0
$${Prove}\:{that}\:{Sgn}\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Commented by Davidtim last updated on 28/May/24
help me
$${help}\:{me} \\ $$
Answered by Frix last updated on 28/May/24
∀c∈C: sign c :=(c/(∣c∣)) ⇒ sign 0 is not defined  The sign is the direction of a vector in the  Complex Plane.  sign (a+bi) =((a+bi)/(∣a+bi∣)); sign (re^(iθ) ) =e^(iθ)   The vector  ((0),(0) ) has no direction.
$$\forall{c}\in\mathbb{C}:\:\mathrm{sign}\:{c}\::=\frac{{c}}{\mid{c}\mid}\:\Rightarrow\:\mathrm{sign}\:\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$$$\mathrm{The}\:\mathrm{sign}\:\mathrm{is}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{Complex}\:\mathrm{Plane}. \\ $$$$\mathrm{sign}\:\left({a}+{b}\mathrm{i}\right)\:=\frac{{a}+{b}\mathrm{i}}{\mid{a}+{b}\mathrm{i}\mid};\:\mathrm{sign}\:\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)\:=\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{The}\:\mathrm{vector}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{has}\:\mathrm{no}\:\mathrm{direction}. \\ $$

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