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Question Number 90457 by M±th+et£s last updated on 23/Apr/20

$$therangeofy=\sqrt{x}is[0,+\mathrm{\infty })ifx⩾0$$$$orjust[0,+\mathrm{\infty })?$$

Commented by MJS last updated on 24/Apr/20

$$\mathrm{talking}\mathrm{of}‘‘{\mathrm{definition}}^{″}\mathrm{and}‘‘{\mathrm{range}}^{″}\mathrm{makes}$$$$\mathrm{no}\mathrm{sense}\mathrm{for}x\notin \mathbb{R}\mathrm{usually}.$$$$y=\sqrt{x}\mathrm{is}\mathrm{defined}\mathrm{for}x⩾0\mathrm{and}\mathrm{its}\mathrm{range}\mathrm{is}[0;+\mathrm{\infty })$$$$\mathrm{for}x\in \mathbb{C}\mathrm{we}\mathrm{get}\mathrm{different}\mathrm{kinds}\mathrm{of}\mathrm{problems}$$$$[\mathrm{try}\mathrm{to}\mathrm{solve}-1=\sqrt{x}]$$

Commented by M±th+et£s last updated on 24/Apr/20

$$thatmeanstherangeis[0,+\mathrm{\infty })because$$$$therangeisyvaluein\mathbb{R}oritsin\mathbb{R}and\mathbb{C}?$$$$andiwasaskingifitswrongifwe{didn}^{\prime }t$$$$wrotex⩾0$$

Commented by MJS last updated on 24/Apr/20

$$\mathrm{first}\mathrm{you}\mathrm{must}\mathrm{specify}\mathrm{the}\mathrm{underlying}\mathrm{set}$$$$\mathrm{for}x\mathrm{and}\mathrm{for}y$$$$\mathrm{i}.\mathrm{e}.f=\{(x,y)\mid x\in \mathbb{N}\wedge y\in \mathbb{N}\wedge y=\sqrt{x}\}$$$$\mathrm{or}f=\{(x,y)\mid x\in \mathbb{R}\wedge y\in \mathbb{Z}\wedge y=\sqrt{x}\}$$$$\mathrm{or}\mathrm{whatever}\mathrm{is}\mathrm{required}$$$$\mathrm{usually},\mathrm{after}\mathrm{having}\mathrm{learned}\mathrm{the}\mathrm{rules}\mathrm{for}$$$$\mathbb{N},\mathbb{Z},\mathbb{Q}\mathrm{and}\mathbb{R}\mathrm{you}\mathrm{learn}\mathrm{about}\mathrm{functions}$$$$\mathrm{with}\mathbb{R}\to \mathbb{R}\mathrm{or}(x,y)\in \mathbb{R}\times \mathbb{R}(={\mathbb{R}}^2)$$$$(x,y)\in {\mathbb{C}}^2\mathrm{is}\mathrm{the}\mathrm{last}\mathrm{step}$$$$\mathrm{so}\mathrm{I}\mathrm{guess}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{in}{\mathbb{R}}^2$$$$\mathrm{in}\mathrm{this}\mathrm{case}y=\sqrt{x}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x<0\mathrm{and}$$$$\mathrm{the}\mathrm{range}\mathrm{is}[0;+\mathrm{\infty })\mathrm{or}\mathrm{simply}y⩾0.\mathrm{if}\mathrm{only}$$$$\mathrm{the}\mathrm{range}\mathrm{is}\mathrm{asked}\mathrm{for},y⩾0\mathrm{is}\mathrm{enough}$$

Commented by MJS last updated on 24/Apr/20

$$\mathrm{if}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{in}{\mathbb{C}}^2{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{getting}\mathrm{more}\mathrm{complicated}$$$$x\in \mathbb{C}\iff x={\mathrm{e}}^{\mathrm{i}\theta }r\mathrm{with}r\in {\mathbb{R}}^{+}\mathrm{and}-\pi ⩽\theta <\pi (?)$$$$\mathrm{here}\mathrm{the}\mathrm{complications}\mathrm{start}.\mathrm{some}\mathrm{use}$$$$0⩽\theta <2\pi .\mathrm{some}\mathrm{say}\sqrt{{\mathrm{e}}^{\mathrm{i}\theta }r}\mathrm{is}\mathrm{not}\mathrm{unique}\mathrm{because}$$$$y^2=x\mathrm{has}2\mathrm{solutions}...\mathrm{but}\sqrt{4}\ne -2\mathrm{even}\mathrm{though}$$$${(-2)}^2=4.\mathrm{which}\mathrm{definitions}\mathrm{to}\mathrm{use}\mathrm{depends}$$$$\mathrm{on}\mathrm{which}\mathrm{kind}\mathrm{of}\mathrm{solution}\mathrm{you}\mathrm{need}.$$$$$$$$y=\sqrt{x}=\sqrt{{\mathrm{e}}^{\mathrm{i}\theta }r}={\mathrm{e}}^{\mathrm{i}\frac{\theta }{2}}\sqrt{r}\mathrm{is}\mathrm{unique}\mathrm{if}\mathrm{we}\mathrm{define}$$$$r⩾0\wedge -\pi ⩽\theta <\pi$$$$\mathrm{then}y=\sqrt{x}\mathrm{is}\mathrm{defined}\mathrm{for}x\in \mathbb{C}$$$$\mathrm{but}\mathrm{you}\mathrm{cannot}\mathrm{get}y\in {\mathbb{R}}^{-}\mathrm{because}$$$$-\pi ⩽\theta <\pi \iff -\frac{\pi }{2}⩽\frac{\theta }{2}<\frac{\pi }{2}$$$$\Rightarrow \mathrm{the}\mathrm{range}\mathrm{is}y={\mathrm{e}}^{\mathrm{i}\varphi }s;s\in {\mathbb{R}}^{+}\wedge -\frac{\pi }{2}⩽\varphi <\frac{\pi }{2}$$$$\mathrm{or}y=a+b\mathrm{i};a\in {\mathbb{R}}^{+}\wedge b\in \mathbb{R}$$$$$$$$\mathrm{change}\mathrm{the}\mathrm{definitions}\mathrm{above}\mathrm{and}\mathrm{you}\mathrm{get}$$$$\mathrm{different}\mathrm{ranges}$$

Commented by M±th+et£s last updated on 24/Apr/20

$$thankyouverrymuchsir$$

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