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Question Number 217772 by yamane last updated on 20/Mar/25

$$Whereistheerror$$$$(-1)={(-1)}^1$$$$={(-1)}^{\frac{2}{2}}$$$$={\left[{(-1)}^2\right]}^{\frac{1}{2}}$$$$={\left[1\right]}^{\frac{1}{2}}$$$$=\sqrt{1}$$$$=1$$

Commented by Ghisom last updated on 21/Mar/25

$${(-1)}^{\frac{2}{2}}={(-1)}^{(2\mathbf{÷}2)}={(-1)}^{\left(1\right)}=-1$$$${(-1)}^{\frac{2}{2}}={(-1)}^{(2\mathbf{÷}2)}\ne {\left({(-1)}^2\right)}^{1/2}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{similar}\mathrm{to}\mathrm{this}\mathrm{mistake}$$$$6/3=6/(2+1)\stackrel{???}{=}6/2+6/1=3+6=9$$

Answered by Marzuk last updated on 20/Mar/25

$$\sqrt{1}=\pm 1notonly1$$$$r_1=1and{r}_2=-1$$$$youcouldalsosay$$$${\left[{(-1)}^{\frac{1}{2}}\right]}^2$$$$=i^2$$$$=-1$$$$sothemainerrorisintheunderstanding$$$$ofapplicationtheproperty{\left(a^m\right)}^n=a^{mn}$$

Commented by Ghisom last updated on 21/Mar/25

$$\sqrt{1}=1$$$$\sqrt{1}=-1\mathrm{is}\mathrm{wrong}$$$$\mathrm{this}\mathrm{has}\mathrm{been}\mathrm{explained}\mathrm{over}\mathrm{and}\mathrm{over}\mathrm{again}.$$$$\mathrm{we}\mathrm{have}\mathrm{exactly}\mathrm{one}\mathrm{rule}\mathrm{of}\mathrm{exponentials}:$$$$z\in \mathbb{C}\Rightarrow z=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{with}r⩾0\mathrm{and}-\pi <\theta ⩽\pi$$$$x\in \mathbb{Q}(\mathrm{also}\mathrm{works}\mathrm{for}x\in \mathbb{R},x\in \mathbb{C})$$$$z^x={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^x=r^x{\mathrm{e}}^{\mathrm{i}x\theta }$$$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}{2}^{\mathrm{nd}}\mathrm{result}.\mathrm{never}.$$$$\mathrm{if}z=1\wedge x=\frac{1}{2}:$$$$z={1\mathrm{e}}^{0\mathrm{i}}$$$${\mathrm{z}}^x={\left({1\mathrm{e}}^{0\mathrm{i}}\right)}^{1/2}=1^{1/2}{\mathrm{e}}^{0\mathrm{i}/2}={1\mathrm{e}}^0=1$$$$\mathrm{how}\mathrm{would}\mathrm{you}\mathrm{get}-1?{}^{(\ast )}$$$$$$$$\mathrm{the}\mathrm{mistake}\mathrm{comes}\mathrm{ftom}\mathrm{solving}\mathrm{equations}$$$$\mathrm{where}\mathrm{we}\mathrm{have}\mathrm{a}\mathrm{different}\mathrm{rule}.$$$$x^2=1\Rightarrow x=\pm 1$$$$\mathrm{because}\mathrm{here}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{looking}\mathrm{for}\mathrm{all}\mathrm{values}$$$$\mathrm{of}x\mathrm{which}\mathrm{give}1\mathrm{if}\mathrm{we}\mathrm{square}x$$$${(+1)}^2=1$$$${(-1)}^2=1$$$$$$$$$$$$(\ast )$$$$\mathrm{the}\mathrm{only}\mathrm{exception}\mathrm{is}\mathrm{being}\mathrm{made}\mathrm{for}$$$$\mathrm{geometric}\mathrm{problems}\mathrm{to}\mathrm{stay}\mathrm{in}\mathbb{R},ignoring$$$$\mathrm{numbers}\in \mathbb{C}(\mathrm{because}\mathrm{they}\mathrm{had}\mathrm{not}\mathrm{been}$$$$\mathrm{invented}\mathrm{yet}\mathrm{in}\mathrm{the}\mathrm{age}\mathrm{of}\mathrm{geometry}):$$$$\sqrt[3]{-8}={(-8)}^{1/3}=-2$$$$\mathrm{we}\mathrm{artificially}\mathrm{take}\mathrm{the}\mathrm{root}\in \mathbb{R}$$$$$$$$\mathrm{but}\mathrm{for}x\in \mathbb{C}\mathrm{we}\mathrm{have}$$$${(-8)}^{1/3}={\left({8\mathrm{e}}^{\mathrm{i}\pi }\right)}^{1/3}=8^{1/3}{\mathrm{e}}^{\mathrm{i}\pi /3}={2\mathrm{e}}^{\mathrm{i}\pi /3}\notin \mathbb{R}$$

Commented by mr W last updated on 21/Mar/25

$$yes.$$$$if{x}^2=4,thenx=\pm \sqrt{4}.$$$$but\sqrt{4}=2,-\sqrt{4}=-2,not\sqrt{4}=\pm 2.$$

Commented by Marzuk last updated on 21/Mar/25

$$Iamapologizingformymisunderstanding.$$$$Iwilltrymybestinfuture.Thanksfor$$$$showingthemistake!$$

Commented by Ghisom last updated on 21/Mar/25

$$\mathrm{no}\mathrm{need}\mathrm{to}\mathrm{apologize}$$$$\mathrm{you}\mathrm{live}-\mathrm{you}\mathrm{learn}$$$$\mathrm{I}\mathrm{might}\mathrm{be}\mathrm{a}\mathrm{fraction}\mathrm{of}\mathrm{a}\mathrm{step}\mathrm{further}\mathrm{than}$$$$\mathrm{you}\mathrm{are}\mathrm{but}\mathrm{still}\mathrm{looking}\mathrm{at}\mathrm{the}\mathrm{countless}$$$$\mathrm{backs}\mathrm{of}\mathrm{those}\mathrm{before}\mathrm{me}$$

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