- #1

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[tex]-1=\sqrt[3]{-1}=(-1)^{1/3} = (-1)^{2/6} = ((-1)^2)^{1/3}=1^{1/3} = 1[/tex]

How can this be?

How can this be?

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- Thread starter gabel
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- #1

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[tex]-1=\sqrt[3]{-1}=(-1)^{1/3} = (-1)^{2/6} = ((-1)^2)^{1/3}=1^{1/3} = 1[/tex]

How can this be?

How can this be?

- #2

HallsofIvy

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It **can't** be. The error is in thinking that [tex]\sqrt[n]{ab}= \sqrt[n]{a}\sqrt[n]{b}[/tex] for complex numbers. It is only true if the roots are real numbers.

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- #3

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I think the assumption that 2*1/3=2/6 might be causing some trouble here, too. ;-)

- #4

HallsofIvy

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I hadn't even noticed that! Though I suspect that was a typo.

- #5

arildno

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[tex]-1=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=((-1)^{2})^{\frac{1}{6}}=1[/tex]

- #6

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I guess the explanation lies in the fact that -1 is one of the complex 6

- #7

arildno

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Thank you for the warm welcome! :shy:gabel and arildno, welcome to PF.

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How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! I haven't welcomed him either Welcome to PF arildno and gabel toogabel and arildno, welcome to PF.

- #9

arildno

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Weclome to yuiop&gabel!How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! I haven't welcomed him either Welcome to PF arildno and gabel too

BTW, Redbelly is a newbie from 2008, so he hasn't been as tardy in welcoming me as you make it out to be..

In fact, you yourself has been even tardier than Redbelly..

- #10

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Be sure to visit the recent https://www.physicsforums.com/showthread.php?t=465551"How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! I haven't welcomed him either Welcome to PF arildno and gabel too

Back on topic -- here is a version of this problem in "lowest terms", so to speak:

-1 = (-1)^{1} = (-1)^{2/2} = [(-1)^{2}]^{1/2} = 1^{1/2} = 1

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- #11

disregardthat

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Itcan'tbe. The error is in thinking that [tex]\sqrt[n]{ab}= \sqrt[n]{a}\sqrt[n]{b}[/tex] for complex numbers. It is only true if the roots are real numbers.

It's not true for real numbers, not the negative ones. And for rational exponentiation of complex numbers we usually define a branch, in which case the equality

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