# Negative one equals one

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

How can this be?

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

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

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

arildno
Homework Helper
Gold Member
Dearly Missed
It is a rather minor flaw, since the fundamental flaw is retained in the following "flawless" argument:
$$-1=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=((-1)^{2})^{\frac{1}{6}}=1$$

Redbelly98
Staff Emeritus
Homework Helper
gabel and arildno, welcome to PF.

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

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

gabel and arildno, welcome to PF.
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

arildno
Homework Helper
Gold Member
Dearly Missed
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
Weclome to yuiop&gabel!

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..

Redbelly98
Staff Emeritus
Homework Helper
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
Be sure to visit the recent https://www.physicsforums.com/showthread.php?t=465551"

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 = 11/2 = 1​

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