# Negative one equals one

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

How can this be?

## Answers and Replies

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.

Last edited by a moderator:
Tinyboss
I think the assumption that 2*1/3=2/6 might be causing some trouble here, too. ;-)

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

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

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.

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

yuiop
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

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..

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​

Last edited by a moderator:
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.

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 is true. For real numbers, rational exponentiation, x^(1/n), is usually defined as a real root of t^n-x = 0 (if such a root exists) in the complex domain, and the positive one is chosen if there are two such roots. In this case the equality does not apply generally, as the example shows.