Is the equality \sqrt[n]{ab}= \sqrt[n]{a}\sqrt[n]{b} true for complex numbers?

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Discussion Overview

The discussion revolves around the equality \(\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}\) in the context of complex numbers. Participants explore whether this equality holds true, particularly contrasting it with its validity for real numbers. The scope includes theoretical implications and mathematical reasoning.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation involving \(-1\) and claims it leads to the conclusion that \(-1 = 1\), raising questions about the validity of the equality.
  • Another participant argues that the equality \(\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b}\) cannot be true for complex numbers, asserting it only holds for real numbers.
  • A participant suggests that an assumption regarding exponentiation may be problematic, indicating potential confusion in the calculations presented.
  • Further contributions highlight that \(-1\) is one of the complex sixth roots of \(1\), suggesting a deeper exploration of roots in the complex plane.
  • Another participant emphasizes that the equality does not apply generally, especially for negative real numbers, and notes that rational exponentiation of complex numbers typically involves defining a branch where the equality may hold.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the equality for complex numbers. Some assert it does not hold, while others explore conditions under which it might be considered true, particularly in the context of defining branches for complex exponentiation.

Contextual Notes

Participants note that the discussion involves assumptions about the nature of roots and exponentiation, particularly in distinguishing between real and complex numbers. The implications of defining branches for complex exponentiation are also mentioned as relevant to the discussion.

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

How can this be?
 
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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. ;-)
 
I hadn't even noticed that! Though I suspect that was a typo.
 
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
 
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.
 
Redbelly98 said:
gabel and arildno, welcome to PF.
Thank you for the warm welcome! :shy:
 
Redbelly98 said:
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! :eek: I haven't welcomed him either :redface: Welcome to PF arildno :smile: and gabel too :wink:
 
yuiop said:
How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! :eek: I haven't welcomed him either :redface: Welcome to PF arildno :smile: and gabel too :wink:
Weclome to yuiop&gabel! :smile:

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
yuiop said:
How come you waited nearly 7 years and over 10,000 posts to welcome arildno to PF? Oh wait! :eek: I haven't welcomed him either :redface: Welcome to PF arildno :smile: and gabel too :wink:
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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  • #11
HallsofIvy said:
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.
 

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