((-8)^2)^(1/3) <> (-8)^(2/3) <> (-8)^(1/3))^2

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Discussion Overview

The discussion revolves around the evaluation of the expressions ((-8)^2)^(1/3), (-8)^(2/3), and (-8)^(1/3))^2, focusing on their equivalence and the nature of their results, particularly in the context of real versus complex numbers. Participants explore the implications of cube roots and the validity of different approaches to these mathematical expressions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant claims that ((-8)^2)^(1/3) equals 4, while (-8)^(1/3))^2 also equals 4, but (-8)^(2/3) results in a complex number.
  • Another participant questions why 2/3 acts as a multiplicative constant instead of an exponent in the context of complex numbers.
  • A participant points out a typographical error in a previous equation and clarifies their intent regarding the use of Euler's formula.
  • Some participants assert that the first two expressions are valid but yield only the real cube root, while acknowledging that there are multiple cube roots, including non-real ones.
  • Another participant emphasizes that the first two expressions are indeed valid and provide one of the possible three answers, specifically the real one.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the first two expressions and their results, with some asserting they are valid while others argue they are restricted to real numbers. The discussion remains unresolved regarding the equivalence of the expressions and the nature of their outputs.

Contextual Notes

Participants highlight the distinction between real and complex numbers, and the discussion reflects varying interpretations of cube roots and their properties. There are unresolved aspects regarding the assumptions made about the nature of the roots.

nomadreid
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((-8)^2)^(1/3)=64^(1/3)=4
(-8)^(1/3))^2 = (-2)^2=4
but
(-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
I understand why the latter one works. Why don't the first two?
Thanks.
 
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nomadreid said:
((2/3)(cos (π) + i*sin(π))
Why does 2/3 act as a multiplicative constant instead of an exponential? Moreover ##\sin \pi =0##.
 
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Your fifth equation isn't one.
 
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Oops, too quickly typed the last equation. I meant to multiply the two-thirds times the angles, not times the whole expression (or equivalently use it as an exponent), using Euler's formula. Anyway, my question is not about Euler's formula, which as I say I do get when I am not typing while my bus is about to leave, but rather why the first two formulas are invalid.
 
nomadreid said:
((-8)^2)^(1/3)=64^(1/3)=4
(-8)^(1/3))^2 = (-2)^2=4
but
(-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
No, it doesn't. sin(π)= 0. The other two cube roots are non-real, of course.
I understand why the latter one works. Why don't the first two?
Thanks.
The first two do also, you just have restricted yourself to real numbers. 64 has three complex cube roots, 4 is one, the other two are non-real. 8 has 3 cube roots, one is 2 and the other two are non-real.
 
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They are valid. They give you one of possible three answers, the real one.
 
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Thanks, all of those who answered, everything clear now, and I apologize for the typos.
 

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