Simple question about nth-roots of negative numbers

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Discussion Overview

The discussion revolves around the nature of nth roots of negative numbers, specifically focusing on cube roots. Participants explore the implications of real and complex solutions, particularly in the context of a math bridging course that has not yet introduced complex numbers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the assertion that \(\sqrt[3]{-125} = -5\), expressing confusion about the nature of roots of negative numbers.
  • Another participant clarifies that every number has three cube roots, with one being real and the other two being complex conjugates.
  • A participant confirms that \(\sqrt[3]{-1} = -1\) is one of the three solutions to the equation \(x^3 = -1\).
  • There is a discussion about the distinction between even and odd roots, with a participant noting that any even root of a negative number has only complex solutions, while odd roots can yield real solutions.
  • One participant reflects on their misunderstanding regarding the multiplication of negative numbers and the nature of roots.

Areas of Agreement / Disagreement

Participants express differing views on the nature of roots of negative numbers, particularly regarding the assertion that all roots of negative numbers are complex. Some agree that cube roots can be real, while others emphasize the complexity of the topic.

Contextual Notes

Participants mention that the course material has not yet covered complex numbers, which may limit their understanding of the topic. There is also a distinction made between "complex" and "non-real" solutions, indicating a potential area of confusion.

Who May Find This Useful

This discussion may be useful for students grappling with the concepts of roots in mathematics, particularly those transitioning from real number systems to complex numbers.

GCH
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I've been doing this online math bridging course from a German uni to which I am planning to go soon, it is going through all high school maths from the beginning(it has been a while since I did it).
The chapter on roots so far ignores complex numbers, which will be introduced in a later chapter... but they claim that \sqrt[3]{-125} = -5 ! Can this be correct? Surely not?
I always thought that any root of a negative number has only complex solutions (I know that positive ones have complex and real solutions)
What is \sqrt[3]{-1} anyway? (I am having a bit of a brain-freeze at the moment.)

Thank you!
 
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GCH said:
I've been doing this online math bridging course from a German uni to which I am planning to go soon, it is going through all high school maths from the beginning(it has been a while since I did it).
The chapter on roots so far ignores complex numbers, which will be introduced in a later chapter... but they claim that \sqrt[3]{-125} = -5 ! Can this be correct? Surely not?
I always thought that any root of a negative number has only complex solutions (I know that positive ones have complex and real solutions)
What is \sqrt[3]{-1} anyway? (I am having a bit of a brain-freeze at the moment.)

Thank you!

Every number has THREE cube roots. Of these three, one will be real, and the other two will be complex conjugates of each other. And yes, every real number has one real cube root. \sqrt[3]{-1} = -1, but that's only one of the three solutions to x^3 = -1.
 
GCH said:
What is \sqrt[3]{-1} anyway? (I am having a bit of a brain-freeze at the moment.)

What is -1 * -1 * -1?
 
GCH said:
I've been doing this online math bridging course from a German uni to which I am planning to go soon, it is going through all high school maths from the beginning(it has been a while since I did it).
The chapter on roots so far ignores complex numbers, which will be introduced in a later chapter... but they claim that \sqrt[3]{-125} = -5 ! Can this be correct? Surely not?
Surely yes. (-5)^3= (-5)(-5)(-5)= (25)(-5)= -125

I always thought that any root of a negative number has only complex solutions (I know that positive ones have complex and real solutions)
Any even root of a negative number has only complex solutions. (Strictly speaking, you mean "non-real", not complex. The real numbers are a subset of the complex numbers.)

What is \sqrt[3]{-1} anyway? (I am having a bit of a brain-freeze at the moment.)

Thank you!
As SteveL27 said, (-1)^3= (-1)(-1)(-1) so one third root of -1 is -1. The other two are 1/2+ i\sqrt{3}/2 and 1/2- i\sqrt{3}/2.
 
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Thank you for clearing this up for me. Somehow I did not think last night of asking myself the question: what number multiplied an uneven number of times yields -1 as a result.
 

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