Simple question about nth-roots of negative numbers

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The discussion centers on the properties of nth roots, particularly for negative numbers. It clarifies that every real number, including negative ones, has one real cube root, while the other two roots are complex conjugates. Specifically, \sqrt[3]{-125} equals -5, confirming that negative numbers can have real roots when the root is odd. The confusion arises from the misconception that negative roots only yield complex solutions, which is true for even roots. Ultimately, the thread emphasizes that \sqrt[3]{-1} equals -1, with two additional complex roots.
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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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