http://www.wolframalpha.com/input/?i=%28-2%29^3+%2C+cube+root+%28-8%29

As demonstrated above, the odd roots of negative numbers are not computed correctly.

For example, (-8)^(2/3) should return 4, as the cube root of -8 is -2 and -2 squared is 4. Here is the link:

http://www.wolframalpha.com/input/?i=%28-8%29+^+%282%2F3%29

## Bugs

### Odd Roots of Negative Numbers

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**1**of**1**• [ 11 posts ]### Odd Roots of Negative Numbers

by » Wed Sep 01, 2010 11:25 pm- Michael3545
**Posts:**5**Joined:**Thu Apr 22, 2010 10:41 pm

### Re: Odd Roots of Negative Numbers

by » Thu Sep 02, 2010 1:05 amMichael3545 wrote:the cube root of -8 is -2

The cube root of -8 = 1+1.73i

You may be thinking of -(8^(1/3)) = -2

Remember if the negative is inside the root, the answer has an imaginary part.

Read the second paragraph here: http://mathworld.wolfram.com/CubeRoot.html

- LachlanArthur
**Posts:**3**Joined:**Wed Sep 01, 2010 2:35 am

### Re: Odd Roots of Negative Numbers

by » Fri Sep 03, 2010 12:52 amBoth of you are correct. As the second paragraph of your article states, the cube root of -8 is negative 2. it is also an imaginary number. Both answers are correct, except mathmatica, and therefore wolfram alpha, gives the result with the imaginary number.

- aFakeName
**Posts:**1**Joined:**Sun Oct 11, 2009 2:25 am

### Re: Odd Roots of Negative Numbers

by » Tue Sep 07, 2010 1:45 pmWould it be possible for WolframAlpha to show both in the future?

- Michael3545
**Posts:**5**Joined:**Thu Apr 22, 2010 10:41 pm

### Re: Odd Roots of Negative Numbers

by » Sat Sep 18, 2010 8:01 pmRemember if the negative is inside the root, the answer has an imaginary part.

- raar78
**Posts:**11**Joined:**Thu Sep 02, 2010 8:04 am

### Re: Odd Roots of Negative Numbers

by » Wed Sep 22, 2010 1:34 amMichael3545 wrote:Would it be possible for WolframAlpha to show both in the future?

Actually not a bad idea.

Perhaps have an option where it says "Did you mean:" instead of "Use as", where it would give common mistakes people made. This would produce a list of possible alternative inputs.

In this example it would be "perhaps you mean (-8^(1/3))?

I think it could be done simply by the types of equations entered. Like anytime -a^b is entered the alternative could be (-a)^b.

-Nuclearman

- Nuclearman
**Posts:**343**Joined:**Sun Sep 13, 2009 6:01 pm

### Re: Odd Roots of Negative Numbers

by » Wed Sep 22, 2010 6:36 amMichael3545 wrote:Would it be possible for WolframAlpha to show both in the future?

Yes it is possible. W|A already mentions assumptions in similar cases and allows you to select from the alternatives and resubmit the query. In addition once it has established these alternatives it is usually possible to specify the alternative you want in your initial query as in:-

Distance London Birminham

gives you the result in miles but can be overridden by saying

Distance London Birmingham in kilometers

However W|A has overlooked it in some instances such as temperature not being offerred in both fahrenheit and centigrade. Such cases should be reported via feedback.

Volunteer Curator http://www.bgls11958.pwp.blueyonder.co. ... /index.htm

- briangilbert
**Posts:**1158**Joined:**Mon May 18, 2009 5:30 pm**Location:**London England

### Re: Odd Roots of Negative Numbers

by » Thu Oct 21, 2010 5:53 pmIt gives you the magnitude of the imaginary number.

r=4.

r=4.

- robgrondahl
**Posts:**2**Joined:**Thu Oct 21, 2010 5:46 pm

### Re: Odd Roots of Negative Numbers

by » Fri Nov 05, 2010 3:46 pmWolfram's default behavior is to included complex numbers and often nonreal complex numbers in the result.

See the graph of x^(1/3) which shows the real part and imaginary part on the same graph.

If I remember correctly, when the range is all complex numbers, there are three roots for cube root x. (See De Moivre's theorem from trig.) Convincing W|A to restrict the range to real numbers, should give you the result your looking for.

The question about negative underneath the radical is irrelevant for odd roots when the answer is restricted to real numbers.

- cuberoot(x) = cuberoot(-x)

same for 5th root, 7th root, etc.

See the graph of x^(1/3) which shows the real part and imaginary part on the same graph.

If I remember correctly, when the range is all complex numbers, there are three roots for cube root x. (See De Moivre's theorem from trig.) Convincing W|A to restrict the range to real numbers, should give you the result your looking for.

The question about negative underneath the radical is irrelevant for odd roots when the answer is restricted to real numbers.

- cuberoot(x) = cuberoot(-x)

same for 5th root, 7th root, etc.

-Nick

______________________________________

Volunteer Curator

If you are interested in tutoring try

nickhbulk-tutorWolfram@yahoo.com

______________________________________

Volunteer Curator

If you are interested in tutoring try

nickhbulk-tutorWolfram@yahoo.com

- nickalh
**Posts:**284**Joined:**Wed Aug 04, 2010 12:34 am

### Re: Odd Roots of Negative Numbers

by » Wed Dec 15, 2010 3:45 amThe multiple y-values for x^(1/3) seem to create recurring confusion and in multiple contexts.

viewtopic.php?f=32&t=51082&p=66772#p66772

See

viewtopic.php?f=32&t=46199

The truth is for negative x, cube root x is better thought of as multivalued because z^3 = -8 has 3 solutions.

http://mathworld.wolfram.com/MultivaluedFunction.html

They are -1, 1/2+√(3)/2 i, and 1/2-√(3)/2 i. So the cube root could be any one of those. The one chosen is 1/2+√(3)/2 i, with credit to Paepok, for the actual numbers.

DeMoivre's theorem justifies the multiple results for x^(-1/3), for those who have taken or remember trig.

The WolframAlpha team posted a method to get the more typical solution taught in middle school, high school and early college classes and not the master's or PhD level understanding.

viewtopic.php?f=32&t=46199#p53662

Basically,

and to be safe include the parentheses only around (1/3)

viewtopic.php?f=32&t=51082&p=66772#p66772

See

viewtopic.php?f=32&t=46199

The truth is for negative x, cube root x is better thought of as multivalued because z^3 = -8 has 3 solutions.

http://mathworld.wolfram.com/MultivaluedFunction.html

They are -1, 1/2+√(3)/2 i, and 1/2-√(3)/2 i. So the cube root could be any one of those. The one chosen is 1/2+√(3)/2 i, with credit to Paepok, for the actual numbers.

DeMoivre's theorem justifies the multiple results for x^(-1/3), for those who have taken or remember trig.

The WolframAlpha team posted a method to get the more typical solution taught in middle school, high school and early college classes and not the master's or PhD level understanding.

viewtopic.php?f=32&t=46199#p53662

Basically,

- Code: Select all
`y = sign(x) abs(x)^(1/3)`

and to be safe include the parentheses only around (1/3)

-Nick

______________________________________

Volunteer Curator

If you are interested in tutoring try

nickhbulk-tutorWolfram@yahoo.com

______________________________________

Volunteer Curator

If you are interested in tutoring try

nickhbulk-tutorWolfram@yahoo.com

- nickalh
**Posts:**284**Joined:**Wed Aug 04, 2010 12:34 am

### Re: Odd Roots of Negative Numbers

by » Wed Feb 23, 2011 7:30 pmThis is my second post and without a change. There are three roots of any negative cube root, true. However, the real root among those is negative. The graph of the x^(1/3) function is wrong on this site because it shows the real cube root of a negative number to positive when no positive cubed is negative.

[quote="nickalh"]The multiple y-values for x^(1/3) seem to create recurring confusion and in multiple contexts.

viewtopic.php?f=32&t=51082&p=66772#p66772

See

viewtopic.php?f=32&t=46199

The truth is for negative x, cube root x is better thought of as multivalued because z^3 = -8 has 3 solutions.

http://mathworld.wolfram.com/MultivaluedFunction.html

They are -1

, 1/2+√(3)/2 i, and 1/2-√(3)/2 i. So the cube root could be any one of those. The one chosen is 1/2+√(3)/2 i, with credit to Paepok, for the actual numbers.

DeMoivre's theorem justifies the multiple results for x^(-1/3), for those who have taken or remember trig.

The WolframAlpha team posted a method to get the more typical solution taught in middle school, high school and early college classes and not the master's or PhD level understanding.

PLEASE CHANGE IMMEDIATELY.

[quote="nickalh"]The multiple y-values for x^(1/3) seem to create recurring confusion and in multiple contexts.

viewtopic.php?f=32&t=51082&p=66772#p66772

See

viewtopic.php?f=32&t=46199

The truth is for negative x, cube root x is better thought of as multivalued because z^3 = -8 has 3 solutions.

http://mathworld.wolfram.com/MultivaluedFunction.html

They are -1

, 1/2+√(3)/2 i, and 1/2-√(3)/2 i. So the cube root could be any one of those. The one chosen is 1/2+√(3)/2 i, with credit to Paepok, for the actual numbers.

DeMoivre's theorem justifies the multiple results for x^(-1/3), for those who have taken or remember trig.

The WolframAlpha team posted a method to get the more typical solution taught in middle school, high school and early college classes and not the master's or PhD level understanding.

PLEASE CHANGE IMMEDIATELY.

- drhettinger
**Posts:**1**Joined:**Wed Jan 12, 2011 2:35 am

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