Is it possible to evaluate x^(2/3) for x<0?

  • Thread starter nietzsche
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In summary, the conversation discusses the confusion around calculating (-1)^{2/3} and whether it is undefined or equal to 1. The use of complex variables and the convention of considering the square root of a square as its absolute value are mentioned. It is also noted that some calculators may restrict the use of fractional powers with negative bases to nonnegative real numbers. Finally, the idea of extending roots to complex values and how it changes the meaning of radicals is discussed.
  • #1
nietzsche
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I am very confused. Here I was thinking I know the basics of algebra, and then I get two questions in my homework today with exponents like this.

[tex](-1)^{\frac{2}{3}} = undefined[/tex]

when I type it into my calculator.

But with manipulation,

[tex](-1)^{2/3} = [(-1)^{2}]^{\frac{1}{3}} = 1[/tex]

Am I missing something here? My textbook solutions manual says that you can graph such a function for negative values, but when I plug it into, for example, my Mac's graphing software, it only graphs it for positive values of x.

Which one is correct?

Thanks!
 
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  • #2
I can't speak for the calculators. However the solution can be obtained using complex variables:
-1=e[pi]i.
We the have 3 possible answers for (-1)2/3:
e2(n+1)[pi]i/3, where n=0,1,2.
 
  • #3
My TI-89 gives an answer of 1 no matter how I use the parenthesis.

(-1)2/3 = 1
((-1)1/3)2 = 1
((-1)2)1/3 = 1

Which should make sense since:
- It is possible to square a negative and then take the cube root of the resulting number.
- It is possible to find the cube root of a negative number and square the resulting number.
 
  • #4
pbandjay said:
My TI-89 gives an answer of 1 no matter how I use the parenthesis.

(-1)2/3 = 1
((-1)1/3)2 = 1
((-1)2)1/3 = 1

Which should make sense since:
- It is possible to square a negative and then take the cube root of the resulting number.
- It is possible to find the cube root of a negative number and square the resulting number.

Thank you, I guess I should buy a new calculator.
 
  • #5
(Details of previous comment)

3 solurions: 1, (-1+(-3)1/2)/2, (-1-(-3)1/2)/2
 
  • #6
When I went to school it was considered conventional to consider, the square root of a square as its absolute value and hence positive. Of course, this is only a convention; but calculators may be constructed to honor this convention.
 
Last edited:
  • #7
I believe what is happening is a restriction of the computing device. Power functions with fractional powers are frequently restricted to nonnegative real numbers (Maple for example).

But the fractional power identity

[tex]x^{m/n}=\sqrt[n]{x^m}={(\sqrt[n]{x})}^m[/tex]

holds provided that the radicals on the right are defined.

However this assumes m and n are in lowest terms.

For example:

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

but

[tex](-1)^{4/6}={(\sqrt[6]{-1})^4=\text{ undefined}[/tex]

This may be part of why calculating devices shy away from fractional powers of negative bases.

The extension of roots to complex values is usually only done if the inputs are also complex. This subtely changes the meaning of some radicals.

[tex]\sqrt[3]{-8}=-2[/tex] in the real number sense.

[tex]\sqrt[3]{-8}=1+i\sqrt{3}[/tex] in the complex number sense.

In general, provided m and n are in lowest terms and n is odd, the expression [itex]x^{m/n}[/itex] is defined for all real numbers and can be graphed.

--Elucidus
 
  • #8
Use x^y = e^(y.log(x)). If x=r.e^(i.t) then log(x)=log(r)+i.(t+2.pi.k) where k is any integer. This will give you all possible values of x^y.
 

1. Can we evaluate x^(2/3) for negative values of x?

Yes, we can evaluate x^(2/3) for negative values of x. However, the result will be a complex number.

2. Is there a real solution to x^(2/3) for x<0?

No, there is no real solution to x^(2/3) for x<0. As mentioned before, the result will be a complex number.

3. How do we evaluate x^(2/3) for negative values of x?

To evaluate x^(2/3) for negative values of x, we can use the complex number system and express the result in terms of imaginary numbers.

4. Can x^(2/3) be simplified for x<0?

Yes, x^(2/3) can be simplified for x<0 by expressing the result in terms of imaginary numbers or using the complex number system.

5. Is it possible to graph x^(2/3) for x<0?

Yes, it is possible to graph x^(2/3) for x<0 by plotting the imaginary numbers on a complex plane.

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