Why is (-1)^(2/3) undefined on calculators?

[RichardParker]

My hand-held calculator (as well as GraphCalc and Wolfram) registers DOMAIN ERROR whenever I try to input this.

Paul's notes (http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet.pdf) says that

a^(m/n)= (a^m)^(1/n) = (a^1/n)^m
However it does not define what set of numbers does a, m, and n represents.

With a, m, n as integers, it follows that

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

or

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

Hence if a, m, and n are integers then (-1)^(2/3) must be an element of the set of real numbers. What am I missing?

 

[Char. Limit]

Well,defining -1=e^{i pi}, we can see that

-1 = e^{\frac{2 i \pi}{3}}

And this is the imaginary number...

-\frac{1}{2} + i \frac{\sqrt{3}}{2}

Which isn't in the domain your calculator is used to using. There should be a setting somewhere to allow imaginaries of the form a+bi.

 

[jbunniii]

There are three different numbers whose cube is -1. They are:

\frac{1}{2} + i\frac{\sqrt{3}}{2}
-1
\frac{1}{2} - i\frac{\sqrt{3}}{2}

Thus

(-1)^{1/3}

can be any of these. Furthermore, the first one listed above is the "principal value", which is what a calculator (or, e.g., Matlab) will return if it supports complex numbers.

If your calculator does not support complex numbers, then it will probably choke. Mine (Microsoft Windows built-in calculator) can't do it either.

 

[HallsofIvy]

Probably your calculator is doing a^b by exp (b ln(a)) and ln(a) is not defined for a negative.

 

[I like Serena]

There's also a few interesting issues with taking roots of negative numbers.

Consider this:

-1 = (-1)^{\frac 2 2} = ((-1)^2)^{\frac 1 2} = 1^{\frac 1 2} = 1

(Just another proof that -1 = 1 )

 

[Mark44]

My hand-held calculator (as well as GraphCalc and Wolfram) registers DOMAIN ERROR whenever I try to input this.

Paul's notes (http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet.pdf) says that

a^(m/n)= (a^m)^(1/n) = (a^1/n)^m
However it does not define what set of numbers does a, m, and n represents.

With a, m, n as integers, it follows that

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

or

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

Hence if a, m, and n are integers then (-1)^(2/3) must be an element of the set of real numbers. What am I missing?

I don't see anything wrong in what you did.

 

[jbunniii]

I don't see anything wrong in what you did.

It's valid, but it is only one of three possible answers.

The other two are

e^{\pm i 2 \pi / 3}

Matlab, for example, gives

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

ans =

-0.5 + 0.866025403784439i

 

[I like Serena]

I don't see anything wrong in what you did.

I concur.
When constrained to how powers are defined for the set of real numbers the OP's calculation is correct.

EDIT: I have to add that your cheat sheet makes no mention of a constraint to real numbers, nor of the extra rules that apply when using fractional powers on negative numbers.
See this section of the wikipedia page for an explanation: http://en.wikipedia.org/wiki/Exponentiation#Real_powers_of_positive_numbers

 

[jbunniii]

I thought about this a bit more in the context of calculators that can only handle real numbers. The problem is that if you perturb the problem even slightly, there might not be any real solutions.

And it can't just check them all to see if one happens to be real: for an irrational exponent, there are a countably infinite number of complex solutions, and none of them are real. Now it's true that every number representable in a calculator is rational, but in lowest terms it might have a huge denominator, and the denominator controls how many complex answers there are.

So while the calculator can in principle calculate the one real-valued root

(-1)^{2/3} = -1

there's generally nothing it can do with

(-1)^{(2/3 + \epsilon)}

and it cannot know that you have given it an exponent of the form m/n (with small n). In fact, you have not done so in this case, since 2/3 can't be expressed exactly with a finite number of binary digits.

 

[Hurkyl]

There are actually several different exponentiation functions (though they usually have the same value in the situations where more than one makes sense).

The real number exponentiation operation is only defined for positive bases. Period. Even (-1)^1 is undefined.

Another common exponentiation operation is defined for any integer exponent. With this operation, (-1)^1 is defined. (But 2^{1/2} is not defined)

Another operation that is sometimes convenient is to allow any positive base with any exponent, and any base with any rational exponent of odd denominator -- this is taking advantage of the fact that, if n is odd, then any real number has a unique n-th root.


Complex exponentiation is also often used, and has the advantage of being defined for any nonzero base. However, it is a multi-valued function. Generally when it is used the principal branch is meant. In this case, the principal branch is (-1)^{2/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i.

 

[vela]

What am I missing?

It depends on how you try to calculate the answer because of the various definitions of exponentiation, as Hurkyl described.

If your calculator has a built-in cube root function, you can get an answer by taking the cube root of -1 and then squaring the result or vice versa. This takes advantage of the fact that all real numbers have a real cube root. On the other hand, if you calculate the exponent first and then try taking -1 to that power, it probably doesn't work because the calculator resorts to using real number exponentiation.

Some calculators are smart enough to look at the exponent and then use the appropriate method to get an answer. Some will even keep track of the fact your calculated exponent is a rational number.

 

[RichardParker]

I see. I am not familiar with the exponential function yet, so I only know of one method of exponentiation.

Thanks everyone.

 

[Ray Vickson]

My hand-held calculator (as well as GraphCalc and Wolfram) registers DOMAIN ERROR whenever I try to input this.

Paul's notes (http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet.pdf) says that

a^(m/n)= (a^m)^(1/n) = (a^1/n)^m
However it does not define what set of numbers does a, m, and n represents.

With a, m, n as integers, it follows that

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

or

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

Hence if a, m, and n are integers then (-1)^(2/3) must be an element of the set of real numbers. What am I missing?

You must be careful: things like a^(m/n) = (a^m)^(1/n) = (a^(1/n))^m are true for integer m, n and for *positive* a. However, they may be false for a < 0, depending on the values of m and n. In fact, for a = -1, m = 2 and n = 3, they *are* false!

RGV

 

[ammontgo]

I think your calculator maybe having trouble taking the root of a negative number?