Homework Help: When are Negative Bases Raised to Rational Powers Undefined?

1. Jan 27, 2017

Saturnine Zero

1. The problem statement, all variables and given/known data
I'm trying to understand negative bases raised to rational powers, when calculating principle roots for real numbers. I'm not worried about complex solutions numbers at this stage. I just can't find a concise explanation I can understand anywhere. I'm self learning as an adult so I don't have a teacher to ask.

2. Relevant equations
When, in general is a negative base raised to a rational power undefined for real numbers?

3. The attempt at a solution
$(-x)^{\frac {odd}{odd}}$ I have this as being a real number but reversing the sign of x
$(-x)^{\frac {even}{odd}}$ I have this as being a real number but reversing the sign of x

$(-x)^{\frac {odd}{even}}$ I have this as being undefined
$(-x)^{\frac {even}{even}}$ I have this as being undefined

But I am still confused. For instance the following example $(-3)^{\frac 2 4}$ I'm not sure how to think about it.

$(-3)^{\frac 2 4}$ is this $(-3^2)^{\frac 1 4}$ which would be $9^{\frac 1 4}$ which would have a real root?

Or would it be $(-3^{\frac 1 4})^{2}$ and since you can't take the 4th root of (-3) you can't square it so it is undefined?

$(-3)^{\frac 3 4}$ I think I understand as it's either $(-3^3)^{\frac 1 4}$ which is trying to take an even root of an odd number, so undefined. Or it's $(-3^{\frac 1 4})^{3}$ which is trying to take an even root of an odd number and then can't be raised to the 3rd power, so is undefined.

Am I on the right track or am I way off?

edit: fixed the latex

2. Jan 28, 2017

Logical Dog

In the real domain, even roots of negative numbers do not exist.

When a negative number is raised to an even number it becomes positive positive. As far as I can see, there is only one case that the root wont be defined.

I know not to put the standard rules for exponents when doing negative numbers, but not why they dont work

Last edited: Jan 28, 2017
3. Jan 28, 2017

Stephen Tashi

We wish to preserve the idea that $\frac{a}{b} = \frac{2a}{2b} = \frac{a/2}{b/2}$. We don't want any special restrictions to be placed on that arithmetic. For example, we don't want a restriction that says "$\frac{a}{b} = \frac{2a}{2b}$ except when the fraction appears in an exponent".

If we preserve the concept that those differently written fractions are equal then we must say $(-x)^{ \frac{a}{b} } = (-x)^{\frac{2a}{2b}} = (-x)^{\frac{a/2}{b/2}}$

So you can't treat the situations $(-x)^{\frac{even}{even}},(- x)^{\frac{even}{odd}}, (-x)^{\frac{odd}{even}}, (-x)^{\frac{odd}{odd}}$ as different cases. For example , if we were to say that $(-3)^{\frac{1}{3}}$ is defined then we would have to apply that same definition to $(-3)^{\frac{2}{6}}$.

You've illustrated the difficulty of defining the situation $(-x)^{\frac{even}{even}}$ unambiguously. All the cases can be turned into that case by multiplying the both the numerator and the denominator of the exponent by 2.

Reducing the fraction in your example gives: $(-3)^{\frac{2}{4}} = (-3)^{\frac{1}{2}}$ and the latter expression is undefined in the arithmetic of real numbers.

By the way, if you wish to write an equation involving a negative value, you don't need to represent the value as "$(-x)$", since a plain "$x$" can take on values like $x = -7$.

4. Jan 28, 2017

Stephen Tashi

To that, I'll add the observation that textbooks aren't consistent in how they handle the rules of real number arithmetic. If you ask an expert a technical question about exponentiation of negative numbers in real number arithmetic, it's amusing how often the expert will begin to talk about the complex number system. The rules and definitions for the complex numbers are standardized and the easiest course for an expert is to take those rules and try to see which of them don't fall apart when applied only to real numbers.

Most texts would agree that an expression like $(-3)^{\frac{1}{3}}$ is another notation for a root $\sqrt[3]{-3}$ and that the cube root of $-3$ exists in the real number system. However they don't define $(-3)$ to a fractional exponent when the numerator of the fraction isn't $1$.

With such a convention, the reason that $(-3)^{\frac{1}{3}}$ is not equal to $(-3)^{\frac{2}{6}}$ is that the former expression is a notation for a number and the latter expression is undefined.

5. Jan 28, 2017

Saturnine Zero

I think I have a better idea now.

If I understand correctly, what needs to happen is that the fractional exponent needs to be expressed or understood in simplest terms first. The exponent should never be a ratio of two even numbers because a factor of 2 can always be factored out.

There is only two possibilities then, a co-prime ratio with an even denominator/root which is undefined for a negative bases, or a co-prime ratio with an odd denominator/root which is defined in the real numbers.

Is that an accurate assessment?

6. Jan 28, 2017

Saturnine Zero

This is exactly what has happened in my book. And why I've been so puzzled! I'm looking forward to getting a handle on it using complex numbers.

7. Jan 28, 2017

Stephen Tashi

You don't fully understand yet. We are dealing with a problem of definition. The question is "How is the notation $x^{\frac{a}{b}}$ defined in the arithmetic of the real numbers?". Definitions are a matter of tradition, not a matter of "true" or "false". We are asking question about how a notation is defined, not about a what a number is equal to - because until a notation is defined , it doesn't represent a particular number.

I'm saying that the tradition for defining the notation $x^{\frac{a}{b}}$ in the arithmetic of the real numbers is not consistent from textbook to textbook. What do your text materials say? If you have two different books, they might say two different things.

I think you are proposing the idea that we can define the notation $x^{\frac {a}{b}}$ when $a$ and $b$ are non-negative integers to mean: Reduce $\frac{a}{b}$ to its lowest terms. Let the reduced fraction be $\frac{p}{q}$. Find the $q$-th root of $x^p$ if it exists in the real number system.

That is a possible definition for the notation. But then we have to worry about how this notation fits-in with other notation. For example, is it consistent with $( x^ {\frac{p}{q}})^2 = (x^{\frac{p}{q}})( x^{\frac{p}{q}}) = x^{(\frac{p}{q} + \frac{p}{q})} = x^{\frac{2p}{q}}$ ? If that notation holds then we would have $( (-3)^{\frac{1}{2}} )^2 = (-3)^{\frac{2}{2}}$ but the notation on left hand side does not define a number in the real number system and the notation on the right hand does represent a real number.

It's not a simple matter to define a system of notation dealing exclusively with real numbers that is self-consistent and implements all the familiar algebraic manipulations that we want to do. It's not surprising that some elementary algebra books "throw up their hands" and just declare that $x^{\frac{a}{b}}$ is undefined when $x < 0$. They may implement the exception that $x^{\frac{1}{q}}$ is notation for $\sqrt[q]{x}$ and point out that this is an exception to the rule, or they may start using that notation and forget to point out that they are making an exception!

The complex numbers are simpler in many ways, but added complexities appear - e.g. "branch points" of functions.

Last edited: Jan 28, 2017
8. Jan 28, 2017

Saturnine Zero

Thanks so much for your help Stephen I think that clears it up. I think it's a case of a book giving particular definitions of the notation and convention to get past a certain point with the assumed knowledge of that level which is "good enough" until such times a deeper explanation can be given using complex numbers.

I guess this happens with a lot of topics, a student is given enough information to progress but there are always some subtleties lurking in the background which can only be understood after a certain point of future topics.

I think this is a good example of the issue at hand and help me get my head around the fact it's an issue of defining notation.