"Complex" Roots, when can we compute result wihout imaginay numbers?

Question

The question is, how can we know if we can solve a n-index root without imaginary numbers when n-index is a non integer number, but a real number?

First, basic theory:

$$\sqrt[2.0]{8} = 2.8284.....$$ or $$\sqrt[3.0]{8} = 2.0$$

In the relationship between roots and powers, this is the same as:

$$8^\frac{1}{2.0} = 2.8284....$$ or $$8^\frac{1}{3.0} = 2.0$$

Nevertheless, as we learn from powers -and this is very very related with final question-, its difficult to imagine (or at least, it is for me), how to represent this, literally:

$$8^{2.0} = 8 * 8$$

but...

$$8^\frac{1}{2.0} = aaa... 8*how much?$$ ...half eight * halft eight (and not 4 by 4)?

Besides, we know when the index of a root is odd, we can compute the result for a negative number without "fight" with complex numbers:

Even: $$\sqrt[2.0]{-8} = 2.8284.....*i = \sqrt[2.0]{8} * \sqrt[2.0]{-1}$$

Odd: $$\sqrt[3.0]{-8} = -2.0$$

The index can't be zero, because, in power form:

$$\sqrt[0.0]{n} = n^ \frac{1}{0.0}$$ results a number powered to infinity, so results infinity.

But there is a little bit more, the index of a root can be negative too, a thing that lead us to "invert" the result (1/result):

$$\sqrt[-2.0]{8} = \frac{1}{2.8284.....}$$

$$\sqrt[-3.0]{8} = \frac{1}{2.0}$$

This put in power form means to invert the base, to let the exponent as a positive numer, in this way:

$$ n^{-exp} = \frac{1}{n^{exp}}$$

as we can see here:

$$8^\frac{1}{-2} = \frac{1}{8^\frac{1}{2}} = \frac{1}{2.8284....} = 0,35355.....$$

$$ 8^\frac{1}{-3} = \frac{1}{8^\frac{1}{3}} = \frac{1}{2.0} = 0.5$$

...and of course, we can combine both "negative" numbers, in randicand and in index:

$$ -8^\frac{1}{-2} = \frac{1}{-8^\frac{1}{2}} = \frac{1}{-2.8284....*i}$$

$$ -8^\frac{1}{-3} = \frac{1}{-8^\frac{1}{3}} = \frac{1}{-2.0} = -0.5$$

Besides, while we are used to roots with integer index (square, cubic, etc..), there is float index roots:

$$\sqrt[2.0]{8} = 2.8284.....$$ $$\sqrt[2.2]{8} = 2,5733.....$$ $$\sqrt[2.4]{8} = 2,3784.....$$ $$\sqrt[2.6]{8} = 2,2250.....$$ $$\sqrt[2.8]{8} = 2,1015.....$$ $$\sqrt[3.0]{8} = 2.0$$

And here is where the point of interest raises: we can compute negative radicand roots while the index stays odd, getting ourselves away from complex numbers (i), but what happens with the sign when the index of the root is a float number?

We know that an index of 2, for the signs, means -by-=+ or +by+=+, never will be negative, and for an index of 3, means -by-by-=- or +by+by+=+, both sign results are possible, but for float index, How can we operate the sign?. We can't do something as "fraction of -" by "fraction of -". We can compute the valor of roots with float index, but, what about the sign?.

A greeting.

Answer 1

For positive real numbers $r$ and real $q$, we may define: $$r^q=e^{q\ln(r)}$$

Since $e^x$ is positive for all real $x$ and $q\ln(r)$ is real, so $r^q$ is positive.

Answer 2

First, some terminology. A floating-point number (sometimes called a "float") is an object defined in computer science. It is a certain limited kind of rational number that can only represent only of a finite number of values allowed by the particular encoding of whichever floating-point representation you are using at the time.

For example, $\sqrt[2]8 = 2.8284\ldots$ is not a "float". In mathematical language, it is a real number.

Arithmetically, taking an $n$th root is the inverse of taking an $n$th power. That means we perform exactly the same kind of operations, but with a known output instead of a known input. For example, to take the square root of $8$, we need to know that there is a positive real number $x$ such that $$ x^2 = 8. $$

To prove that such a number exists, you need a definition of the real numbers, which I think is outside the scope of this question. Given a good definition of "real number," however, it is possible to prove that such a number $x$ exists. We call that number $\sqrt[2]8.$

Similarly, if we want $\sqrt[5]{11},$ we need a positive real number $y$ such that $$ y^5 = 11. $$ There is only one positive real number that solves that equation, and one if its names is $\sqrt[5]{11}.$

Rational powers of positive real numbers can be computed by taking integer powers of integer roots (or integer roots of integer powers). Specifically, if $x$ is real and $m$ and $n$ integers that have no common factor greater than $1$, and $n > 1,$ then $$ x^{m/n} = \left( \sqrt[n]x \right)^m = \sqrt[n]{x^m}. $$

It is conventional to extend this further to cover some cases where $x$ is negative. An example of such a definition is cited in What are the Laws of Rational Exponents? In short, if $x$ is positive and $n$ is odd, then we find that $$ \left(-\sqrt[n]x\right)^n = - \left(\sqrt[n]x\right)^n = -x, $$ and therefore we are able to define $$ \sqrt[n]{-x} = -\sqrt[n]x. $$

This does not work when $n$ is even. In order to define $\sqrt[2]{-8}$ we have to define the complex numbers (using symbols such as $i$), and for the equation $z^n = x$ (with $x$ known) there are generally then $n$ distinct roots $z$ from which we may decide to select one as the value of $\sqrt[n]x.$

Considering your examples: \begin{align} \sqrt[2.2]{8} &= 8^{1/2.2} = 8^{5/11} = (\sqrt[11]8)^5,\\ \sqrt[2.4]{8} &= 8^{1/2.4} = 8^{5/12} = (\sqrt[12]8)^5,\\ \sqrt[2.6]{8} &= 8^{1/2.6} = 8^{5/13} = (\sqrt[13]8)^5,\\ \sqrt[2.8]{8} &= 8^{1/2.8} = 8^{5/14} = (\sqrt[14]8)^5. \end{align}

If we replace $8$ by $-8$, we could write \begin{align} \sqrt[2.2]{-8} &= (-8)^{1/2.2} = (-8)^{5/11} = (\sqrt[11]{-8})^5 = -(\sqrt[11]{8})^5,\\ \sqrt[2.4]{-8} &= (-8)^{1/2.4} = (-8)^{5/12},\\ \sqrt[2.6]{-8} &= (-8)^{1/2.6} = (-8)^{5/13} = (\sqrt[13]{-8})^5 = -(\sqrt[13]{8})^5,\\ \sqrt[2.8]{-8} &= (-8)^{1/2.8} = (-8)^{5/14}, \end{align} but to work out the cases $\sqrt[2.4]{-8}$ and $\sqrt[2.8]{-8}$ we require some kind of agreement about how to take rational powers of a negative number when the power is not an integer divided by an odd integer. One convention says that $$ (-8)^{5/12} = \left(8\cos\pi + i8\sin\pi\right)^{5/12} = 8^{5/12}\cos\tfrac{5\pi}{12} + i8^{5/12}\sin\tfrac{5\pi}{12} $$ and $$ (-8)^{5/14} = \left(8\cos\pi + i8\sin\pi\right)^{5/14} = 8^{5/14}\cos\tfrac{5\pi}{14} + i8^{5/14}\sin\tfrac{5\pi}{14}. $$

But if we're going to do this for $(-8)^{5/12}$ and $(-8)^{5/14},$ it raises the question why we do not apply the same kind of arithmetic to $(-8)^{5/11}$ and $(-8)^{5/13}.$ Many authors avoid this issue by saying that you simply cannot take the $5/12$ or $5/14$ power of a negative number.

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When the exponent $r$ is a real number, not necessarily rational, if $x > 0$ we could define $x^r$ as the smallest number that is greater than $x^p$ for every rational number $p$ such that $p < r,$ or the largest number that is less than $x^q$ for every rational number $q$ such that $q > r.$ Another alternative is to construct the exponential function $\exp(x) = e^x$ and the natural logarithm function $\log(x),$ then define $$ x^r = \exp(r \log(x)). $$

The answers to What does $2^x$ really mean when $x$ is not an integer? discuss both of these approaches as well as some other ideas.

Although floating-point numbers in a computer are actually rational numbers, so that in theory we could raise a number to a floating-point power using the techniques I described for rational powers, this is not really a practical method and computer math libraries tend to calculate $\exp(r \log(x))$ instead. (They would implement $\exp$ and $\log$ functions for other purposes anyway, so it is simply easier and more efficient to use those functions for this purpose too.)

But the last few paragraphs apply only for positive numbers $x.$ When $x < 0,$ there are two answers one typically finds to the question of raising $x$ to an irrational power:

• Don't do it.
• Generalize to complex powers of complex numbers.

The approach for complex powers of complex numbers is similar to the formula $\exp(r \log(x)),$ but not quite as simple. The difficulty is that there are many possible values of $\log(z)$ when $z$ is a complex number. Therefore it is necessary to come up with a rule for picking just one value of $\ln(z)$ for any complex number $z.$ If we say $\operatorname{Log}(z)$ is the value of $\log(z)$ selected according to the rule, then we can write $$ z^r = \exp(r \operatorname{Log}(z)). $$

The "rule" typically involves something called a branch cut, for example as discussed in the answers to Determination of complex logarithm and How do I formally define complex exponentiation?

The way I attempted to define numbers such as $(-8)^{5/12}$ earlier amounted to a certain kind of branch cut; using such a branch cut, one could define a general real power of a negative number so that (for example) $$ (-8)^r = \left(8\cos\pi + i8\sin\pi\right)^r = 8^r\cos(r\pi) + i8^r\sin(r\pi). $$

This leads to the conclusion that $$ (-8)^{1/3} = 8^{1/3}\cos\tfrac\pi3 + i8^{1/3}\sin\tfrac\pi3 = 1 + i\sqrt3. $$

This does not look much like $-2,$ the result we might have liked to see for $\sqrt[3]{-8},$ so if you really want to define general real powers of negative numbers you may have to decide that $\sqrt[3]x$ is a different function than $x^{1/3}$ after all, giving up on the idea that $\sqrt[r]x$ could be a general function of real numbers $r$ and $x.$

These limitations carry over in to computer arithmetic. In C++, for example, the function std::pow for floating-point powers of floating-point numbers rules out taking any non-integer power of a negative number, so you cannot use std::pow to compute $n$th roots of negative numbers. But there is another function, std::cbrt, that allows you to take cube roots of negative numbers. You could extend that idea to other odd integer powers safely enough.