Why we can't use rational numbers as a root degree?

Question

My class on math lesson interested in strange question. Why we can't use rational numbers as a root degree? For example: $\sqrt[0.25]{2} = 2^4$

Answer 1

Roots are usually introduced a little after integer powers, as solutions of

$$y=\sqrt[n]x\iff y^n=x.$$

This is the first encounter with irrational numbers and paves the way to non-integer powers. But in the initial setting, only natural $n$ are "accessible".

After more theory, rational powers will come and the generalization

$$\sqrt[p/q]x=\sqrt[p]{x^q}=x^{q/p}$$ where $p,q$ are integer can be introduced.

After even more theory, you will get to real powers with

$$\sqrt[r]x=x^{1/r}$$ holding.

Answer 2

Well, we can. But, since it is not a standard notation, you would have to explain every time that$$\sqrt[\alpha]\beta\quad\text{ means }\quad\beta^{1/\alpha}.$$

Answer 3

Development

I think that "can't" is really the wrong word here. It isn't that we can't have non-natural indices on a radical expression, but rather that we usually don't—there simply isn't usually a good reason to consider radical expressions with a non-natural index. When I teach this topic, the flow of ideas is usually something similar to the following:

1. Exponential notation is introduced as repeated multiplication. If $a$ is a real number and $n$ is a natural number (excluding zero), then $$ a^n = \underbrace{a \cdot a \cdot \dotsb \cdot a}_{\text{$n$-times}}. $$

2. For any natural number $n$, the function $x \mapsto x^n$ is invertible on the positive real axis (and is invertible on the entire real line if $n$ is odd). The inverse function is the $n$-th root function, $$ \sqrt[n]{\cdot} : [0, \infty) \to \mathbb{R} : x \mapsto \sqrt[n]{x}, $$ which has the property that $$ \sqrt[n]{x^n} = \sqrt[n]{x}^n = x $$ for all $x \ge 0$. In short, $\sqrt[n]{a}$ is the unique nonnegative solution to the equation $x^n = a$, where $a$ is a nonnegative number.

3. The exponential function can be extended to non-natural exponents. First, insist that a few basic properties are retained: $$ a^m a^n = a^{m+n} \qquad\text{and}\qquad (a^m)^n = a^{mn}. $$ These properties are immediate when exponentiation is treated as repeated multiplication, and extend to the integers quite readily by observing that $$ 1 = a^0 = a^{n-n} = a^n a^{-n} \iff a^{-n} = \frac{1}{a^n}, $$ assuming that $a\ne 0$. To extend to the rationals, observe that $$ a = a^{1} = a^{n/n} = (a^{1/n})^n. $$ Hence $x = a^{1/n}$ solves the equation $a = x^n$. If we assume that $a$ is nonnegative, then there is only one solution, and we already have notation for that solution: $$ a^{1/n} = \sqrt[n]{a}. $$ Note the goal of the previous section: I am trying to understand the exponential function; I am extending the exponential function $n \mapsto a^n$ from the natural numbers to the integers. That $a^{1/n} = \sqrt[n]{a}$ is a nice bit of notational convenience, but at this point in the exposition, I don't really care about radical expressions.

4. In a precalculus class, this is basically where the discussion ends. In further classes, the exponential function can be developed further: we can extend to real exponents by continuity, and to complex exponents by choosing a branch of the logarithm. Notice, again, that the object of interest is the exponential function. Radicals are, in this context, a secondary notation for expressing exponential functions.

Punch Line

As the theory is usually developed, radical expressions are not really the objects being described. Rather, radicals are encountered early on in the solutions to certain equations, and relate to exponential functions via $$ \sqrt[n]{a} = a^{1/n}. $$ We typically don't bother to define $\sqrt[x]{a}$ for non-natural $x$ because by the time we might want to, we are already working with exponential functions. However, the fact that we don't usually define this notation doesn't mean that we can't. We could very easily define $$ \sqrt[x]{a} = a^{1/x} $$ for $a \ge 0$ and $x\ne 0$; we just don't generally have a reason to do so.