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My high school math teacher stated that roots with non-natural indexes are meaningless, just like $\frac{\infty}{\infty}$ or $0^0$, because "mathematicians decided so and so it is unless we change axioms".

It doesn't make sense to me.

Isn't $\sqrt[\frac 1a]{x}$ just $x^{a}$ for every $a\in\mathbb{R}$ (or even $\mathbb{C}$) or am I missing something?

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If $\alpha\ne 0$, the $\alpha$-th root of a non-negative real certainly exists. One is more likely to use the notation $x^{1/\alpha}$ for it than $\sqrt[\alpha]{x}$. – André Nicolas Mar 19 '14 at 19:25
Is there a way I can show my teacher that she's wrong? Even just citing a theorem based on that – mattecapu Mar 19 '14 at 19:55
Your teacher is undoubtedly aware that $x^\beta$ is defined for positive $x$ and non-zero $\beta$, in particular for $\beta=1/\alpha$. So it would be a question of finding a paper or other publication that used the notation $\sqrt[\alpha]{x}$. I have no immediate example. – André Nicolas Mar 19 '14 at 20:01
Yeah, she knows it. But she denies that $\sqrt[\frac13]{x}=x^3$ while she accepts that $\sqrt[2]{x}=x^{\frac12}$... – mattecapu Mar 20 '14 at 17:08
up vote 2 down vote accepted

We can define exponentiation to every real number as follows:

Let $\alpha\in\mathbb{R}$ and $x\in \mathbb{R}_+$ be a non negative real number. We define $x^{\alpha}$ as $$x^{\alpha}:=\text{exp}(\alpha\ln(x))$$ with this definition one can prove all the 'exponent laws', for example $$x^{\alpha}x^{\beta}=x^{\alpha+\beta}$$

Another approach is to start with exponentiation of natural numbers, defined recursively as follows:

For $n\in\mathbb{N}$, and $x\in\mathbb{R}_+$; $$x^n=x^{n-1}\cdot x$$ $$x^1=x$$ Then we can extend this definition, allowing the exponent to be an integer number putting $$x^{-1}=\frac{1}{x}$$

To allow rational exponents, we define the $n$-th root as the (unique positive) solution to $y^n=x$ and we denote it as $$x^{\frac{1}{n}}=y$$

The final step needs the supreme axiom;

Let $\alpha\in\mathbb{R}$ and $x\in\mathbb{R}_+$, $$x^{\alpha}:=\text{sup}\lbrace x^r \mid r\in\mathbb{Q} \,\,\text{ and }\,\, r\leq\alpha\rbrace$$

One can prove that the definition given at the beginning of this post and this last one are equivalent.

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Thank you! What is the purpose of the last statement? – mattecapu Mar 20 '14 at 17:11
@mattecapu Just to be reassured that both approaches are the same i.e. they give the same answer. – Chazz Mar 20 '14 at 22:17
Otherwise $x^\alpha$ wouldn't be a crescent function? – mattecapu Mar 21 '14 at 17:19
@mattecapu You can prove any property of exponentiation with any of the definitions given above. When one is defining exponentiation it is more natural to proceed first defining it for $\mathbb{N}$, then for $\mathbb{Z}$, then for $\mathbb{Q}$ and finally for $\mathbb{R}$ as we outline above. But this is a little bit complicated to work with, so one proves $x^{\alpha} =\text{sup} \lbrace x^{r}\mid \alpha\in\mathbb{Q}\,\text{ and }\,r\leq\alpha \rbrace=\text{exp}(\alpha\ln(x))$, as the latter is easier to work with. – Chazz Mar 21 '14 at 17:54
Get it, thanks! – mattecapu Mar 21 '14 at 22:23

You are correct. Natural powers are easy to define (repeated multiplication). If negative powers are to work nicely with positive integer powers, they must represent reciprocals. If rational powers are to work nicely with integer powers, they must represent natural roots. The way to define irrational powers is trickier, but perfectly well-defined with a bit of calculus (continuous extension of a function defined on a dense subset).

As an easy counter-example to "no non-natural roots", $\sqrt[\frac{1}{2}]{2} = 2^\frac{1}{\frac{1}{2}} = 2^2 = 4$. However, even irrational roots make sense. If you have a sequence of rational numbers ($(q_n)\in\mathbb{Q}^\mathbb{N}$) that converges to $\pi$, then $\sqrt[\pi]{a} = \lim_{n\rightarrow\infty} \sqrt[q_n]{a} = \lim_{n\rightarrow\infty} a^{\frac{1}{q_n}}$. Admittedly, proving that statement makes sense is complicated, but I think it makes intuitive sense.

From Wolfram|Alpha: .

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