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We suppose that $a\in\Bbb R^{+}_0$ we know that

$$a^{\frac{m}{n}}=\sqrt[n]{a^m} \tag 1$$

Is $(1)$ provable or is it a given definition. Many years ago I remember that perhaps there was a proof of such an equality. Does anyone remember it? Thank you all.

Sebastiano
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  • The way I've learned it, if you look at it in the sense of exponentiation $\text{pow}:(0,\infty)\times \Bbb Q\to\Bbb R$, then it's a definition (which may or may not require you to prove that $\sqrt[kn]{a^{km}}=\sqrt[n]{a^m}$ for all $k\ne 0, a>0, n\ne 0$ and $m$). If you look at it in the sense of exponentiation $\Bbb (0,\infty)\times \Bbb R\to\Bbb R$, $a^b=\lim\limits_{q\to b} \operatorname{pow}(a,q)$, then it's a theorem. Surely it isn't the only approach. – Sassatelli Giulio Oct 31 '23 at 21:43

1 Answers1

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I would say that this is a definition for the left hand side. What you are thinking of is probably this line of reasoning.

$(a^{\frac{m}{n}})^n = a^m\iff a^{\frac{m}{n}}=\sqrt[n]{a^m} $.

This shows that if we want the exponential identity $a^{bc}=(a^b)^c$ to hold for rational numbers as well as integers, we need to use this definition for $a^{\frac{m}{n}}$.

Andrew
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