Proof (i) When $\beta=0$. In this case, we have $m=n \alpha$ and $a=p^{n \alpha}$, hence we get $\sqrt[n]{a}=p^{\alpha}$ is an integer.
(ii) When $\alpha=0$ and $1 \leqslant \beta \leqslant n-1$. In this case, we have $m=\beta$ and $a=p^{\beta}$. If there exist two positive integers $b, c$ such that $\sqrt[n]{a} \Rightarrow \frac{b}{c}$ holds, let $(b, c)=d$, then we have $b=b_{1} d, c=\dot{c}_{1} d$ and $\left(b_{1}, c_{1}\right)=1$. From $\sqrt[n]{a}=\frac{b}{c}$ we have $\sqrt[n]{p^{\beta}}=\sqrt[n]{a}=\frac{b}{c}=\frac{b_{1}}{c_{1}}, p^{\beta}=\left(\frac{b_{1}}{c_{1}}\right)^{n}$, thus we have
$$b_{1}^{n} \doteq p^{\beta} c_{1}^{n}$$

By $\left(b_{1}, c_{1}\right)=1$ and (29), we have $p \mid b_{1}$. Let $b_{1}=p^{l} b_{2}$, where $l$ is a positive integer, $\left(b_{2}, p\right)=1$. From (29) we have
$$c_{1}^{n}=p^{n l-\beta} b_{2}^{n}$$

Since $1 \leqslant \beta \leqslant n-1, l \geqslant 1$ and (30), we have $p \mid c_{1}$. Since $p\left|b_{1}, p\right| c_{1}$, this contradicts $\left(b_{1}, c_{1}\right)=1$, hence there do not exist two positive integers $b, c$ such that $\sqrt[n]{a}=\frac{b}{c}$.
(iii) When $\alpha$ is a positive integer and $1 \leqslant \beta \leqslant n-1$. In this case, if there exist two positive integers $b, c$ such that $\sqrt[n]{a}=\frac{b}{c}$, then from $\frac{b}{c}=\sqrt[n]{a}=$ $\sqrt[n]{p^{n \alpha+\beta}}=p^{\alpha} \sqrt[n]{p^{\beta}}$ we get, $\sqrt[n]{p^{\beta}}=\frac{b}{c p^{\alpha}}$ can be expressed as a fraction, which contradicts the proof in (ii) that $\sqrt[n]{p^{\beta}}$ cannot be expressed as a fraction. Hence the lemma is proved.