23. Proof: We use proof by contradiction, assuming that $\sqrt[n]{A}$ is a rational fraction, that is,
$$\sqrt[n]{A}=\frac{p}{q}, \quad q>1, \text { and }(p, q)=1 .$$

This leads to a contradiction. Raising both sides to the $n$-th power, we get
$$A=\frac{p^{n}}{q^{n}}$$

Since $(p, q)=1$, it follows that $\left(p^{n}, q^{n}\right)=1$, and $q^{n}>1$, so $\frac{p^{n}}{q^{n}}$ is not an integer, while $A$ is an integer. Therefore, $A=\frac{p^{n}}{q^{n}}$ cannot hold. This contradiction arises from the assumption that $\sqrt[n]{A}$ is a rational fraction, so $\sqrt[n]{A}$ cannot be a rational fraction.