# Find the relationship between $n$ and $m$ (both natural numbers) such that $m^{1/n}$ is a rational number.

I know how to show that specific numbers such as $2^{1/2}, 2^{1/3}, 3^{1/2}, etc.,$ are irrational, but what about the general form $m^{1/n}$?

## 2 Answers

Such numbers are either natural or irrational.

Suppose $m^{\frac{1}{n}}=\frac{p}{q} \quad \Rightarrow m=\frac{p^n}{q^n}$.

If $m$ is not a $n$-th power of any natural number, there is a prime factor such that the highest power dividing $m$ is not a multiplie of $n$.

Now, what can we say about how often that prime occurs in the factorization of the left hand side of this equation? What about the right hand side?

Hint  By the Rational Root Test, $\,m^{1/n}\,$ is rational iff it is integral. If so then $\,m^{1/n}= a\,$ for an integer $\,a\,$ so $\,m = a^n\,$ is a perfect $n$'th power; equivalently, the power of each prime in $\,a\,$ is a multiple of $\,n,\,$ i.e. $\,a = 2^{e_1} \cdots p_k^{e_k}\,$ $\Rightarrow\, n\mid e_i,\,$ for all $\,i.\,$ Or, said contrapositively, $\,m^{1/n}\,$ is irrational if some $\,e_i$ is not a multiple of $\,n.\,$ The prior inferences depend crucially on the Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations.