1
$\begingroup$

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 $m^{1/n}$?

$\endgroup$

2 Answers 2

3
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .