Rational number with rational exponent becomes rational

Question

I'm looking for a proof to show when $p^q$ for $p,q \in \mathbb{Q}$ is again in $\mathbb{Q}$, without factoring. I'm not sure, if it's possible, given these two numbers to say if the result is again rational and if so, calculate the result efficiently as well.

Thanks!

If $q = \frac{m}{n}$ in lowest form, then $p$ must be an $n$-th power of some other rational number for $p^q$ to be rational. — Arthur, Jun 04 '15 at 10:51

score 5 · Answer 1 · answered Jun 04 '15 at 10:57

5

it is not correct for eg: $$ 2\in \mathbb{Q},\frac{1}{2}\in \mathbb{Q}\;\;\;but :\sqrt{2}=2^{\frac{1}{2}}\notin \mathbb{Q}$$

answered Jun 04 '15 at 10:57

mounir ben salem

434

Yes, I want to have a general proof without factorization. – Robert Eisele Jun 04 '15 at 11:03
1

i think taking eg is enough to show that it's not right – mounir ben salem Jun 04 '15 at 11:07
I don't want to show that the statement is true. The question is for what $p, q$ the statement holds. – Robert Eisele Jun 05 '15 at 07:18
If $x$ is rational, $p \in \mathbb Z$ and $0 \ne q \in \mathbb Z$, then $x^{\frac pq}$ is rational if and only if $x = y^q$ for some rational number $y$. – Steven Alexis Gregory Jul 25 '17 at 10:03

score 1 · Answer 2 · answered Jun 04 '15 at 11:12

Expanding slightly on Arthur's comment. Let $p,q\in \mathbb{Q}$ and let $p=\frac{r}{s}$ and $q=\frac{m}{n}$ where $\gcd(r,s)=1$ and $\gcd(m,n)=1$. Then $p^q\in \mathbb{Q}$ iff $\exists l,k\in\mathbb{Z}$ such that $r=l^n$ and $s=k^n$.

I think this can be proved without resorting to using a prime number decomposition of $r,s$ just by using $\gcd$ but I will have to think about it for a little bit. It's certainly easier to prove using decomposition.

Rational number with rational exponent becomes rational

2 Answers2