# Is there a real number r such that

Is there a real number $r$ such that all ${r}^{n}$ are irrational for all integers $n\ge 1$ but it is not transcendental?
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behk0
Yes: $1+\sqrt{2}$, for instance.
$\left(1+\sqrt{2}{\right)}^{n}=1+2\left(\genfrac{}{}{0}{}{n}{2}\right)+{2}^{2}\left(\genfrac{}{}{0}{}{n}{4}\right)+\cdots +\left(\left(\genfrac{}{}{0}{}{n}{1}\right)+2\left(\genfrac{}{}{0}{}{n}{3}\right)+{2}^{2}\left(\genfrac{}{}{0}{}{n}{5}\right)+\cdots \right)\sqrt{2}.$

pipantasi4
A number is transcendental if it is not root of any polynomial having coefficients in $\mathbb{Q}$. A number such that ${r}^{n}$ is irrational $\mathrm{\forall }n$ is only not a root of ${x}^{n}-q$, $\mathrm{\forall }n\mathrm{\forall }q\in \mathbb{Q}$, but it could be not transcendental anyway, by being root of a polynomial coprime with the last one.