Prove or give a counterexample If $a \in \Bbb R$\ $\Bbb Q$ exists $n \in \Bbb N$ such that $a^n \in \Bbb Q$

Question

1)If $a \in \Bbb R$\ $\Bbb Q$ exists $n \in \Bbb N$ such that $a^n \in \Bbb Q$

2)If $a \in \Bbb R$\ $\Bbb Q$ , $_n\sqrt a \in \Bbb R$ \ $\Bbb Q $ $\forall n \in \Bbb N$

For the second one: [by contradiction] Let $a \in \mathbb{R} \backslash \Bbb Q $ .

$_n\sqrt a \in \Bbb Q \Rightarrow \exists p \in \Bbb Z, q \in \Bbb N$ such that $p/q= _n\sqrt a $

$\Rightarrow {(p/q)}^n = {_n\sqrt a}^n \Rightarrow p^n/q^n=a \Rightarrow p^n = q^n a $.

As we know the product between irrational and rational is irrational $\Rightarrow p^n $ is irrational. And this is te contradiction because $p \in \Bbb Z\subseteq \Bbb Q$. Hence $_n\sqrt a \in \Bbb R$\ $\Bbb Q$.

Is it correct (or why not)?

Any ideas for the first one?

Answer 1

You showed that any real not rational isn't the power of a rational (and that's also obvious since the power of a rational is a rational). So the 2) is false.

For the 1), take $a=1+\sqrt{2}$. You have that $$(1+\sqrt{2})^n=a_n+b_n\sqrt{2}$$ with $a_n>0$, $b_n>0$, so that it's not rational. This proves that 1) is false

Answer 2

1) take any transcendental all powers are not integers by definition eg $\pi.$

2) if $x^{1/n}=q$ is rational then $$ x= (x^{1/n})^n = q^n \in {\Bbb Q} $$ so 2 is false.