I'm in my final year of secondary school in Ireland and was thinking about powers of irrational numbers with rational solutions and very quickly noticed that $(x^\frac{1}{a})^a$ where $x\in Q$ and $a \notin Q$ was always rational.
I wanted to prove that $x^\frac{1}{a} \notin Q$
And this was as far as I got
$$ f(x) = x^b,\ \ x\in Q,\ \ b\notin Q, \ \ x\ne0\\ f'(x) = bx^{b-1}\\ \therefore \ \ f'(x) \notin Q $$
if I can deduce from that $f(x)$ is also irrational then I can deduce that $x^\frac{1}{a} \notin Q$ where $x\ne0$
In my head this should be fine as if the function always changes by an irrational number it should ultimately be irrational, but I cant find a way too prove it, or a source verifying this.
Am I allowed make that deduction and is there anything else flawed in my reasoning?
The statement that $x \in \mathbb{Q}$, and $a \notin \mathbb{Q}$ implies $x^{\frac{1}{a}} \notin \mathbb{Q}$ is false. Here's a counter-example:
Let $$ x = 2, a = \frac{1}{\log_2(3)} $$
Then $x \in \mathbb{Q}$ and $a \notin \mathbb{Q}$, but $$ \begin{eqnarray} x^{\frac{1}{a}} =& 2^{\frac{1}{\frac{1}{\log_2(3)}}}\\ =& 2^{\log_2(3)} \\ =& 3 \\ \in& \mathbb{Q} \end{eqnarray} $$